Karlsruhe Institute of Technology - International ...kremer/ph10/tutsol3.pdf · Karlsruhe Institute...

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Karlsruhe Institute of Technology - International Department GmbH Pre-Semester 2010 - Physics Course - Extra Tutorial St´ ephane Ngo Dinh Solution 3 [email protected] 24.08.2010 1. Pendulum A pendulum of length l = 80 cm with a bob of mass m =0.6 kg is released from rest at initial angle of θ 0 with the vertical. At the bottom of the swing, the speed of the bob is v b =2.8 m/s. (a) What was the initial angle θ 0 of the pendulum? Solution: The simplest way to determine θ 0 is via energy considerations: The total energy of the system is E tot = E kin + E pot = 1 2 mv 2 + mgz. We choose a coordinate system such that the bottom of the swing is at z = 0. Then, the angle θ with the vertical satisfies z = l(1 - cos θ). Initially, the bob is at rest and E kin = 0. At the bottom of the swing, E pot = 0. Due to energy conservation 1 2 mv 2 b = E tot = mgz 0 = mg (1 - cos θ 0 ) cos θ 0 =0.51 θ 0 59 . (b) What angle θ does the pendulum make with the vertical when the speed of the bob is v =1.4 m/s? Solution: Now energy conservation reads 1 2 mv 2 b = E tot = mgz + 1 2 mv 2 = mg (1 - cos θ 0 )+ 1 2 mv 2 cos θ =1 - (v 2 b - v 2 ) /(2gl) 0.63 θ 51 . 2. Loop Consider a bob of mass m gliding “down” a frictionless slide which contains a loop of radius R. The ball starts at rest at a height h 2R above the bottom of the loop. (a) Compute the ball’s speed v top at the top of the loop. Solution: The total energy is E tot = 1 2 mv 2 + mgz. Since in the beginning we have z = h and v = 0, it is E tot = mgh. At the top of the loop, z =2R, energy conservation gives mgh = 1 2 mv 2 top + mg 2R v top = 2g (h - 2R) (b) What is the path which the bob describes in the case h =2R? Solution: In this case, the bob reaches the top of the loop with a velocity v top =0 (that means, without gravity it would just stay there). Due to its weight the bob then falls downward along a straight line toward the bottom of the loop. (c) What is the minimal height h min from which the bob needs to start in order to make it through the loop? Solution: To have the bob performing a circular motion of radius R with (instan- taneous) velocity v, there has to be a centripetal force F cp = mv 2 /R acting on it which points towards the circle’s center. Where does it come from? It is the sum of the normal force F N exerted by the slide and the radial component F g,r of the weight (the tangential component F g,t changes the velocity | v|).

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Page 1: Karlsruhe Institute of Technology - International ...kremer/ph10/tutsol3.pdf · Karlsruhe Institute of Technology - International Department GmbH Pre-Semester 2010 - Physics Course

Karlsruhe Institute of Technology - International Department GmbH

Pre-Semester 2010 - Physics Course - Extra Tutorial

Stephane Ngo Dinh Solution 3

[email protected] 24.08.2010

1. Pendulum

A pendulum of length l = 80 cm with a bob of mass m = 0.6 kg is released from rest atinitial angle of θ0 with the vertical. At the bottom of the swing, the speed of the bob isvb = 2.8m/s.

(a) What was the initial angle θ0 of the pendulum?

Solution: The simplest way to determine θ0 is via energy considerations: The totalenergy of the system is Etot = Ekin + Epot = 1

2mv2 + mgz. We choose a coordinate

system such that the bottom of the swing is at z = 0. Then, the angle θ with thevertical satisfies z = l(1 − cos θ). Initially, the bob is at rest and Ekin = 0. At thebottom of the swing, Epot = 0. Due to energy conservation 1

2mv2

b = Etot = mgz0 =mg(1 − cos θ0) ⇒ cos θ0 = 0.51 ⇒ θ0 ≈ 59◦.

(b) What angle θ′ does the pendulum make with the vertical when the speed of the bobis v′ = 1.4m/s?

Solution: Now energy conservation reads 12mv2

b = Etot = mgz′ + 12mv′2 = mg(1 −

cos θ0) + 12mv′2 ⇒ cos θ′ = 1 − (v2

b − v′2) /(2gl) ≈ 0.63 ⇒ θ′ ≈ 51◦.

2. Loop

Consider a bob of mass m gliding “down” a frictionless slide which contains a loop ofradius R. The ball starts at rest at a height h ≥ 2R above the bottom of the loop.

(a) Compute the ball’s speed vtop at the top of the loop.

Solution: The total energy is Etot = 12mv2 + mgz. Since in the beginning we have

z = h and v = 0, it is Etot = mgh. At the top of the loop, z = 2R, energyconservation gives mgh = 1

2mv2

top + mg 2R ⇒ vtop =√

2g(h − 2R)

(b) What is the path which the bob describes in the case h = 2R?

Solution: In this case, the bob reaches the top of the loop with a velocity vtop = 0(that means, without gravity it would just stay there). Due to its weight the bobthen falls downward along a straight line toward the bottom of the loop.

(c) What is the minimal height hmin from which the bob needs to start in order to makeit through the loop?

Solution: To have the bob performing a circular motion of radius R with (instan-taneous) velocity v, there has to be a centripetal force Fcp = mv2/R acting on itwhich points towards the circle’s center. Where does it come from? It is the sumof the normal force FN exerted by the slide and the radial component Fg,r of theweight (the tangential component Fg,t changes the velocity |~v|).

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The slide takes care that the distance of the bob to the circle’s center can never be> R. If Fg,r < Fcp, then FN = Fcp − Fg,r > 0. However, since the bob is not fixedto the slide, the normal force exerted by the slide cannot point away from circle’scenter ⇒ FN ≤ 0. In the case Fg,r > Fcp the net force Fg,r + FN towards the centeris larger than the one needed (Fcp) to keep the bob on the R-circle. The bob ispulled away from the slide. Hence, we require Fg,r < Fcp.

If we denote the position of the bob along the circle by the angle ϕ between the hori-zontal and the line connecting bob and circle’s center, then z = R(1+sin ϕ), Fg,r =mg sin ϕ, and due to energy conservation v2 = 2g(h − z) = 2gR(h/R − 1 − sin ϕ).Then the condition for the bob to stay on the slide is

mg sin ϕ ≤ mv2/R = 2mg(h/R − 1 − sin ϕ) ⇒ h/R ≥ 1 +3

2sin ϕ.

This is satisfied for all ϕ if and only if

h/r ≥5

2⇒ hmax =

5

2R.

(In fact, we just have proven that it would have been sufficient to require Fg,r < Fcp

only at the top of the loop.)

3. Spring

A block of mass m is dropped onto the top of a vertical spring whose force constant isk. The block is released from a height h above the top of the spring.

(a) What is the maximum kinetic energy Ekin,max of the block?

Solution: In this problem there are 3 contributions to the total energy Etot =Egrav + Ekin + Ecomp:

• Egrav = mgz, the potential energy due to gravity. Coordinate z is chosen suchthat the top of the unrelaxed spring is at z = 0 and the initial position of theblock is z = h. Thus, if the block compresses the spring by s > 0, its positionis z = −s < 0, and Egrav < 0.

• Ekin = 12mv2, the kinetic energy of the block.

• Ecomp = 12ks2, the potential energy of the compressed spring. The spring is

initially relaxed, its compression then is s = 0.

Since the total energy is conserved we may calculate it in the initial state, z = h,where the block is at rest, v = 0, and the spring is not compressed, s = 0: Etot =Egrav = mgh.

As long as the block does not touch the spring it holds z > 0, s = 0, and thusEkin ≤ Etot = mgh.

Let us consider the case z < 0. Then s = −z and Ekin = Etot − Egrav − Ecomp =mg(h − z) − 1

2Dz2. The maximal kinetic energy is reached at z = −mg/D (just

solve dEkin)/dz = 0), where it assumes the value Ekin,max = mgh+ 12m2g2/D > Etot

(remember that Egrav < 0!).

(b) What is the maximum compression smax of the spring?

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Solution: The spring is maximally compressed when the block has reached itsreversal point z = −smax where 0 = Ekin = mg(h + smax)−

12Ds2

max. This quadraticequation has two solutions one of which is negativ. The positive one is

smax =mg

D

(

1 +

1 + 2hD

mg

)

.

(c) At what compression s is the block’s kinetic energy half its maximum value?

Solution: We have to solve

Ekin =1

2Ekin,max

⇔ mg(h + s) −1

2Ds2 =

1

2mgh +

1

4m2g2/D.

The two solutions of this quadratic equation are

s =mg

D

(

1 ±

1 +D

mg

(

h −1

2

mg

D

)

)

.

4. 2D Collision

A particle with mass m1 has initial speed v0. It collides with a second particle with massm2 that is at rest, and is deflected through an angle θ1. Its speed after the collision is v.The second particle recoils, and its velocity makes an angle θ2 with the initial directionof the first particle.

(a) Show that

tan θ2 = −v sin θ1

v0 − v cos θ1

.

Solution: Momentum conservation parallel and perpendicular to the initial direc-

tion of the first particle reads,

m1v0 = m1v cos θ1 + m2v2 cos θ2 ⇒ m2v2 sin θ2 = −m1v sin θ1, (1)

0 = m1v sin θ1 + m2v2 sin θ2 ⇒ m2v2 cos θ2 = m1 (v0 − v cos θ1) , (2)

where v2 = |~v2| is the final speed of the second particle. Since tanx = sin x/ cos x,deviding (1) and (2) gives

tan θ2 = −v sin θ1

v0 − v cos θ1

.

(b) Show that if the collision is elastic and m1 = m2, then v = v0 cos θ1.

Solution: In elastic collisions the kinetic energy is conserved, thus

1

2m1v

20 =

1

2m1v

2 +1

2m2v

22 ⇒ v2

2 = v20 − v2.

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Using tan x = sin x/ cos x, cos2 x + sin2 x = 1, and the result of (a), we obtain

sin2 θ2 =sin2 θ2

cos2 θ2 + sin2 θ2

=tan2 θ2

1 + tan2 θ2

=v2 sin2 θ1

v20 + v2 − 2v0v cos θ1

.

And, hence, the square of (2) is

m1v2 sin2 θ1 = m2v

22 sin2 θ2 = m2

1

(

v20 − v2

)

sin2 θ2 = m21

(

v20 − v2

) v2 sin2 θ1

v20 + v2 − 2v0v cos θ1

⇒ 2(

v2 − v0v cos θ1

)

= 0 ⇒ v = v0 cos θ1.