Chapter 9 Rotation of rigid bodies - d.umn.edud.umn.edu/~vvanchur/2015PHYS2013/Chapter9.pdf ·...

Chapter 9

Rotation of rigid bodies

9.1 Angular Velocity and Acceleration

Angular quantities. Angular position is most conveniently describe interms of radians defined by

2π rad ≡ 360◦ = 1 rev (9.1)

or

1 rad ≡360◦

2π=

1

2πrev. (9.2)

One can convert from one angular unit to another using,

1 =

!

2π rad

360◦

"

1 =

!

1rev

2π rad

"

1 =

!

360◦

1rev

"

For example,

180◦ = 180◦!

2π rad

360◦

"

= π rad =1

2rev

90◦ = 90◦!

2π rad

360◦

"

=π

2rad=

1

4rev

45◦ = 45◦!

2π rad

360◦

"

=π

4rad=

1

8rev

60◦ = 60◦!

2π rad

360◦

"

=π

6rad =

1

12rev. (9.3)

113

CHAPTER 9. ROTATION OF RIGID BODIES 114

Similarly to how linear average (velocity and acceleration)

vavg ≡∆x

∆t

aavg ≡∆v

∆t(9.4)

and instantaneous (velocity and acceleration)

v ≡ lim∆t→0

∆x

∆t=

dx

dt

a ≡ lim∆t→0

∆v

∆t=

dv

dt=

d2x

dt2(9.5)

was defined from linear position replaced, the average angular (velocity andacceleration)

ωavg ≡∆θ

∆t

αavg ≡∆ω

∆t(9.6)

and the instantaneous angular (velocity and acceleration)

ω ≡ lim∆t→0

∆θ

∆t=

dθ

dt

α ≡ lim∆t→0

∆ω

∆t=

dω

dt=

d2θ

dt2. (9.7)

Example 9.1-9.2. The angular position θ of a 0.36m-diameter flywheelis given by

θ(t) =#

2.0 rad/s3$

t3. (9.8)

a) Find θ, in radians and in degrees, at t1 = 2.0 s and t2 = 5.0 s.b) Find the total distance (not displacement) that a particle on the flywheelrim moves over the time interval from t1 = 2.0 s to t2 = 5.0 s.c) Find the average angular velocity, in rad/s and in rev/min over that in-terval.d) Find the instantaneous angular velocity at t1 = 2.0 s and t2 = 5.0 s.e) Find the average angular acceleration between t1 = 2.0 s and t2 = 5.0 s.f) Find the instantaneous angular acceleration at t1 = 2.0 s and t2 = 5.0 s.


a) The angular position in degrees is

θ(2.0 s) =#

2.0 rad/s3$

(2.0 s)3 = 16 rad

!

360◦

2πrad

"

≈ 920◦ = 2× 360◦ + 200◦

θ(5.0 s) =#

2.0 rad/s3$

(5.0 s)3 = 250 rad

!

360◦

2πrad

"

≈ 14330◦ = 39× 360◦ + 290◦. (9.9)

b) The distance that a particle moves is

s =d

2(θ(t2)− θ(t1))

= (0.18m) (14000◦ − 920◦)

!

2πrad

360◦

"

= 42m (9.10)

c) The average angular velocity is

ωavg =θ(t2)− θ(t1)

t2 − t1

=

!

14000◦ − 920◦

5.0 s− 2.0 s

"!

2πrad

360◦

"

= 78 rad/s (9.11)

or

ωavg = 78 rad/s

!

1 rev

2π rad

"!

60 s

1min

"

= 740 rev/min. (9.12)

d) The instantaneous velocity is

ω =dθ

dt=#

6.0 rad/s3$

t2 (9.13)

or

ω(2.0 s) =#

6.0 rad/s3$

(2.0 s)2 = 24 rad/s

ω(5.0 s) =#

6.0 rad/s3$

(5.0 s)2 = 150 rad/s. (9.14)

e) The average angular acceleration is

αavg =ω(5.0 s)− ω(2.0 s)

5.0 s− 2.0 s

=126 rad/s

3.0s= 42 rad/s2. (9.15)


f) The instantaneous acceleration is

α =dω

dt=#

12.0 rad/s3$

t (9.16)

or

α(2.0 s) =#

12.0 rad/s3$

(2.0 s) = 24 rad/s2

α(5.0 s) =#

12.0 rad/s3$

(5.0 s) = 60 rad/s2. (9.17)

9.2 Rotation with constant angular accelera-tion

The angular velocity and acceleration can be also thought of as vector quan-tities pointing along the axis of rotation with direction determined by theright-hand rule. Then, at the level of magnitudes there is the following equiv-alence of linear and angular quantities

x → θ

vx → ωz

ax → αz. (9.18)

and so for motion with constant acceleration we have

a(t) = a → α(t) = αz

v(t) = v0 + axt → ω(t) = ω0z + αzt

x = x0 + v0xt+1

2axt

2 → θ(t) = θ0 + ω0zt +1

2αzt

2 (9.19)

as well useful relations

vx(t)2 = v20x + 2ax(x− x0) → ωz(t)

2 = ω20z + 2αz (θ(t)− θ0)

x(t)− x0 =1

2(v0x + vx(t)) t → θ(t)− θ0 =

1

2(ω0z + ωz(t)) t. (9.20)

Example 9.3. You have finished watching a movie Blu-ray and the diskis slowing to a stop. The disc’s angular velocity at t = 0 is 27.5 rad/s, andits angular acceleration is a constant −10.0 rad/s2. A line PQ on the disc’ssurface lies along the +x-axis at t = 0 s.


a) What is the disc’s angular velocity at t = 0.300 s?b) What angle does the line PQ make with the +x-axis at this time?a) The disc’s angular velocity is

ωz(t) = ω0z + αzt

ωz(0.300 s) = 27.5 rad/s +#

−10.0 rad/s2$

(0.300 s) = 24.5 rad/s(9.21)

b) The angular displacement is given by

θ(t) = θ0 + ω0zt+1

2αzt

2

θ(t) = 0 + (27.5 rad/s) (0.300 s) +1

2

#

−10.0 rad/s2$

(0.300 s)2 = 7.80 rad(9.22)

and so the line PQ makes angle

(7.80 rad− 2π rad )360◦

2πrad= 87◦ (9.23)

with +x-axis.Example. Supposed the disc in Example 9.3 was initially spinning at

twice the rate (i.e. 55.0 rad/s) and slowed down at twice the rate (i.e. −20.0 rad/s2).a) Compared to the situation in Example 9.3 how long would it take the

disc to come to a stop?(i) the same amount of time;(ii) twice as much time;(iii) four times as much time;(iv) one half as much time;(v) one fourth as much time


b) Compared to the situation in Example 9.3 through how many revolu-tions would the disc rotate before coming to a stop?(i) the same number of revolutions;(ii) twice as many revolutions;(iii) four times as many revolutions;(iv) one half as many revolutions;(v) one fourth as many revolutions.

9.3 Relating linear and angular kinematics

The linear displacement of a given point P on rotating body at distancer from the axis of rotation is related to angular displacement through thefollowing relation

s = rθ, (9.24)

where θ must be measured in radians.

This suggests that the (magnitude of) linear velocity of the point P isgiven by

%

%

%

%

ds

dt

%

%

%

%

= r

%

%

%

%

dθ

dt

%

%

%

%

(9.25)


orv = rω. (9.26)

and the tangential (or parallel) component of linear acceleration is given by

a|| =dv

dt= r

dω

dt= rα. (9.27)

and the centripetal (or perpendicular) component of linear acceleration isgiven by

a⊥ =v2

r= ω2r. (9.28)

Example 9.4. An athlete whirls a discus in a circle of radius 80.0 cm.At a certain instant, the athlete is rotating at 10.0 rad/s and the angularspeed is increasing at 50.0 rad/s2. At this instant, find the tangential andcentripetal components of the acceleration of the discus and the magnitude ofthe acceleration.

Components of linear acceleration are given by

a|| = rα = (0.800m)#

50.0 rad/s2$

= 40.0m/s

a⊥ = ω2r = (10.0 rad/s)2 (0.800m) = 80.0m/s (9.29)

and so the magnitude is

a =&

a2|| + a2⊥ =&

(40.0m/s)2 + (80.0m/s)2 = 89.4m/s. (9.30)

Example 9.5. You are designing an airplane propeller that is to turnat 2400 rpm (or revolutions per minute). The forward airspeed of the planeis to be 75.0m/s, and the speed of the tips of the propeller blades through theair must not exceed 270m/s. (This is about 80% of the speed of the soundin air. If the speed of the propeller tips were greater than this, they wouldproduce a lot of noise.) (a) What is the maximum possible propeller radius?(b) With this radius, what is the acceleration of the propeller tip?


(a) The tangential component of the velocity of the tip must at most

vmax =&

v2tip − v2plane (9.31)

which is also given byv = rω (9.32)

and thusrmaxω =

&

v2tip − v2plane (9.33)

or

rmax =

&

v2tip − v2plane

ω

=

&

(270m/s)2 − (75.0m/s)2

(2400 rev/min) (1min/60 s) (2π rad/1 rev )

=259m/s

251 rad= 1.03m. (9.34)

(b) The centripetal acceleration of the tip is

a⊥ = ω2rmax

= (251 rad)2 (1.03m) = 6.5× 10m/s2 = 6600 g. (9.35)

Example. Information is stored on a disc in a coded pattern of tiny pits.The pits are arranged in a track that spirals outwards towards the rim of thedisc. As the disc spins inside a player, the track is scanned at a constantlinear speed. How must the rotation speed of the disc change as the player’sscanning head moves over the track?(i) The rotation speed must increase.(ii) The rotation speed must decrease.(iii) The rotation speed must stay the same.


9.4 Energy in Rotational Motion

When a rigid body rotates around a fixed axis different partsof it move with different linear velocities and so have differentkinetic energies energies.

Ki =1

2miv

2i =

1

2mir

2iω

2. (9.36)

The total (rotational) kinetic energy is then

K ='

i

Ki

=' 1

2mir

2iω

2

=1

2

(

'

mir2i

)

ω2

=1

2Iω2 (9.37)

whereI ≡

'

mir2i (9.38)

is known as the moment of inertia. Note that the rotational ki-netic energy looks very similar to the translational kinetic energy

1

2Iω2 ↔

1

2mv2. (9.39)


Example 9.6. A machine part consists of three discs linked by light-weightstruts.

(a) What is this body’s moment of inertia about an axis through the centerof the disc A, perpendicular to the plane of the diagram?(b) What is its moment of inertia about an axis through the centers of disksB and C?(c) What is the body’s kinetic energy if it rotates about the axis through Awith angular speed ω = 4.0 rad/s?

a) The moment of inertia about an axis through the center of disk A is

IA = mAr2A|A +mBr

2B|A +mCr

2C|A

= (0.30 kg) (0)2 + (0.10 kg) (0.50m)2 + (0.20 kg) (0.40m)2 = 0.057 kg ·m2(9.40)

b) The moment of inertia about an axis through the center of disk B is

IBC = mAr2A|BC +mBr

2B|BC +mCr

2C|BC

= (0.30 kg) (0.4m)2 + (0.10 kg) (0)2 + (0.20 kg) (0)2 = 0.048 kg ·m2(9.41)

c) The kinetic energy is

KA =1

2IAω

2 =1

2

#

0.057 kg ·m2$

(4.0 rad/s)2 . (9.42)

Form more complicated objects the moment of inertia must be obtainedby volume integration. Here we summarize some useful results


Example 9.7. We wrap a light, nonstretching cable around a solid cylin-der of mass 50 kg and diameter 0.120m, which rotates in frictionless bearingsabout a stationary horizontal axis. We pull the free end of the cable with aconstant 9.0−N force for a distance of 2.0m; it turns the cylinder as it un-winds without slipping. The cylinder is initially at rest. Find its final angularspeed and the final speed of the cable.

The total work done by the pulling force is

W = Fs = (9.0N) (2.0m) = 18 J.


According to the work-energy theorem

W = Kf −Ki

W =1

2Iω2

f − 0

ωf =

*

2W

I

where

I =1

2mR2 =

1

2(50kg) (0.060m)2 = 0.090 kg·m2

and thus the final angular speed of the cylinder is

ωf =

+

2 · 18 J0.090 kg·m2 = 20 rad/s.

and the final linear speed of the cable is

vf = ωfR = (20 rad/s) (0.060m) = 1.2m/s.

Example 9.8. We wrap a light, non-stretching cable around a solidcylinder with mass M and radius R. The cylinder rotates with negligiblefriction about a stationary horizontal axis. We tie the free and of the cableto a block of mass m and release the block from rest at distance h above thefloor. As the block falls, the cable unwinds without stretching or slipping.Find expression for the speed of the cylinder as the block strikes the floor.


Since there are no other (non-gravitational) forces acting on the systemthe total (mechanical) energy must be conserved

mgh =1

2mv2 +

1

2Iω2 (9.43)

whereω =

v

R(9.44)

and the moment of inertia for the disc is

I =1

2MR2. (9.45)

By plugging Eqs. (9.44) and (9.45) into (9.43) we get

mgh =1

2mv2 +

1

2

!

1

2MR2

"

( v

R

)2

mgh =

!

m

2+

M

4

"

v2 (9.46)

or

v =

+

mghm2 + M

4

. (9.47)

9.5 Parallel-Axis Theorem

There are (infinitely) many axis passing through a given object, and thusthere are infinitely many moments of inertia. However, some of them arerelated to each other. Consider two axis of rotation:

• axis O: passing through the center of mass and

• axis D: parallel to the first one, but displaced by distance d.

Let’s choose z-axis to point along the direction of the first axis and withorigin at the center of mass. Then the O axis of rotation has coordinates

(x, y, z) = (0, 0, z) (9.48)

and the D axis of rotation has coordinates

(x, y, z) = (a, b, z) (9.49)

whered =

√a2 + b2. (9.50)


Then the two moments of inertia are given by

IO ='

i

mi

#

x2i + y2i

$

(9.51)

andID =

'

i

mi

#

(xi − a)2 + (yi − b)2$

. (9.52)

The second equation can be expanded as

ID ='

i

mi

#

(xi − a)2 + (yi − b)2$

(9.53)

='

i

mi

#

x2i − 2xia + a2 + y2i − 2yib+ b2

$

='

i

mi

#

x2i + y2i

$

− 2a'

i

mixi − 2a'

i

mixi +#

a2 + b2$

'

i

mi

The first term is the moment of inertia around center of mass (9.51), thesecond and third terms a proportional to center of mass and thus vanishin the chosen coordinate system, and the forth term can be rewritten usingtotal mass

M ='

i

mi (9.54)

and (9.50). The resulting expression is known as the parallel-axis theorem:

ID = IO +Md2. (9.55)

Example 9.9. A part of mechanical linkage has a mass of m = 3.6 kg.Its moment of inertia IP about an axis l = 0.15m from its center of mass isIP = 0.132 kg · m2. What is the moment of inertia Icm about a parallel axisthrough the center of mass?

From the parallel-axis theorem we get

Icm = Ip −ml2

= 0.132 kg ·m2 − (3.6 kg) (0.15m)2 = 0.051 kg ·m2 (9.56)


9.6 Moment-of-Inertia Calculations

In general moment-of-inertia is defined by replacing summation with inte-gration, i.e.

I ='

i

r2mi → I =

,

r2dm. (9.57)

In the case when the density (or mass per unit volume)

ρ =dm

dV(9.58)

is independent of position the moment of inertia

I =

,

r2ρdV (9.59)

or

I =

,

r2ρ dx dy dz (9.60)

where the limits of integration are set by the geometry of the body.Example 9.10. Consider a hollow cylinder of uniform mass density ρ

with length L, inner radius R1, and outer radius R2. Find its moment ofinertia about its axis of symmetry.

Integration on the x-y plane is most convenient done in polar coordinates

x(r, θ) = r cos θ

y(r, θ) = r sin θ (9.61)


and so one needs to multiply by determinant of Jacobian

dxdy = det

!

∂x∂r

∂y∂r

∂x∂θ

∂y∂θ

"

drdθ

= det

!

cos θ sin θ−r sin θ r cos θ

"

drdθ

=#

r cos2 θ + r sin2 θ$

drdθ

= rdrdθ. (9.62)

Then the moment of inertia is given by

I =

, L

0

, R2

R1

, 2π

0

r2ρ (rdrdθdz)

= 2πρL

, R2

R1

r3dr

= 2πρL

!

R41

4−

R42

4

"

=πρL

2

#

R41 −R4

2

$

. (9.63)

Similarly on can calculate the volume of the cylinder

V =

, L

0

, R2

R1

, 2π

0

r2ρ (rdrdθdz)

= 2πL

, R2

R1

rdr

= 2πL

!

R21

2−

R22

2

"

= πL#

R21 − R2

2

$

(9.64)

and so the total mass

M = V ρ = πLρ#

R21 −R2

2

$

. (9.65)

From (9.63) and (9.65) we get

I =πρL

2

#

R21 − R2

2

$ #

R21 +R2

2

$

=1

2M#

R21 +R2

2

$

. (9.66)


Example 9.11. Find the moment of inertia of a solid sphere of uniformmass density ρ (like a billiard ball) about an axis through its center.

Then the moment of inertia is given by

I =

, R

−R

, 2π

0

,

√R2−z2

0

r2ρ (rdrdθdz)

= 2πρ

, R

−R

,

√R2−z2

0

r3drdz

= 2πρ

, R

−R

-

(R2 − z2)2

4

.

dz

=πρ

2

, R

−R

#

R4 − 2R2z2 + z4$

dz

=πρ

2

!

R4 (2R)− 2R

!

2R3

3

"

+2R5

5

"

=8πρR5

15. (9.67)

Chapter 9 Rotation of rigid bodies - d.umn.edud.umn.edu/~vvanchur/2015PHYS2013/Chapter9.pdf ·...

Documents

Transcript of Chapter 9 Rotation of rigid bodies - d.umn.edud.umn.edu/~vvanchur/2015PHYS2013/Chapter9.pdf ·...