Physics 1 - Problems With Solutions

Physics 1. Mechanics

Problems with solutions

1 Coordinates and vectors.

Problem 1.1. Find the distance between two points P1 = (r, ϕ1) and P2 = (r, ϕ2) (polar

coordinates).

Solution.

s2 = (x2 − x1)2 + (y2 − y1)

2

= (r cos ϕ2 − r cos ϕ1)2 + (r sin ϕ2 − r sin ϕ1)

2

= 2r2(1− cos ϕ1 cos ϕ2 − sin ϕ1 sin ϕ2)

= 2r2[1− cos(ϕ2 − ϕ1)]

= 4r2 sin2

(ϕ2 − ϕ1

2

)

Problem 1.2. Find the distance between P1 = (R1, θ1, ϕ1) and P2 = (R2, θ2, ϕ2) (spherical

coordinates).

Solution.

s2 = (x2 − x1)2 + (y2 − y1)

2 + (z2 − z1)2

= (R2 sin θ2 cos ϕ2 −R1 sin θ1 cos ϕ1)2

+ (R2 sin θ2 sin ϕ2 −R1 sin θ1 sin ϕ1)2 + (R2 cos θ2 −R1 cos θ1)

2

= R21 + R2

2 − 2R1R2 cos θ1 cos θ2 − 2R1R2 sin θ1 sin θ2 cos(ϕ2 − ϕ1)

Problem 1.3. We define elliptical coordinates as follows: r =√

x2/a2 + y2/b2 and ϕ as in polar

coordinates. Find x, y as functions of r, ϕ.

Solution.

y = x tan ϕ →

1

Physics 1. Mechanics Problems

x2

a2+

x2 tan2 ϕ

b2= r2 →

x = abr cos ϕ(a2 sin2 ϕ + b2 cos2 ϕ)−1/2

y = abr sin ϕ(a2 sin2 ϕ + b2 cos2 ϕ)−1/2

Problem 1.4. In elliptical coordinates (r =√

x2/a2 + y2/b2 and ϕ as in polar coordinates) find

distance between two points P1 = (r, ϕ1) and P2 = (r, ϕ2).

Solution.

s2 = (x2 − x1)2 + (y2 − y1)

2

= (abr)2

(cos ϕ2√

a2 cos2 ϕ2 + b2 sin2 ϕ2

− cos ϕ1√a2 cos2 ϕ1 + b2 sin2 ϕ1

)2

+ (abr)2

(sin ϕ2√


− sin ϕ1√a2 cos2 ϕ1 + b2 sin2 ϕ1

)2

=(abr)2


+(abr)2


− 2(abr)2 cos(ϕ2 − ϕ1)√


√a2 cos2 ϕ2 + b2 sin2 ϕ2

Problem 1.5. Two different coordinate systems are established on a straight line, x and x′, which

are related as follows: x/a = (x′/b)3, where a and b are constants. The distance element in terms of

x is given by ds2 = dx2. Find the expression for the distance element in terms of x′.

Solution.

ds2 = dx2 = (dx

dx′)2dx′

2= (3ax′2/b3)2dx′

2

Problem 1.6. Two different coordinate systems are established on a plane, (x, y) and (x′, y′),

which are related as follows:

x′ = x cos θ − y sin θ, y′ = x sin θ + y cos θ,

where θ = const. Find the distance element in terms of coordinates (x′, y′), if (x, y) are Cartesian.

Solution. Inverting (solving for x and y) we get

x = x′ cos θ + y′ sin θ, y = −x′ sin θ + y′ cos θ

2


Now

ds2 = dx2 + dy2 = (∂x

∂x′dx′ +

∂x

∂y′dy′)2 + (

∂y

∂x′dx′ +

∂y

∂y′dy′)2

= (cos θdx′ + sin θdy′)2 + (− sin θdx′ + cos θdy′)2 = dx′2+ dy′

2

Problem 1.7. Same as above but the relation reads

x′ = x cos θ + y sin θ, y′ = x sin θ + y cos θ,

.

Solution. Inverting one gets

x =x′ cos θ − y′ sin θ

cos2 θ − sin2 θ

y =−x′ sin θ + y′ cos θ

cos2 θ − sin2 θ

Now (cos2 θ − sin2 θ = cos 2θ)

ds2 = dx2 + dy2 = (∂x

∂x′dx′ +

∂x

∂y′dy′)2 + (

∂y

∂x′dx′ +

∂y

∂y′dy′)2

=1

cos2 2θ[(cos θdx′ − sin θdy′)2 + (− sin θdx′ + cos θdy′)2]

=1

cos2 2θ[dx′

2+ dy′

2 − 2 sin 2θdx′dy′]

Problem 1.8. New coordinates (ρ′, ϕ′) are introduced which are related to the ordinary polar

coordinates (ρ, ϕ) as follows: ϕ′ = ϕ, ρ′ = 1/ρ. Find the distance element.

Solution.

ϕ′ = ϕ, ρ′ = 1/ρ → ϕ = ϕ′, ρ = 1/ρ′

ds2 = dρ2 + ρ2dϕ2 = (∂ρ

∂ρ′dρ′ +

∂ρ

∂ϕ′dϕ′)2 + (1/ρ′)2(∂ϕ

∂dρ′dρ′ +

∂ϕ

∂ϕ′dϕ′)2

= (1/ρ′)4dρ′2+ (1/ρ′)2dϕ′2

Problem 1.9. Starting with cartesian coordinates we define new ones as follows: x′ = ax + by,

y′ = cx + dy, where a, b, c, d are some constant parameters. What conditions on these parameters

should be satisfied in order that the new coordinates also be cartesian and the measure of the distance

remain the same ?

3


Solution.

ds2 = dx2 + dy2 = dx′2+ dy′

2 →

(adx + bdy)2 + (cdx + ddy)2 = dx2 + dy2 →

(a2 + b2)dx2 + (c2 + d2)dy2 + 2(ab + cd)dxdy = dx2 + dy2 →

a2 + c2 = b2 + d2 = 1, ab + cd = 0

d = −ab/c → b2 + (ab/c)2 = 1 → (b/c)2 = 1 →

c = b or c = −b

if c = b → d = −a

if c = −b → d = a

and a2 + b2 = 1

Problem 1.10. Four vertices of the tetrahedron are given by four radius-vectors ri, i = 1, 2, 3, 4.

Find the volume.

Solution.

V = |(r4 − r1)× (r3 − r1) · (r2 − r1)|

Problem 1.11. Let r1 be the vector from the center of the Earth to New-York and r2 is the

vector from the center of the Earth to Jerusalem. Find the angle between the two vectors (find the

lacking data in a geographical atlas).

Problem 1.12. Find the distance element ds2 in the elliptical coordinates r =√

x2/a2 + y2/b2,

tan ϕ = y/x. Define unit vectors tangential and normal to r = const curves and derive their relation

to ex and ey.

Problem 1.13. Are the coordinates x′ = xy, y′ = y/x legitimate ? If yes, find the distance

element.

Solution. Not good: x, y and −x,−y have the same x′, y′. Let us limit ourselves with x > 0,

y > 0, then

y = (x′y′)1/2, x = (x′/y′)1/2

ds2 = dx2 + dy2 = (∂x

∂x′dx′ +

∂x

∂y′dy′)2 + (

∂y

∂x′dx′ +

∂y

∂y′dy′)2

= (1

2√

x′y′dx′ −

√x′

2√

y′3dy′)2 + (

√y′

2√

x′dx′ +

√x′

2√

y′dy′)2

=1 = y′2

4x′y′dx′

2+

x′(1 + y′2)

4y′3dy′

2+ 1

2(√

x′y′ −√

x′

y′2)dx′dy′

4


Problem 1.14. Calculate the length of the parabola y = ax2, a = const, from x = −a to x = a.

Solution.

ds2 = dx2 + dy2 =

(1 + (

dy

dx)2)

)dx2 = (1 + 4a2x2)dx2

s =

∫ds =

∫ a

−a

√1 + 4a2x2dx

Problem 1.15. Ellipse is given by the equation x2/a2 + y2/b2 = 1. Calculate the circumference.

Solution. Let x = a cos t, y = b sin t, 0 ≤ t < 2π. Then

ds2 = dx2 + dy2 =

[(dx

dt)2 + (

dy

dt)2

]dt2 = (a2 sin2 t + b2 cos2 t)dt2

s =

∫ds =

∫ 2π

0

√a2 sin2 t + b2 cos2 tdt

Problem 1.16. A plane passes through the point r0 and its normal is n. Write down the equation

of the plane.

Solution. All vectors in the plane are perpendicular to the normal, therefore the plane equation

is

(r − r0) · n = 0

Problem 1.17. Find the relation between the unit vectors of 3D Cartesian coordinates and unit

vectors of spherical coordinates.

Solution. First of all, we have

r = sin θ cos ϕx + sin θ sin ϕy + cos θz

ϕ = − sin ϕx + cos ϕy

θ = ϕ× r = cos θ cos ϕx + cos θ sin ϕy − sin θz

Let now x = a1r + a2 + θ + a3ϕ, then

a1 = x · r = sin θ cos ϕ

a2 = x · θ = cos θ cos ϕ

a3 = x · ϕ = − sin ϕ

5


Similarly for y and z.

Problem 1.18. Decompose unit vectors of cylindrical coordinates using unit vectors of spherical

coordinates.

Solution. We know that

ρ = cos ϕx + sin ϕy −ϕ = − sin ϕx + cos ϕy

Let, for example, ρ = a1r + a2θ + a3ϕ. Like it was in the previous problem,

a1 = ρ · r = sin θ

a2 = ρ · θ = cos θ

a3 = ρ · ϕ = 0

and similarly for ϕ and z.

Problem 1.19. Let ω = ωz (cylindrical coordinates). Derive ω×r, ω×(ω×r), and r×(ω×r).

Solution.

ω × r = ωz × (xx + yy + zz) = −ωyx + ωxy

ω × (ω × r) = ω(ω · r)− ω2r = ω2zz − ω2r = −ω2(xx + yy)

r × (ω × r) = ωr2 − r(ω · r) = ω(x2 + y2)z − ωz(xx + yy)

Problem 1.20. Two cities on Earth surface (assume it is a sphere with the radius R) have the

coordinates (α1, β1) and (α2, β2), where α is the latitude, and β is the longitude. Find the shortest

distance between the cities. Hint: Arc of the circle passing through the center of the Earth.

Solution. The distance l = Rγ, where γ is the angle between the vectors ~OP1 and ~OP2 (O is

the center). One has

~OP1 = R(cos α1 cos β1, cos α1 sin β1, sin α1)

~OP2 = R(cos α2 cos β2, cos α2 sin β2, sin α2)

cos γ = ~OP1 · ~OP2/| ~OP1| · | ~OP2| = cos α1 cos α2 cos(β1 − β2) + sin α1 + sin α2

Problem 1.21. Four vertices of the tetrahedron are given by four radius-vectors ri, i = 1, 2, 3, 4.

Find the volume.

6


Solution.

V = (1/6)|(r2 − r1) · ((r3 − r1)× (r4 − r1))|

Problem 1.22. Find the distance element ds2 in the elliptical coordinates r =√

x2/a2 + y2/b2,

tan ϕ = y/x. Define unit vectors tangential and normal to r = const curves and derive their relation

to x and y.

Solution. As we found earlier,

x =abr cos ϕ√

a2 sin2 ϕ + b2 cos2 ϕ

y =abr sin ϕ√


We have further

ds2 =

(∂x

∂rdr +

∂x

∂ϕdϕ

)2

+

(∂y

∂rdr +

∂y

∂ϕdϕ

)2

The derivatives are

∂x

∂r=

ab cos ϕ√a2 sin2 ϕ + b2 cos2 ϕ

∂y

∂r=

ab sin ϕ√a2 sin2 ϕ + b2 cos2 ϕ

∂x

∂ϕ= − abr sin ϕ√


(1 +

cos2 ϕ(a2 − b2)


)∂y

∂ϕ=

abr cos ϕ√a2 sin2 ϕ + b2 sin2 ϕ

(1− cos2 ϕ(a2 − b2)


)Please substitute.

Calculation of the tangential unit vector t. Let us consider two close points on the ellipse:

(x, y) and (x + dx, y + dy). The vector connecting these two points is a tangential vector, that is,

dr = dx · x + dy · y ‖ t. However, we have to take into account that dx and dy are not independent,

since they have to be calculated with r = const, that is,

dx = (∂x

∂ϕ)dϕ, dy = (

∂y

∂ϕ)dϕ

Using this we find

t =

((∂x

∂ϕ)x + (

∂y

∂ϕ)y

)/

((∂x

∂ϕ)2 + (

∂y

∂ϕ)2

)

7


Please substitute. After t is found, the normal unit vector can be found as n = z × t.

Problem 1.23. Simplify (a× b) · (c× d).

Solution.

(a× b) · (c× d) = a · (b)× (c× d)) = a · (c(b · d)− d(b · c)) = (a · c)(b · d)− (a · d)(b · c)

Problem 1.24. Given two non-parallel unit vectors e1 and e2, find another unit vector e3 such

that all three are in the same plane and e3 divides the angle between the first two in the way

ˆe1e3 = 2 ˆe2e3.

Solution. Let e3 = ae1 + be2. Since e3 · e3 = 1 we have

a2 + b2 + 2abe1 · e2 = 1

Let α be the angle between e1 and e3, and β be the angle between e2 and e3. We have

cos α = e1 · e3 = a + be1 · e2

cos β = e2 · e3 = b + ae1 · e2

Since α = 2β we have cos α = cos(2β) = 2 cos2 β − 1 so that

e1 · e3 = a + be1 · e2 = 2(e2 · e3 = b + ae1 · e2)2 − 1

Since e1 · e2 is known we get two equations for a and b which have to be solved. Please solve.

Problem 1.25. What is the projection of the vector a onto the unit vector e ?

Problem 1.26. What is the projection of the vector a onto the vector b ?

Problem 1.27. Given two vectors a and b, represent vector a as a sum of two vectors, a‖ and

a⊥, a = a‖ + a⊥, such that a‖ ‖ b and a⊥ ⊥ b.

Problem 1.28. On the Earth a man is at the point 20 east longitude and 40 latitude moves

in the north-east direction. Express the unit vector along the velocity in spherical and cartesian

coordinates.

Problem 1.29. Show that a straight line is given by the relation r = λa + b, where a and a are

constant vectors and −∞ < λ < ∞.

Problem 1.30. Find the intersection line of the two planes r · a1 = d1 and r · a2 = d2.

8


Solution. Two planes intersect by a straight line (we assume that a1 and a2 are not parallel).

First let us write the equation of a straight line which passes through the point r0 in the direction

given by the vector l. Since for each point r on this straight line the vector (r − r0) ‖ l we have

(r − r0)× l = 0

This is the vector equation for a straight line. Now, our line has to be perpendicular to a1 and a2,

thus we may choose l = a1 × a2. The point r0 belongs two the both planes simultaneously, that is,

r0 · a1 = d1 and r0 · a2 = d2. Let us try to find is as r0 = k1a1 + k2a2. The conditions give

k1a21 + k2a1 · a2 = d1

k1a1 · a2 + k2a22 = d2

This set of equations always has a unique solution since a21a

22 > (a1 · a2)

2.

Problem 1.31. Write in the vector form the equation for a circle with the radius R, normal

direction n (unit vector !), and center at r0.

Solution. Set of equations:

(r − r0) · n = 0

(r − r0)2 = R2

Problem 1.32. Write in the vector form the equation for a cylindrical surface with the radius R

and the axis parallel to a and crossing the point r0.

Solution. Let a = a/|a|. Then the equation is

(r − r0)2 − (a · (r − r0))

2 = R2

Problem 1.33. Prove that four different points ri, i = 1, 2, 3, 4, are always on a sphere and find

the center and the radius of the sphere.

Solution. Let r0 be the center of a sphere and R be its radius. We have

(ri − r0)2 = R2

where i = 1, 2, 3, 4. Subtracting, say, equation for i = 1 from others, we have

r0 ·X2 = r22 − r2

1

r0 ·X3 = r23 − r2

1

9


r0 ·X4 = r24 − r2

1

where Xj = rj − r1, j = 2, 3, 4. We assume that Xj are not all in the same plane. Then we can

decompose r0 = k2X2 + k3K3 + k4K4 and substituting one has

k2X2 ·K2 + k3X3 ·X2 + k4X4 ·X2 = r22 − r2

1

k2X2 ·K3 + k3X3 ·X3 + k4X4 ·X3 = r23 − r2

1

k2X2 ·K4 + k3X3 ·X4 + k4X4 ·X4 = r24 − r2

1

These equations always have a solution (why ?).

Problem 1.34. Calculate a)∑

i δii, b)∑

ij δij, c)∑

j δijδjk, d)∑

jk δijδjkδkl.

Problem 1.35. Prove a)∑

ijk εijkεijk = 6, b)∑

jk εijkεmjk = 2δim, c)∑

k εijkεmnk = δimδjn −δinδjm.

Problem 1.36. Express A× (B ×C) with the use of εikj and δij.

Problem 1.37. Angular momentum is defined as J = mr × v. Express in terms of angular

velocity for a circularly moving particle.

Problem 1.38. Given two nonparallel vectors a and b build three mutually perpendicular unit

vectors.

Solution. For example, we may choose e1 = a/|a|, e2 = a× b/|a× b|, and e3 = e1 × e2.

2 Velocity, acceleration, trajectory

Problem 2.1. A particle moves in the x− y plain according to the law: x = k1t, y = k2t2. Find:

a) velocity, b) acceleration, c) distance from the coordinate origin.

Problem 2.2. A particle moves along the ellipse x2/a2 + y2/b2 = 1 so that dϕ/dt = const. Find

the velocity and acceleration. Find the radius of curvature.

Solution.

vx = x = k1, vy = y = 2k2t

ax = vx = 0, ay = vy = 2k2

r = (x2 + y2)1/2 = (k21t

2 + k22t

4)1/2

10


Problem 2.3. Trajectory is given by y = kx2, k = const, and x = at2, k = const and a = const.

Find v and a.

Problem 2.4. Trajectory is given by r = kϕ and ϕ = ωt, k = const and ω = const. Find v and

a. What is the angle between the velocity and acceleration as a function of time ?

Solution.

x = kωt cos ωt

y = kωt sin ωt

vx = kω(cos ωt− ωt sin ωt)

vy = kω(sin ωt + ωt cos ωt)

ax = kω2(−2 sin ωt− ωt cos ωt)

ay = kω2(2 cos ωt− ωt sin ωt)

cos(av) = a · v/|a| · |v|

= (axvx + ayvy)/(a2x + a2

y)1/2(v2

x + v2y)

1/2

Solution. Second method: polar coordinates.

r = rr, r = cos ϕx + sin ϕy

v = r = rr + r ˙r

= rr + r(− sin ϕx + cos ϕy)ϕ

= rr + rϕϕ

˙ϕ = (− cos ϕx− sin ϕy)ϕ

= −ϕr

˙r = ϕϕ

a = v = rr + r ˙r + rϕϕ + rϕϕ + rϕ ˙ϕ

= (r − ϕ2r)r + (rϕ + 2rϕ)ϕ

We have ϕ = ωt so that ϕ = ω, ϕ = 0. We have r = kωt, so that r = kω, r = 0. Now

v = kωr + ωrϕ

a = −ω2rr + 2kω2ϕ

Problem 2.5. Two spacecraft are orbiting Earth. The orbit radii and angular velocities are the

11


same but one (A) orbit is always above the equator, while the other (B) passes above the poles.

When B is above the equator the spacecraft A is on the opposite side of the diameter. Find the

vectors connecting A and B as a function of time.

Problem 2.6. Express the velocity vector (in general) in terms of spherical coordinates and unit

vectors eR, eθ, eϕ.

Solution.

dr = drr + rdθθ + r sin θdϕϕ

v = rr + rθθ + r sin θϕϕ

Problem 2.7. A particle moves according to the law: x = k1t cos(ωt), y = k2t sin(ωt), z = bt2.

Find the velocity and acceleration.

Solution.

vx = k1 cos ωt− k1ωt sin ωt

vy = k2 sin ωt + k2ωt cos ωt

vz = 2bt

ax = −2k1ω sin ωt− k1ω2t cos ωt

ay = 2k2ω cos ωt− k2ω2t sin ωt

az = 2b

Problem 2.8. Derive the expression for the trajectory length for the motion with constant

acceleration.

Problem 2.9. A particle moves along the trajectory r = a/(1− ε cos ϕ) in cylindrical coordinates

so that r2(dϕ/dt) = l. Here a, ε, and l are constant parameters. Find (dr/dt) as a function of r.

Same as a function of ϕ.

Solution.

r =dr

dϕϕ = − aε sin ϕ

(1− ε cos ϕ)2

l

r2

(1− ε cos ϕ) = a/r

cos ϕ = (1− a/r)ε

sin ϕ = ±(1− (1− a/r)2/ε2)1/2

r = ∓ l(ε2 − (1− a/r)2)1/2

a

12


Problem 2.10. Given x(t) = R cos(ωt), y(t) = R sin(ωt). Find the angle between the vectors of

velocity and acceleration as a function of time.

Solution.

vx = −γ1R exp(−γ1t) cos(ω1t)− ω1R exp(−γ1t) sin(ω1t)

vy = −γ2R exp(−γ2t) sin(ω2t) + ω2R exp(−γ2t) sin(ω2t)

ax = (γ21 − ω2

1)R exp(−γ1t) cos(ω1t) + 2ω1γ1R exp(−γ1t) sin(ω1t)

ay = (γ22 − ω2

2)R exp(−γ1t) sin(ω1t)− 2ω2γ2R exp(−γ2t) cos(ω2t)

cos(av) = a · v/|a| · |v|

= (axvx + ayvy)/(a2x + a2

y)1/2(v2

x + v2y)

1/2

Problem 2.11. Given x(t) = R exp(−γ1t) cos(ω1t), y(t) = R exp(−γ2t) sin(ω2t). Find the angle

between the vectors of velocity and acceleration as a function of time.

Problem 2.12. Given r(t) = kt, ϕ(t) = ωt. Find the tangential and normal acceleration as

functions of time.

Solution.

x = kt cos ωt, y = kt sin ωt

vx = k cos ωt− kωt sin ωt

vy = k sin ωt + kωt cos ωt

ax = −2kω sin ωt− kω2t cos ωt

ay = 2kω cos ωt− kω2t sin ωt

v = v/|v| = vx√v2

x + v2y

x +vy√

v2x + v2

y

y

at = (a · v)v, an = a− at

Solution. Second method: polar coordinates. As in Problem 0.2

v = rr + rϕϕ = kr + rωϕ

a = (r − ϕ2r)r + (rϕ + 2rϕ)ϕ

= −ω2rr + 2kωϕ

|v| = (k2 + ω2r2)1/2

at = (a · v)v/|v|2

13


an = a− at

Problem 2.13. Given vx(t) = vd + v0 cos(ωt), vy(t) = v0 sin(ωt). Find x(t) and y(t). What are

the conditions on the parameters for the absence of self-intersection.

Problem 2.14. Given vx(t) = vd + v0 cos(ωt), vy(t) = v0 sin(ωt), vz(t) = at. Find the distance

from the coordinate origin to the particle as a function of time if r(t = 0) = 0.

Solution.

x = x0 +

∫ t

t0

vx(t′)dt′ = vdt +

v0

ωsin ωt

y =v0

ω(1− cos ωt)

z = at2/2

r = (x2 + y2 + z2)1/2

Problem 2.15. In a universe all bodies move away from the coordinate origin with the velocities

v = Kr, where K = const. What would see an observer at an arbitrary position r0 ?

Solution.

r′ = r − r0

v′ = v − v0 = K(r − r0) = Kr′

Problem 2.16. * A rabbit starts to run at t = 0 from the point (x0, 0) in the positive direction of

axis y with the velocity v0 (magnitude !). A fox starts to run from (0, 0) at the same moment and its

velocity v > v0 always points towards the rabbit. How much time does it take to catch the rabbit.

Problem 2.17. Given ax = a0 exp(−γt), ay = a1 sin(ωt). Find r(t).

Solution.

a = a0 exp(−γt)x + a1 sin ωty

v = v0 +

∫ t

t0

a(t′)dt′

= v0 −a0

γ[exp(−γt)− exp(−γt0)]x

− a1

ω(cos ωt− cos ωt0)y

r = r0 +

∫ t

t0

v(t′)dt′

14


= r0 + v0(t− t0) +a0 exp(−γt0)

γ(t− t0)x

+a1 cos ωt0

ω(t− t0)y

+a0

γ2[exp(−γt)− exp(−γt0)]x

− a1

ω2[sin ωt− sin ωt0]y

Problem 2.18. Given vx = v0x exp(−γt), vy = v0y sin(ωt), vz = v0z + at. Write down the

expression for the path length (integral).

Problem 2.19. Given x = x0 cos(ωt), y = y0 sin(ωt), x0 6= y0, find the tangential and normal

components of the acceleration.

Problem 2.20. * A body starts from the equator of the sphere (“Earth”) with the radius R and

moves all the time in the north-east direction so that the velocity magnitude v remains constant.

Where does it stop and how much time does it take ?

3 Inertial and noninertial reference frames

Problem 3.1. A plane takes off with the acceleration 0.5|g| at the angle 30 to the horizon. What

is the weight of the 75 kg passenger ?

Solution.

W = m|g + a| = m√

(g + a sin θ)2 + a2 cos θ2

Problem 3.2. What should be the length of the day on Earth to compensate the gravity at the

equator ?

Solution.

g = ω2R, T =2π

ω= 2π

√R

g

Problem 3.3. A body starts moving with the velocity v from the center of the rotating disk

(angular velocity ω). There are no external forces. Describe the motion from the point of view of

the rotating observer.

Solution.

x = vt, y = 0

x′ = x cos ωt− y sin ωt = vt cos ωt

15


y′ = x sin ωt + y cos ωt = vt sin ωt

Problem 3.4. What is the weight of a standing 1000 kg car on the equator ? What is its weight

if it is moving in the east direction with the velocity 300 km/hour ?

Solution.

W = mg −mω2R

W = mg −mω2R− 2mvω

Problem 3.5. A biker enters a quarter-circle turn of the radius R with the velocity v. What is

the angle between the biker’s body and the vertical ?

Solution.

tan θ = a/g =v2

gr

Problem 3.6. A body hangs on a rope from the ceiling in a standing train. The train starts

moving with the acceleration a. What is the angle between the rope and the vertical ?

Problem 3.7. A body hangs on a rope from the ceiling in a rotating (angular velocity ω) cell.

The distance from the rotation center is r. What is the angle between the rope and the vertical ?

Problem 3.8. A horizontal carousel rotates with the angular velocity ω. What is the weight of

a person who sits at the radius R ?

Problem 3.9. A project of a space station suggests rotation in order to produce artificial gravity.

If the diameter of the station is 20 m, what should be the rotation period in order to produce the

gravity equivalent to 0.5g ?

Problem 3.10. A body is moving along x axis with constant velocity vx in the inertial (standing)

frame. Write down x′(t) and y′(t) in the rotating frame. What is the direction of acceleration as a

function of time in the rotating frame ?

Solution.

x′ = vxt cos ωt

y′ = vxt sin ωt

a′x = x′ = −2ωvx sin ωt− ω2vxt cos ωt

a′y = y′ = 2ωvx cos ωt− ω2vxt sin ωt

16


Problem 3.11. A body falls with the velocity v = gτ [1 − exp(−t/τ)] (because of the air drag

force). Write down the second Newton law in its frame.

Solution.

A = v = g exp(−t/τ)

ma = F −mA

Problem 3.12. A river flows from the north to the south in the northern hemisphere at the

latitude θ. The flow velocity is v and the river width is L. What is the difference of the water level

at the western and eastern coasts ? (Hint: Coriolis.)

Solution.

ax = 2ω cos θv

ay = g

h/L = ax/ay =2ωv cos θ

g

4 Particle dynamics, Newton laws

Problem 4.1. A particle is moving so that ϕ = at2, r = r0 exp(bt), where a, b, and r0 are

constants. Find the force.

Problem 4.2. A body (mass m) starts falling. The air friction force is Ff = −kv, where k = const

and v is the body velocity. Find v(t) and r(t).

Solution.

v = vz

mv = mg − kv

lnmg − kv

mg=

kt

m

v =mg

k(1− exp(−kt/m))

Problem 4.3. A body (mass m) is thrown horizontally with the initial velocity v0. The air

friction force is Ff = −kv, where k = const and v is the body velocity. Find v(t) and r(t).

17


Solution.

mvx = −kvx

mvy = mg − kvy

vx = v0 exp(−kt/m)

vy =mg

k(1− exp(−kt/m))

Problem 4.4. Force Fx = F0 sin2(ωt) acts on a particle (mass m) which is initially at rest. Find

v(t) and r(t).

Solution.

mvx = F0 sin2(ωt) = 12F0[1− cos(2ωt)]

x =F0

2m

[t− 1

2ωsin(2ωt)

]

Problem 4.5. At high speeds the air drag force (friction) is Ff = −kvv. A body is falling

vertically in the air with the initial velocity v0. Find v(t) and r(t).

Solution.

mv = mg − kv2∫ v

0

dv′

mg/k − v′2=

kt

m

ln|mg/k − v2|

mg/k= 2

√mg

k

kt

m

Problem 4.6. A charged particle (charge q, mass m) is accelerated by the electric field E =

E1 sin(ω1t)ex + E2 sin(ω2t)ey. Find the trajectory.

Problem 4.7. A charged particle moves with constant velocity v ⊥ B (B - magnetic field). Find

the electric field.

Problem 4.8. A charged particle (mass m, charge q, velocity v ) enters a cylinder with the length

l. The entry point is at the cylinder axis, and the particles enters at the angle α to the axis. There

is a homogeneous magnetic field along the axis inside the cylinder. At what distance from the axis

18


the particle leaves the cylinder ?

Problem 4.9. A charged particle (mass m, charge q) is at rest in an homogeneous magnetic field

B = (0, 0, B). Suddenly, at t = 0 an electric field E = (0, E, 0) is switched on. The electric field

is suddenly switched off at t = T/2, where T = 2π/(|q|B/m). Describe the motion of the particle.

What is its final energy ?

5 Potential energy, conservation laws

Problem 5.1. For each of the forces given below check whether it is conservative and find the

potential energy, if possible:

a) Fx = 2yz(1− 6xyz), Fy = 2xz(1− 6xyz), Fz = 2xy(1− 6xyz)

b) Fx = y2 + z2 +2(xy + yz + zx), Fy = x2 + z2 +2(xy + yz + zx), Fz = y2 +x2 +2(xy + yz + zx)

Problem 5.2. Potential energy is given by U = a/r2− b/r. At what r a particle is in equilibrium

?

Problem 5.3. Potential energy is given by U(x) = k|x| (one-dimensional motion). Find the

period of the bound motion of the particle with the energy E > 0.

Solution.

vx = ±√

2

m(E − k|x|)

dt =dx

vx

T = 2

∫ E/k

−E/k

dx√2m

(E − k|x|)

= 4

∫ E/k

0

dx√2m

(E − kx)

=8

k

√m

2

√E − kx|0E/k

=8

k

√mE

2

Problem 5.4. Force is given by Fρ = a/ρ2, Fϕ = b sin ϕ/ρ2 (cylindrical coordinates). Is the force

conservative ? If yes, find the potential. What is conserved in this force ?

19


Solution. Let the force be potential, then

Fρ = −∂U

∂ρ

Fϕ = −1

ρ

∂U

∂ϕ∂Fρ

∂ϕ=

∂ρFϕ

∂ρ

0 6= −b sin ϕ

ρ2

Not potential.

Problem 5.5. Potential energy is given in polar coordinates by U = a cos ϕ/ρ. Find the force. Is

angular momentum conserved ? Find the torque at (ρ, ϕ).

Solution.

F = −∂U

∂ρρ− 1

ρ

∂U

∂ϕϕ

=a cos ϕ

ρ2ρ +

a sin ϕ

ρ2ϕ

τ = r × F

= ρρ× (a cos ϕ

ρ2ρ +

a sin ϕ

ρ2ϕ)

=a sin ϕ

ρz

Problem 5.6. A particle orbit is r = a(1− cos ϕ). Find the central force.

Solution.

U(r) = E − m

2r2 − J2

2mr2

r = a sin ϕϕ

r2 = a2 sin2 ϕϕ2

= [a2 − (a− r)2]

(J

mr2

)2

Problem 5.7. A bead (mass m) is moving on a circularly shaped wire (r = const) without friction

and is connected to the two points, P1 = (0,−r/2) and P2 = (0, r/2), with identical springs (spring

constant k) of initially zero length (so that |F | = kl where l is the length of the spring). a) Write

down the force vectors. b) Derive the potential energy (if the forces are conservative). c) Find the

20


velocity as a function of angle ϕ (for a given energy). c) Find the angular momentum relative to the

coordinate origin as a function of ϕ.

Solution. Let r1 = (0,−r/2), r2 = (0, r/2) = −r1

F1 = −k(r − r1)

F2 = −k(r − r2)

U =k

2

[(r − r1)

2 + (r − r2)2]

=k

2

[2r2 + 2r2

1

]

Problem 5.8. A bead (mass m) is moving on an elliptically shaped (r = p/(1 − ε cos ϕ) wire

without friction. The bead is attracted to the focus (0, 0) by the force inversely proportional to the

distance r1 squared between the bead and the focus, |F |1 = k1/r21. The bead is attracted to the

center of the ellipse by the force proportional to the distance r2 between the bead and the center,

|F |2 = k2r2. a) Write down the force vectors. b) Derive the potential energy (if the forces are

conservative). c) Find the velocity as a function of angle ϕ (for a given energy). c) Find the angular

momentum relative to the coordinate origin as a function of ϕ.

Problem 5.9. In a galaxy the gravitational potential (potential energy) is U = −k/rα, 0 < α < 1.

Find the relation between the total energy and angular momentum for circular orbits. Find the

dependence of the orbit period on the radius.

Solution.

Ueff = − k

rα+

J2

2mr2

r = 0 ⇒ dUeff

dr= 0

E = Ueff

Problem 5.10. A particle moves under the influence of the body O which is in the coordinate

origin. In the beginning the particle is at a very large distance from O, moves with the velocity v and

would pass at the distance l from O if there were no interaction (this is called impact parameter).

What is the minimal distance between the particle and O for the potential energy is U = k/rα

(analyze k > 0 and k < 0).

Solution.

J = mvl

21


E =mv2

2= U +

J2

2mr2

Problem 5.11. Find x(t) for a particle with E = 0 in the potential energy U(x) = −ax2/2+bx4/4.

(Hint: x → 0 for t → −∞.)

Solution.

x = ±√

2(E − U)/m

= ±√

2(ax2/2− bx4/4)/m

t =

∫dx√

2(ax2/2− bx4/4)/m

Problem 5.12. A satellite of the mass m, moving in the Earth potential U(r) = −k/r, has

the total energy E and angular momentum J . Find the maximum (apogee) and minimum (perigee)

distance from the Earth.

Problem 5.13. A particle (mass m) is moving in the central field U = −k/r on a circular orbit

r = r0. The energy and angular momentum suddenly are changed by ∆E and ∆J . What are the

maximal and minimal distances from the attracting body on the new orbit ?

6 Oscillations

Problem 6.1. Find the frequency of small oscillations of a particle (mass m) near the equilibrium

in the potential U(x) = a/x13 − b/x7.

Solution. Equilibrium: dU/dx = 0 → −13a/x14 + 7b/x8 = 0 → x0 = (13a/7b)1/6. Taylor

expansion: d2U/dx2 = 182a/x150 − 56b/x9

0. Now ω2 = (182a/x150 − 56b/x9

0)/m.

Problem 6.2. A particle is in the stable equilibrium in the potential energy U(x) = U0[1 −l2/(l2 +x2)]. Suddenly it gets a small addition of energy E ′. Assuming that the oscillations are small

find the frequency and amplitude.

Problem 6.3. A particle moves in a well of the shape y = ax2 without friction (potential energy

U = mgy). Show that the motion can be described as a harmonic oscillation and find the frequency.

Solution. Potential energy U = mgy = mgax2. Kinetic energy

K = (m/2)(x2 + y2) =m

2(1 + 4a2x2)x2

22


When ax 1 we have K ≈ m2x2 and ω2 = 2ga.

Problem 6.4. A body with the mass m is attached to a spring (spring constant k). The other

end of the spring is brought into the motion according to the law x = x0 cos(ωt). The friction acting

on the body is Fx = −bvx. Show that the body can oscillate with a constant amplitude and find this

amplitude.

Solution. Let y be the coordinate of the body. Then F = −k∆l = −k(y − x). We have

my = −k(y − x)− by ⇒ my + by + ky = kx0 cos(ωt)

Therefore, y = A cos(ωt− φ). Substitute and complete.

Problem 6.5. Find the motion of an oscillator with the natural frequency ω0 and mass m under

the force F = F0 + F1 cos(ωt), ω 6= ω0.

Solution. If F = F0 the solution is x1 = F0/mω2. When F = F1 cos(ωt) the solution is

x2 = A cos(ωt− φ). The total solutions is x1 + x2.

Problem 6.6. Find the frequency of small radial oscillations of a particle with a mass m near a

circular orbit r = r0 in a central potential U(r) = −k/r.

Solution. Ueff = −k/r + J2/2mr2. Equilibrium: k/r2 − J2/mr3 = 0 ⇒ r0 = J2/km. Taylor:

d2Ueff/dr2 = −2k/r30 + 3J2/mr4

0 and ω2 = (−2k/r30 + 3J2/mr4

0)/m.

Problem 6.7. Find the frequency of small radial oscillations of a particle with a mass m near a

circular orbit r = r0 in a central potential U(r).

Problem 6.8. Find the average power of the external force F = F0 cos(ωt) for the oscillator:

mx + Γx + mω20x = F (t).

Solution. Substitute x = C1 cos(ωt) + C2 sin(ωt):

(ω20 − ω2)(C1 cos(ωt) + C2 sin(ωt)) + ωΓ(−C1 sin(ωt) + C2 cos(ωt)) = F0 cos(ωt)/m

(ω20 − ω2)C1 + ωΓC2 = F0/m

(ω20 − ω2)C2 − ωΓC1 = 0

C1 = (ω20 − ω2)C2/ωΓ

C2 =F0ωΓ

m[(ω20 − ω2)2 + Γ2ω2]

P = Fvx = F0 cos(ωt)ω(−C1 sin(ωt) + C2 cos(ωt))

〈P 〉 = 12F0C2ω

〈P 〉 =F 2

0 ω2Γ

2m[(ω20 − ω2)2 + Γ2ω2]

23


Problem 6.9. A bead of the mass m can move on a straight wire along y axis without friction.

The bead is connected to two springs (spring constant k, length a). The springs are connected to

the points (−l, 0) and (l, 0), l > a, respectively. Initially the bead starts moving from (0, 0) with the

velocity v. Assuming that the the oscillations are small, find their frequency and amplitude.

Solution.

E =my2

2+ 2 · 1

2k(√

l2 + y2 − a)2

≈ my2

2+ 1

2· 2k[(l − a)2 +

y2(l − a)

l]

ω2 =2k(l − a)

ml

Problem 6.10. A particle is moving in the magnetic field B = (0, 0, B), B = const, and electric

field E = (E cos(ωt), 0, 0), ω 6= |q|B/m. Find v(t).

mvx = qE cos(ωt) + qvyB

mvy = −qvxB ⇒ vx = −(m/qB)vy

−(m2/qB)vy = qE cos(ωt) + qvyB

vy + Ω2vy = F cos(ωt), Ω = qB/m, F = −Ω2E/B

Substitute vy = A cos(ωt− φ).

Problem 6.11. A particle with the mass m is moving in x − y plane with the potential energy

U = 12k(x2 + y2). Initially the particle is in the position (a, 0) and its velocity is (0, v0). Find the

trajectory.

7 Rotation of a rigid body

Problem 7.1. A disk of the mass m and radius r is connected to two parallel identical springs

(k, l) as shown in the figure.

24



≈ my2

2+ 1

2· 2k[(l − a)2 +

y2(l − a)

l]

ω2 =2k(l − a)

ml

Problem 6.9. A particle is moving in the magnetic field B = (0, 0, B), B = const, and electric

field E = (E cos(ωt), 0, 0), ω #= |q|B/m. Find v(t).

mvx = qE cos(ωt) + qvyB

mvy = −qvxB ⇒ vx = −(m/qB)vy

−(m2/qB)vy = qE cos(ωt) + qvyB

vy + Ω2vy = F cos(ωt), Ω = qB/m, F = −Ω2E/B

Substitute vy = A cos(ωt − φ).

Problem 6.10. A particle with the mass m is moving in x − y plane with the potential energy

U = 1

2k(x2 + y2). Initially the particle is in the position (a, 0) and its velocity is (0, v0). Find the

trajectory.

7 Rotation of a rigid body

Problem 7.1. A disk of the mass m and radius r is connected to two parallel identical springs

(k, l) as shown in the figure.

dd

ll

Find the frequency of rotational oscillations around the center of the disk.

22

Find the frequency of rotational oscillations around the center of the disk.

Solution. Let the disk rotate by a small angle θ 1. The total energy is E = Iθ2/2+2 · 12k(dθ)2,

so that ω2 = 2kd2/I, where I = mr2/2.

Problem 7.2. A ball of the radius r rotates with the angular velocity ω around the horizontal

axis passing through the center-of-mass. The ball is carefully put on a horizontal surface with the

friction coefficient µ. Find vcm(t).

Solution. The ball starts to move with sliding so that the friction force F = µmg accelerates

the ball but the torque N = Fr = µmgr decelerates the rotation:

mv = µmg ⇒ v = µgt

Iω = −µmgr ⇒ ω = ω0 − (µmgr/I)t

where I = (2/5)mr2. When v = ωr the sliding stops, the friction force vanishes and the ball continues

to roll without sliding so that v and ω no longer change. This happens when

µgt = ω0r − (µmgr2/I)t ⇒ t = ω0r/µg(1 + mr2/I)

and the velocity is

v = ω0r/(1 + mr2/I)

Problem 7.3. A cylinder of the radius a is rolling without sliding inside a larger cylinder of the

radius R as shown in the figure (vertical cross-section).

25



Solution. Let the disk rotate by a small angle θ ! 1. The total energy is E = I θ2/2+2 · 1

2k(dθ)2,

so that ω2 = 2kd2/I, where I = mr2/2.

Problem 7.2. A ball of the radius r rotates with the angular velocity ω around the horizontal

axis passing through the center-of-mass. The ball is carefully put on a horizontal surface with the

friction coefficient µ. Find vcm(t).

Solution. The ball starts to move with sliding so that the friction force F = µmg accelerates

the ball but the torque N = Fr = µmgr decelerates the rotation:

mv = µmg ⇒ v = µgt

Iω = −µmgr ⇒ ω = ω0 − (µmgr/I)t

where I = (2/5)mr2. When v = ωr the sliding stops, the friction force vanishes and the ball continues

to roll without sliding so that v and ω no longer change. This happens when

µgt = ω0r − (µmgr2/I)t ⇒ t = ω0r/µg(1 + mr2/I)

and the velocity is

v = ω0r/(1 + mr2/I)

Problem 7.3. A cylinder of the radius a is rolling without friction inside a larger cylinder of the

radius R as shown in the figure (vertical cross-section).

a) Find the minimal angular frequency in the lowest point which allows to reach the highest point.

b) Find the frequency of small oscillations near the equilibrium.

Solution. When the center-of-mass of the small cylinder deviates by θ from the vertical position,

23

a) Find the minimal angular frequency in the lowest point which allows to reach the highest point.

b) Find the frequency of small oscillations near the equilibrium.

Solution. When the center-of-mass of the small cylinder deviates by θ from the vertical position,

it rotates by ϕ = Rθ/a− θ = (R− a)θ/a around its center of mass. The energy is therefore given as

follows

E =m

2(R− a)2θ2 +

I

2

(R− a)2θ2

a2+ mg(R− a)(1− cos θ)

Complete the solution.

Problem 7.4. A homogeneous cube is rotating around the axis passing through the center-of-

mass. Describe qualitatively the motion of the axis depending on the angle of the axis with the

normal to the cube side.

Solution. The axis does not move.

Problem 7.5. A hollow cylinder and a solid cylinder of the same radius start rolling simultaneously

without sliding down the same slope from the same height. What is the ratio of the final velocities

? Which one comes to the end of the slope earlier and what is the ratio of times ?

Solution. For the rolling without sliding one has

mgr sin α = (Icm + mr2)a/r

that is,

a =g sin α

1 + Icm/mr2

Since v2 = 2al, where l is the length of the slope, one has v ∝√

a. Respectively, t ∝ 1/√

a.

Problem 7.6. Two identical masses m connected by a massless rod of the length l are moving on

a circular orbit r = const around the Earth. The attraction force between the Earth and a point mass

26


is |F | = GMm/r2, where M is the Earth mass and G is a universal constant. Find the frequency of

small rotational oscillations of the system (masses on the rod) around the center-of-mass.

Solution. The equilibrium position is along the radius, when the center of mass moves on the

circular orbit. The angular velocity ω0 on the orbit is obtained from mω20r = GMm/r2, that is,

ω0 =√

GM/r3. Since the graviational force and the centrifugal force depend on the radius, different

forces are applied to the two masses. The total force on a mass at the radius R, rotating with the

rod, is F = −GMm/R2 + mω2R, so that the difference is ∆F = (dF/dR)l = (2GMm/r3 + ω20)l =

(3GMm/r3)l. If the rod rotates by a small angle θ the resulting torque is N = −∆Flθ, so that one

has

Iθ = −(3GMml2/r3)θ

where I = 2m(l/2)2 = ml2/2. Thus,

ω2 = 6GM/r3 = 6ω20

Problem 7.7. A car engine is applying a torque to a wheel. The wheel mass is m, radius is r and

the moment of inertia with respect to the center is I. The coefficient of the static friction with the

road is µ. What is the maximum torque N which can be applied without making the wheel slide ?

Solution. Let the friction force be f , then f = ma and N − fr = Iα = Ia/r. Excluding a one

gets f = Nrm/(I + mr2) ≤ µmg ⇒ N ≤ µg(1 + I/mr2).

Problem 7.8. A bobbin is moved by pulling a thread which is winded on the inner cylinder. The

outer radius is R, the inner radius is r, the bobbin mass is m, the friction coefficient is µ. What is

the maximal force F for which the bobbin rolls without friction ? What is the bobbin velocity after

it moves by the distance l from the rest ?


Problem 7.7. A car engine is applying a torque to a wheel. The wheel mass is m, radius is r and

the moment of inertia with respect to the center is I. The coefficient of the static friction with the

road is µ. What is the maximum torque N which can be applied without making the wheel slide ?

Solution. Let the friction force be f , then f = ma and N − fr = Iα = Ia/r. Excluding a one

gets f = Nrm/(I + mr2) ≤ µmg ⇒ N ≤ µg(1 + I/mr2).

Problem 7.8. A bobbin is moved by pulling a thread which is winded on the inner cylinder. The

outer radius is R, the inner radius is r, the bobbin mass is m, the friction coefficient is µ. What is

the maximal force F for which the bobbin rolls without friction ? What is the bobbin velocity after

it moves by the distance l from the rest ?

Solution. See previous problem. Here N = Fr.

Problem 7.9. Six identical point masses m are in the positions r1 = (a, 0, 0), r2 = (−a, 0, 0),

r3 = (a, a, 0), r4 = (−a,−a, 0), r5 = (0, 0, a), r6 = (0, 0,−a). The anglular velocity vector ω =

(ω, 0, 0). Find J.

Problem 7.10. Three identical disks with the mass m and radius r each are connected so that

they have the common center and their planes are mutually perpendicular. Find the moment of

inertia relative to an arbitrary axis passing through the center.

Solution. Let us choose axes x, y, z along the disks axes. Then for one of them Ixx ≡ I0 = 1

2mr2,

Iyy = Izz = I0/2 (other components). For the others one just has to change x to y or to z. Thus,

the total tensor of inertia is diagonal Ixx = Iyy = Izz = 2I0 = mr2, that is, Iij = mr2δij. This means

that the moment of inertia with respect to arbitrary axis passing through the common center is mr2.

25

Solution. See previous problem. Here N = Fr.

Problem 7.9. Six identical point masses m are in the positions r1 = (a, 0, 0), r2 = (−a, 0, 0),

r3 = (a, a, 0), r4 = (−a,−a, 0), r5 = (0, 0, a), r6 = (0, 0,−a). The anglular velocity vector ω =

27


(ω, 0, 0). Find J .

Problem 7.10. Three identical disks with the mass m and radius r each are connected so that

they have the common center and their planes are mutually perpendicular. Find the moment of

inertia relative to an arbitrary axis passing through the center.

Solution. Let us choose axes x, y, z along the disks axes. Then for one of them Ixx ≡ I0 = 12mr2,

Iyy = Izz = I0/2 (other components). For the others one just has to change x to y or to z. Thus,

the total tensor of inertia is diagonal Ixx = Iyy = Izz = 2I0 = mr2, that is, Iij = mr2δij. This means

that the moment of inertia with respect to arbitrary axis passing through the common center is mr2.

Problem 7.11. Two identical particles of the mass M are connected to the two ends of a

rigid massless rod of the length a. The system initially rotates around the center-of-mass with the

angular velocity ω. One of the particles encounters a third one (with the same mass) at rest, which

momentarily sticks to it. What is the angular velocity of the rotation around the center-of-mass after

the collision ? (No gravity.)

Typical problems for exam

Problem 0.1. A particle is moving so that ϕ = at2, r = r0 exp(bt), where a, b, and r0 are

constants. Find the force.

Problem 0.2. A particle (mass m) is moving in the central field U = −k/r on a circular orbit

r = r0. The energy and angular momentum suddenly are changed by ∆E and ∆J . What are the

maximal and minimal distances from the attracting body on the new orbit ?

Problem 0.3. A particle is in the stable equilibrium in the potential energy U(x) = U0[1 −

l2/(l2 +x2)]. Suddenly it gets a small addition of energy E ′. Assuming that the oscillations are small

find the frequency and amplitude.

Problem 0.4. Two identical particles of the mass M are connected to the two ends of a rigid

massless rod of the length a. The system initially rotates around the center-of-mass with the angular

velocity ω. One of the particles encounters a third one (with the same mass) at rest, which momen-

tarily sticks to it. What is the angular velocity of the rotation around the center-of-mass after the

collision ? (No gravity.)

Problem 0.5. A homogeneous ball (mass m, radius r) is struck by a horisontal force F in the

point which is above the center by the distance l < r. The time t of force action is very small, but

Ft is nonzero. Find the velocity of the center-of-mass and the angular velocity of the ball around

1

Problem 7.12. A homogeneous ball (mass m, radius r) is struck by a horisontal force F in the

point which is above the center by the distance l < r. The time t of force action is very small, but

Ft is nonzero. Find the velocity of the center-of-mass and the angular velocity of the ball around

the axis which goes through the center-of-mass, if a) there is no friction with the floor, and b) if the

friction prevents sliding. (ICM = 25mr2)

the axis which goes through the center-of-mass, if a) there is no friction with the floor, and b) if the

friction prevents sliding. (ICM = 2

5mr2)

Problem 0.6. A source of light is moving with the angular velocity ω on the circle with the

radius R in the plane x− y. The light frequency in the source frame is f . A distant observer (in the

same x− y plane) measures the frequency of received light as a function of time. Find this function.

2

28


8 Gravitation

Problem 8.1. What force acts on a star inside a spherically symmetric galaxy of the mass M

and radius R. The star has a mass m and is at bin the radius r < R from the center of the galaxy.

Solution. As if all stars within r were in the center: F = GM(r3/R3)m/r2.

Problem 8.2. A binary stellar system consists of two identical stars rotating around the center-

of-mass of the system on circular orbits. The period of rotation T and the velocity of the stars v are

known. Find the masses and the distance between the stars.

Solution.

l = v/2πT

F = Gm2/l2 = mv2/(l/2)

Problem 8.3. Saturn rings consist of football ball size particles which are moving on circular

orbits around the planet. What is the maximal ratio of the ring width to its inner radius if the

velocities at the inner and outer edge should not differ by more than 0.5% ?

Solution. mv2/r = GMm/r2 → r = GM/v2 → ∆r/r = 2∆v/v.

Problem 8.4. Three identical stars with the mass m are rotating so that they form an equilateral

triangle (side length a). What is the angular velocity ? What is the ratio J/E ? Is this configuration

stable ?

Solution.

F = 2(Gm2/a2) cos 30

F = mω2(2a cos 30/3)

ω =√

3Gm/r3

J = Iw, I = 3m(2a cos 30/3)2

K = Iw2/2

U = −3Gm2/a

Problem 8.5. A particle is moving along the axis of a homogeneous ring (mass M , radius R).

The particle velocity at infinity is zero. What is its velocity when it passes through the center of the

ring ?

29


Solution.

U(z) = −GMm/√

R2 + z2

E = 0 ⇒ z = 0 → mv2/2 = GMm/R

Problem 8.6. The space between the two concentric spheres with the radii a and b, a < b, is

filled with a matter with the constant density ρ. Find the gravitational field g as a function of radius

r in the whole space.

Solution. g = F/m

r < a ⇒ g = 0

a < r < b ⇒ M(r) = 43πρ(r3 − a3)

g = GM(r)/r2

r > b ⇒ M = 43πρ(b3 − a3)

g = GM/r2

Problem 8.7. A binary system consists of two stars with the masses M and 2M . The distance

between them is R. Find the period of the orbital motion.

Solution.

F = 2GM2/R2

µ = m1m2/(m1 + m2) = 2M/3

F = µω2R

9 Relativity

Problem 9.1. A box has the volume V in its rest frame. What is its volume in the frame moving

with the velocity v0 ?

Solution. One dimension is contracted by γ, so that V ′ = V/γ.

Problem 9.2. Two bodies are moving with the same speed v. The angle between their velocities

is α. What is the relative velocity ?

Solution.

v0 = (−v, 0, 0)

30


v′ = (v cos α, v sin α, 0)

γ = 1/√

1− v2/c2

vx =v′x + v0x

1 + v′xv0x/c2=

v(cos α− 1)

1− v2 cos α/c2

vy =v′y

γ(1 + v′xv0x/c2)=

v sin α

γ(1− v2 cos α/c2)

Problem 9.3. The momentum of a particle is p and the energy is E. Find the velocity v.

Solution. v = pc2/E

Problem 9.4. A star is moving on a circular orbit with the radius R and period T . The star

emits at the wavelength λ in its rest frame. The line of sight of a distant observer (distance R) is

in the orbit plane (makes the angle 90 with the normal to the orbit plane). Find λ′max− λ′min/λ (λ′

is the wavelength measured by the distant observer).

Solution.

v = 2πR/T

λmax

λ=

√1 + v/c

1− v/c

λmin

λ=

√1− v/c

1 + v/c

Problem 9.5. Same as in the previous problem but the angle between the line of sight and the

orbit plane is θ < 90.

Solution.

v = 2πR/T

λmax

λ= γ(1 + v cos θ/c)

λmin

λ= γ(1− v cos θ/c)

γ = 1/√

1− v2/c2

Problem 9.6. A hot gas of identical atoms emitting at the same wavelength λ0 (in the rest frame)

is observed by a distant observer. The atoms are moving and the number of atoms moving with the

velocity v (parallel to the line of sight) is given but the relation N(v) = C exp(−v2/2v2T ), where C

and vT are constant. Each atom emits the energy E. How much energy obtaines the distant observer

31


at the wavelength λ.

Problem 9.7. In the rest frame of the particle A, the particle B moves with the velocity vB, and

the particle C moves with the velocity vC . What is the angle between the velocities of A and C in

the rest frame of B ?

Problem 9.8. In the process of β-decay a neutron n decays in into a proton p, electron e− and

antineutrino ν: n → p + e− + ν. The rest masses obey the following inequality mn > mp + me, the

rest of the antinuetrino mν = 0. What is the energy of the antineutrino if the proton momentum is

negligible ?

Solution.

pν = −pe → |pν | = |pe| = p

mnc2 = mpc

2 +√

mec4 + p2c2 + pc

Problem 9.9. What should be the electron Lorentz factor γ in order that the gyroradius of the

gyration around the magnetic field B = 10−3T be equal to the Earth radius R = 6400 km? Find

yourself all relevant data.

Solution.

p = qv ×B

p = mvγ, γ = const

mγv = qv ×B ⇒

ω = qB/mγ

r = v/ω ≈ c/ω

Problem 9.10. A charged particle of the charge q and mass m is moving in the electric field

E = E0 sin(ωt). Find γ(t) if the particle is in the rest at t = 0.

Solution.

p = qE0 sin(ωt)

p =qE0

ω(1− cos(ωt))

γ =√

p2/m2c2 + 1

Problem 9.11. A source of light is moving with the angular velocity ω on the circle with the

32


radius R in the plane x− y. The light frequency in the source frame is f . A distant observer (in the

same x− y plane) measures the frequency of received light as a function of time. Find this function.

10 Problems given at various tests, including midterms and

finals

Problem 10.1. A particle of the mass m is moving on the elliptic orbit r = p/(1− ε cos ϕ). The

particle is attracted to the focus O1 by the force |F1| = k1/r21 (where |r1| is the distance between

the focus O1 and the particle). In addition, it is attracted to the ellipse center O2 with the force

|F2| = k2r2. The third force acting on the particle is always perpendicular to the particle velocity.

When the particle is in A, its velocity is v1 = (0, v1). What is its angular momentum in B ?

x

y

O1 O2A

B

Solution. Points:

First we need the coordinates of the points A, B, O1, and O2:

1. O1 is the coordinate origin, so that rO1 = xO1 = yO1 = 0.

2. A corresponds to ϕA = π, so that rA = p/(1 + ε), xA = −p/(1 + ε), yA = 0.

3. It is convenient to use the point C: ϕC = 0, rC = p/(1− ε), xC = p/(1− ε), yC = 0.

4. O2 is in the middle between A and C, so that xO2 = (xA + xC)/2 = εp/(1 − ε2), yO2 =

(yA + yC)/2 = 0.

5. B is just above O2, so that xB = xO2 = εp/(1 − ε2). The equation r = p(1 − ε cos ϕ) can be

written as

r − εr cos ϕ = p → r = p + εx, x = r cos ϕ

so that rB = p/(1− ε2). Now

yB =√

r2B − x2

B =p√

1− ε2

Force vectors:

We denote the position of the particle with r = (x, y), respectively, other points rO1 = (0, 0),

rO2 = (xO2 , 0), rA = (xA, 0), rB = (xB, yB).

33


Now we introduce the vectors r1 = r − rO1 = r and r2 = r − rO2 . With this notation the forces

(both attractive) will be written as follows

F1 = −k1

r21

r1,

F2 = −k2r2r2

where r1 and r2 are unit vectors, r1 = |r1| and r2 = |r2|.Potential energy:

Both forces are central forces: the direction of F1 is to O1 and the magnitude depends only on

|mbr1|, the direction of F2 is to O2 and the magnitude depends only on |mbr2|. Central forces are

conservative, and the relation to the potential energy is the following:

F1 = −∂U1

∂r1

r1, F2 = −∂U2

∂r2

r2

which gives∂U1

∂r1

=k1

r21

,∂U2

∂r2

= k2r2

and finally

U1 = −k1

r1

, U2 =k2r

22

2

Energy conservation:

Since the third (unknown) force is always perpendicular to the velocity, F3 ⊥ v, it does not

produce work, so that the energy conservation gives

mv2

2+ U1 + U2 =

mv2

2− k1

r1

+k2r

22

2= E = const

When the particle is in the point A, we have v = (0, v1), r1 = p/(1 + ε), r2 = p/(1− ε2).

When the particle is in the point B, we have v = (0, v) (this v is the unknown we are looking

for), r1 = p/(1− ε2), r2 = p/√

1− ε2.

Therefore, we have

mv21

2− k1(1 + ε)

p+

k2p2

2(1− ε2)2=

mv2

2− k1(1− ε2)

p+

k2p2

2(1− ε2)

Finally

v =

[v2

1 −2k1ε(1− ε)

mp+

k2ε2p2

(1− ε2)2

]1/2

Angular momentum:

34


J = mr × v. In the point B the velocity is v = (v, 0, 0), so that

Jz = −mvyB = − mvp√1− ε2

Problem 10.2. A particle, initially resting in the coordinate origin, suddenly breaks up into three

particles with the masses m1, m2, and m3. The particle m1 has the charge q > 0. It starts moving

into negative x-direction in the homogeneous magnetic field B = (0, 0, B). After having completed

half a circle the particle finds itself at the distance l1 from the starting point. The particle m2 has

the charge −q < 0. It starts moving in the positive x-direction, and after having completed half a

circle is at the distance l2 from the starting point. What is the velocity of the third particle ? The

magnetic force is F = qv ×B.

x

y

m1

m2

Solution. For each particle in the magnetic field ma = qv × B. Since v ⊥ B, we can write

m|a| = |q||v||B|, ormv2

r= |q|B

which gives

v =|q|Br

m

From this expression we immediately find (l = 2r for a semicircle !)

v1 =qBl12m1

, v2 =qBl22m2

Momentum conservation gives

m1v1 + m2v2 + m3v3 = 0

35


or

v3 = −m1v1 + m2v2

m3

Substituting v1 = v1ex, v2 = −v2ex, we find

v3 =m2v2 −m1v1

m3

ex =qB(l2 − l1)

2m3

ex

Problem 10.3. A particle of the mass m is connected with the point O with the use of a massless

rod of the length L. The rod can freely rotate around O. When the angle between the rod and the

vertical is θ0 the particle velocity is v0 = v1θ + v2ϕ. a) What is conserved ? b) Find vθ and vϕ (or

θ and ϕ) as functions of θ; c) Find θ and ϕ as functions of θ; d) Find the maximum speed |v|max of

the particle; e) Find the tension in the rod as a function of θ.

O

θg ↓

1

Solution. Preparations:

F = −mgz + T r

v = vθθ + vϕϕ

= rθθ + r sin θϕϕ

r = 0

a):

E =mv2

2+ U, U = mgz = mgr cos θ

E =m

2(v2

θ + v2ϕ) + mgr cos θ

=m

2(r2θ2 + r2 sin2 θϕ2) + mgr cos θ

36


E = T · v = T r · (vθθ + vϕϕ) = 0

E = const =m

2(v2

1 + v22) + mgL cos θ0

p = F 6= 0

J = r ×mv = mrr × (vθθ + vϕϕ)

= mrvθϕ−mrvϕθ

J = r × F = −rmgr × z

Jz = 0

Jz = J · z = mr sin θvϕ = mr2 sin2 θϕ

Jz = const = mv2L sin θ0

b):

vϕ =Jz

mr sin θ

ϕ =vϕ

r sin θ=

Jz

mr2 sin2 θm

2(v2

θ + v2ϕ) + mgr cos θ = E

vθ = ±√

2E

m− v2

ϕ − 2gr cos θ

= ±√

2E

m− J2

z

m2r2 sin2 θ− 2gr cos θ

θ =vθ

r= ±1

r

√2E

m− J2

z


c):

θ =dθ

dθ· θ

=d

dθ

(±1

r

√2E

m− J2

z


)

·

(±1

r

√2E

m− J2

z


)

= ±1

r

(2E

m− J2

z


)−1/2

·(

J2z cos θ

m2r2 sin3 θ+ gr sin θ

)θ

=1

r2

(J2

z cos θ

m2r2 sin3 θ+ gr sin θ

)

37


ϕ =dϕ

dθ· θ

= ∓1

r

√2E

m− J2

z


(2Jz cos θ

mr2 sin3 θ

)d):

v2 =2

m(E −mgr cos θ)

d

dθv2 = 2gr sin θ > 0

v = vmax → θ = θmax > π/2

θ = θmax ⇒ θ = 0 ⇒2E

m− J2

z

m2r2 sin2 θmax

− 2gr cos θmax = 0

e):

(T + mg) · r = −mv2

r

T −mg cos θ = −mv2

r= −2E − 2mgr cos θ

r

T = 3mg cos θ − 2E

r

Problem 10.4. A particle moves according to r = r0 exp(ϕ) in the central force. Find U(r).

Solution. In the central force energy and angular momentum are conserved. Energy conservation

gives

E = 12mr2 + 1

2mr2ϕ2 + U = const.

Angular momentum conservation gives:

mr2ϕ = J = const.

and therefore

ϕ =J

mr2.

From the given relation r = r0 exp(ϕ) we have

r = (d/dt)(r0 exp(ϕ)) = r0 exp(ϕ)ϕ = rϕ =J

mr.

38


Substituting (??) and (??) into the energy conservation (??) we finally obtain

E =J2

mr2+ U

and

U = E − J2

mr2

Problem 10.5. A particle moves according to

x = A exp(−γt) cos(ωt)

y = A exp(−γt) sin(ωt)

Find tangential acceleration.

Solution. The problem is solved by direct differentiating: vx = x, vy = y, ax = vx, ay = vy, so

that the velocity and acceleration are:

vx = x = −γA exp(−γt) cos(ωt)− ωA exp(−γt) sin(ωt) = −γx− ωy,

vy = −γy + ωx,

ax = −γvx − ωvy = (γ2 − ω2)x + 2γωy,

ay = −γvy + ωvx = (γ2 − ω2)y − 2γωx.

Tangential acceleration is parallel to the velocity vector, so that a‖ = a · v/|v|:

a‖ = (axvx + ayvy)/(v2x + v2

y)1/2

= −γ√

γ2 + ω2A exp(−γt)

Problem 10.6. A particle of the mass m moves in the potential energy U(x) = −Ax2/2+Bx3/3,

A > 0, B > 0. Find the frequency of small oscillations.

Solution. Equilibrium is where Fx = −(dU/dx) = 0:

−dU

dx= −Ax + Bx2 = 0 → x1 = 0, x2 = A/B.

Whether the equilibrium is stable or not is determined by the second derivative (d2U/dx2) = −A +

2Bx. If (d2U/dx2) < 0 the point is a maximum and the equilibrium is unstable, if (d2U/dx2) > 0

the point is a minimum and the equilibrium is stable. In x = x1 = 0 we have (d2U/dx2) = −A < 0

so that this is an unstable equilibrium. In x = x2 = A/B we have (d2U/dx2) = A > 0 and this is a

39


stable equilibrium.

Near the equilibrium point the potential energy can be Tayor expanded:

U(x) ≈ U(x2) + 12A(x− x2)

2.

Let us denote X = x− x2 then vx = x = X and the energy conservation is written as

12mX2 + 1

2AX2 = const

which immediately gives (according to the general rule)

ω2 = A/m.

Problem 10.7. A hollow cylinder is sliding without friction (no rolling) with the velocity v. The

cylinder comes to a surface with friction. What is the final velocity of the cylinder ?

Solution. When the cylinder comes to the surface with friction it is decelerated by the friction

force and at the same time its rotation is acceletrated until the cylinder begins to roll without sliding.

In the rolling state the friction force is zero and the velocity does not change. Let the friction force

magnitude be Fs. Then the deceleration is

mV = −Fs → V (t) = v − Fs

mt,

whereV (t) is the velocity of the center-of-mass in the moment t. The moment of the friction force

(torque) accelerates the rotation around the axis passing through the center-of-mass, as follows:

Iω = Fsr → ω(t) =Fsr

It,

where ω(t) is the angular velocity of rotation and I is the moment of inertia around the axis passing

through the center-of-mass. According to the definition I =∑

mir2i , where ri is the distance from

the rotating mass (small part of the body) from the rotation axis. All points of the cylinder are at

the same distance r from the rotation axis, so that we have

I =∑

mir2i =

∑i

mir2 = (

∑i

mi)r2 = mr2.

From (??) and (??) we find

ω(t) =Fs

mrt.

The velocity and angular velocity stop changing when sliding stops. The condition of the rolling

40


without sliding is V = ωr which gives

v − Fs

mt =

Fs

mrtr =

Fs

mt,

and therefore, in the moment when the velocity stops changing

Fs

mrt = v/2.

Substituting into (??) we have

Vfinal = v/2.

Problem 10.8. A rod of the mass M and length l can rotate in the vertical plane around the

axis which passes through the point at a < l/2 from its upper end. A bullet of the mass m M

with the horizontally directed velocity v strikes the rod at the upper end and remains in it. What is

the rotation angle of the rod ?

v

m, v

E ! mc2

λ

µ

µ " m " M M m

R

2

Solution. Angular momentum is conserved during the collision, which means

Iω = mva → ω = mva/I,

where ω is the angular velocity of the rotation just after the collision, and I is the moment of inertia.

After the collision the energy is conserved, that is, the initial kinetic energy goes into the potential

energy as the bar rotates to the angle θ and its center-of-mass rises. Therefore, for the maximum

angle we have

Iω2/2 = MgL(1− cos θ) → cos θ = 1− Iω2/2MgL.

where L = l/2 − a is the distance between the axis and center-of-mass of the bar (we neglect the

mass of the bullet).

Calculation of I: Let us choose coordinate origin in the axis, and the positive direction downwards.

41


Then

I =

∫ l−a

−a

x2(M/l)dx =M

3l[(l − a)3 + a3]

Combine all calculations.

Problem 10.9. An electron moving with the energy E mc2 toward the coordinate origin emits

a photon with the wavelength λ forward. What wavelength measures a non-moving observer in the

coordinate origin ?

Solution. We shall use directly the expression for the Doppler effect

λ′

λ=

1

γ(1− v cos θ/c).

In our case the source is moving towards the receptor, so that θ = 180 and

λ′

λ=

√1− v/c

1 + v/c.

We need velocity which can be found as follows: γ = E/mc2 and v = c√

1− 1/γ2.

Problem 10.10. Potential energy (2D) is given by U = A cos ϕ/ρ2. A particle of the mass m is

in the point r = (a, a) and its velocity is v ⊥ r. Find the normal acceleration.

Solution. By definition an ⊥ v. Since r ⊥ v we have an ‖ r. Thus, the magnitude an = |a · r|.a) The long way. r = (ax + ay)/

√2a2 = (x + y)/

√2. Express the potential energy in Cartesian

coordinates x = r cos ϕ, y = sin ϕ: U = Ax(x2 + y2)−3/2. Then

F = −∂U

∂xx− ∂U

∂yy

= −(

A

r3− 3Ax2

r5

)x +

(3Axy

r5

)y

=A(3 cos2 ϕ− 1)

r3x +

3A sin ϕ cos ϕ

r3y.

Now a = F /m, r = a√

2, and ϕ = 45, so that an = A/2ma3.

b) The short way. F = −(∂U/∂r)r−(1/r)(∂U/∂ϕ)ϕ, so that an = |(∂U/∂r)|/m = |2A cos ϕ/r3|/m =

A/2ma3.

Problem 10.11. A satellite of the mass m is moving on an elliptical orbit round the Earth (mass

M m), so that rmax = 2rmin. The satellite energy E < 0 is known. Find the angular momentum.

Solution. When r = rmin or r = rmax the radial velocity vanishes r = 0, so that in these points

42


the energy and momentum conservation give

E =J2

2mr2− GMm

r.

Thus,

r1,2 =GMm±

√(GMm)2 − 2|E|J2/m

2|E|

(here we used E = −|E| < 0) and

GMm +√

(GMm)2 − 2|E|J2/m

GMm−√

(GMm)2 − 2|E|J2/m= 2.

Solving this equation we get

J =2GMm

√m

3√|E|

.

Another way of solving: from the energy expression we have

J2/2m = r2E + GMmr = (2r)2E + GMm(2r),

so that r = −GMm/3E and

J =√

2m(r2E + GMmr)

same as above.

Problem 10.12. Two disks of the masses m1 and m2 and radii R1 and R2 are connected to a

massless rod which can rotate in the vertical plane. The disk centers are at the distances l1 and l2

from the rotation axis. Find the frequency of small oscillations.

m1, R1

m2, R2

l1

l2

m R M

h r

B = (0, B, 0) x − y

q > 0 m

2

43


Solution. If the system moved to the angle θ 1 and the angular velocity is θ, the energy is

E =Iθ2

2+ mgL(1− cos θ) =

Iθ2

2+

mgLθ2

2,

where

I = (m1R

21

2+ m1l

21) + (

m2R22

2+ m2l

22)

is the moment of inertia relative to the axis, and

L =m1l1 −m2l2

m1 + m2

is the distance of the center of mass from the axis. The frequency is ω2 = mgL/I.

Problem 10.13. To a disk of the mass M and radius R a smaller disk of the mass mv and radius

r is connected. At what height a horisontal force should be applied in order that the body start (in

the very first moment) to roll without sliding. There is no friction.

m1, R1

m2, R2

l1

l2

m R M

h r

B = (0, B, 0) x − y

q > 0 m

2

Solution. Let the force be F . The acceleration of the center of mass is a = F/(M + m). The

angular acceleration around the contact point is α = N/I = Fh/I, where the moment of inertia is

I = (MR2/2 + MR2) + (mr2/2 + mr2) = (3/2)(MR2 + mr2). If there is no sliding, the linear and

angular acceleration are related by a = αL, where L = (MR + mr)/(M + m) is the distance of the

center of mass from the rotation axis (contact point). Thus, we have F/(M + m) = (Fh/I)L, so

that h = I/L(M + m).

Problem 10.14. There is a homogeneous magnetic field B = (0, B, 0) between two plates,

parallel to the x − y plane. A particle of the mass m and electric charge q > 0 enters the space

between the planes through the lower plate, when its velocity is v = (v cos 45, 0, v sin 45). What

is the minimal distance between the plates for which the particle will come back to the lower plate

44


(without touching the upper one) ? Gravity is negligible.

v = (v cos 45, 0, v sin 45)

x

z

v

v1 = (0.5c, 0.5c, 0)

|vrel| v2 = (−0.5c, 0.5c, 0)

R M

3

Solution. The particle moves on a circle of the radius r which is obtained from F = qvB =

mv2/r, that is, r = mv/qB, as shown in the figure.

Physics 1. Mechanics Solutions 2

x

z

v

From the figure it is clear that the distance between the two planes should be L > r + r/√

2.

Problem 1.6. The easiest way is to rotate the coordinates so that the new x axis will be along v1,

then the new y axis will be along v2. In the new coordinates v1 = (0.5√

2c, 0, 0), v2 = (0, 0.5√

2c, 0).

Taking v1 ≡ v0 as the velocity of the moving frame (S ′), and v2 as the velocity of a body we are

looking at, we have

v′

x =vx − v0

1 − vxv0/c2= −v0 = −0.5

√2c, (11)

v′

y =vy

γ(1 − vxv0/c2)=

vy

γ=

0.5√

2c

γ. (12)

where we used vx = (v2)x = 0, and

γ = (1 − v2

1/c2)−1/2 =

√2. (13)

Thus, we have v′

x = −0.5√

2c, v′

y = 0.5c, and v′ =√

v′

x2 + v′

y2 = 0.5

√3c.

For advanced only !

Problem 1.7. Let us consider a small part of the mass of the length dr and cross section dA at

the radius r. This part has a mass dm = ρdAdr, where the density ρ = 3M/4πR3. Is is attracted to

the center by the mass M ′ = ρ(4πr3/3) inside the radius r. This force is balanced by the pressure

forces. From the inside the pressure pushed it outwards with the force p[r]dA. From the outside the

pressure pushes it inward with the force p[r + dr]dA. Thus, we have

p[r]dA − p[r + dr]dA =GM ′dm

r2=

GMrρ

R3dAdr. (14)

3

From the figure it is clear that the distance between the two planes should be L > r + r/√

2.

Problem 10.15. An observer on Earth sees one galaxy moving with the velocity v1 = (0.5c, 0.5c, 0)

and another with v2 = (−0.5c, 0.5c, 0). Find their relative velocity |vrel|.Solution. The easiest way is to rotate the coordinates so that the new x axis will be along v1,

then the new y axis will be along v2. In the new coordinates v1 = (0.5√

2c, 0, 0), v2 = (0, 0.5√

2c, 0).

Taking v1 ≡ v0 as the velocity of the moving frame (S ′), and v2 as the velocity of a body we are

looking at, we have

v′x =vx − v0

1− vxv0/c2= −v0 = −0.5

√2c,

v′y =vy

γ(1− vxv0/c2)=

vy

γ=

0.5√

2c

γ.

where we used vx = (v2)x = 0, and

γ = (1− v21/c

2)−1/2 =√

2.

Thus, we have v′x = −0.5√

2c, v′y = 0.5c, and v′ =√

v′x2 + v′y

2 = 0.5√

3c.

Problem 10.16. Potential energy is given as U = −A/r4, A > 0. When the particle (mass m)

is in the point (a, 0, 0) its velocity is v = (0, v, 0) and its energy is negative. What is the maximum

45


distance between the particle and the coordinate origin ?

Solution. Central force means that J = mr×v = const and E = mv2r/2+J2/2mr2+U = const.

From the initial conditions we have J = J z, J = mav, E = mv2/2 − A/a4. Since E < 0 we have

v2 < 2A/ma4. In the closest and farthest points vr = 0 so that we have

J2

2mr2− A

r4= E

where J , m, E < 0, and A are known. The equation is easily solved to give

r2 = − J2

4m|E|±

√(

J2

4m|E|)2 +

A

|E|

The solution with − is not physical (r2 < 0) which means that there is no minimum radius: the

particle reaches the maximum radius, bounces back and falls into r = 0.

Problem 10.17. A particle moves according to x = r0 cos(ωt), y = r0 sin(ωt), z = kt2/2. Find

the normal acceleration.

Solution.

v = ωr0(− sin(ωt)x + cos(ωt)y) + ktz

a = −ω2r0(cos(ωt)x + sin(ωt)y) + kz

an = |a× v|

= ωr0

√ω2k2t2 + k2 + ω4r2

0√ω2r2

0 + k2t2

Problem 10.18. Potential energy is U(x) = A(x − a)2(x − b)2, A > 0. Find the frequency of

small oscillations.

Solution. Equilibrium: dU/dx = 0 → x = a or x − b. It is easy to find that U = A(b − a)2X2

where X = x− a or X = x− b. Thus, ω2 = 2A(b− a)2/m.

Problem 10.19. Three identical cylinders are in equilibrium (see figure). Find the minimum

friction coefficient between the cylinders.

dve‘zd z‘ ‘vn . z = at2/2 , y = r0 sin(ωt) , x = r0 cos(ωt) itl rp wiwlg

.zilnxepd

Solution .

v = ωr0(− sin(ωt)x + cos(ωt)y) + atz

a = −ω2r0(cos(ωt)x + sin(ωt)y) + az

an = |a × v|

= ωr0

√ω2a2t2 + a2 + ω4r2

0√ω2r2

0 + a2t2

3. dl‘y

ly zeiexicz ‘vn . U(x) = A(x − a)2(x − b)2 i’’r dpezp zil‘ivphet dibxp‘

.(cg‘n xzei) lwyn ieeiy cil zephw zecepz

Solution . Equilibrium: dU/dx = 0 → x = a or x − b. It is easy to find that

U = A(b − a)2X2 where X = x − a or X = x − b. Thus, ω2 = 2A(b − a)2/m.

4. dl‘y

mcwn z‘ ‘va .xeiva d‘xpy itk lwyn ieeiya mi‘vnd midf mililb dyely

.mililbd oia ilnipind jekigd

Solution . Let the normal force acting on the lower cylinder from the upper is

N , and the friction force between the upper and lower cylinders is fs, fs/N ≤ µ.

From the torque balance on the lower cylinder the friction force between the lower

2

Solution. Let the normal force acting on the lower cylinder from the upper is N , and the friction

46


force between the upper and lower cylinders is fs, fs/N ≤ µ. From the torque balance on the lower

cylinder the friction force between the lower cylinder and the table is fs. The force balance on the

lower cylinder in the horisontal direction is

fs + fs cos 30 = N cos 60 → µ ≥ fs

N=√

3/3

Problem 10.20. A body is built of two small identical balls connected with a massless rod of

the length l. Initially the body is at rest. Another identical ball, moving with the velocity v, collides

elastically with one of these two (see figure). Find the angular velocity of the body rotation after

the collision.

v

N yig‘ ote‘a mibltzn (dgepn zkxrna) a, b, c zecin zlra daiza6. dl‘y

v zexidn rpd dtevd cecni (gtp zcigil xtqn) miwiwlgd zetitv efi‘ .miwiwlg

? a rlv jxe‘l

Solution . a′ = a/γ, V ′ = V/γ → n′ = nγ

4

Solution. Elastic collision means that the momentum, angular momentum, and energy are

conserved. Let the velocity of the third ball after the collision be v′ (it will be in the same direction

as the initial velocity or in the opposite direction - in the last case v′ < 0). Let the velocity of the

center of the rod be V and the angular velocity of the rotation around this center (counterclockwise)

be ω. Momentum conservation gives

mv = mv′ + 2mV

Angular momentum conservation (relative to the collision point):

0 = 2mV (l/2)− 2m(l/2)2ω

Energy conservation givesmv2

2=

mv′2

2+

2mV 2

2+

2m(l/2)2ω2

2

Solving these equations we obtain ω = v/l.

Problem 10.21. In a box of the size a× b× c (in its rest frame) N particles are homogeneously

distributed. Find the particle density as viewed by an observer moving with the velocity v along a.

Solution. a′ = a/γ, V ′ = V/γ → n′ = nγ

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Physics 1 - Problems With Solutions

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