Question

A sphere of radius r0 = 22.0 cm and mass m = 1.20kg starts from rest and rolls without slipping down a 34.0 degree incline that is 12.0 m long.

Part A: Calculate its translational speed when it reaches the bottom. (m/s)

Part B: Calculate its rotational speed when it reaches the bottom. (rad/s)

Part C: What is the ratio of translational to rotational kinetic energy at the bottom? (Ktr/Krot)

Part D: Does your answer in part A depend on mass or radius of the ball?

a. Depends on mass

b. doesnt depend on mass and radius

   c. depends on mass and radius

   d. depends on radius

Part E: Does your answer in Part B depend on mass or radius of the ball?

a. Depends on mass

b. doesnt depend on mass and radius

   c. depends on mass and radius

   d. depends on radius

Part F: Does your answer in part C depend on mass or radius of the ball?

a. Depends on mass

b. doesnt depend on mass and radius

   c. depends on mass and radius

   d. depends on radius

Answer 1

A.

Use law of conservation of energy

MEi =MEf

KEi+PEi =KEf+PEf

½*mvi2+mghi=½*mvfTrans2+=½*IsphereωRot2+mghf

0+mghi=½*mvfTrans2+½*2/5*mr02*(vfTrans/r0)2+mghf

0+ghi=½*vfTrans2+½*2/5*mr02*(vfTrans/r0)2+ghf

[ghi-ghf] ==½*vfTrans2+½*2/5*(vfTrans)2

[ghi-ghf] ==7/10*vfTrans2

vfTrans=sqrt[10/7*(ghi-ghf)]

vfTrans=sqrt[10/7*g(hi-hf)]……(1)

Plug values,

vfTrans=sqrt[10/7*9.8(12.0m)]

vfTrans=12.96m/s

B.

ω=vfTrans /R=(12.96m/s)/(0.22m)=58.91rad/s

C.

Ktr/Krot=(1/2*mvfTrans2)/(IsphereωRot2)= (1/2*mvfTrans2)/(½*2/5*mr02*(vfTrans/r0)2)

Ktr/Krot=5/2

D.

From eqn(1),

vfTrans=sqrt[10/7*g(hi-hf)]

Thus, vfTrans does not depend on mass and radius