Question

A sphere of radius r0 = 22.0 cm and mass m = 1.20kg starts from rest and rolls without slipping down a 38.0 degree incline that is 11.0 mm long.

A. Calculate its translational speed when it reaches the bottom.

B. Calculate its rotational speed when it reaches the bottom.

C. What is the ratio of translational to rotational kinetic energy at the bottom?

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

E. Does your answer in part B depend on mass or radius of the ball?

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

Answer 1

here,

the radius of sphere , r = 22 cm = 0.22 m

mass , m = 1.2 kg

theta = 38 degree

s = 11 mm = 0.011 m

a)

let the translational speed at the botto of the incline be v

using conservation of energy

0.5 * I * w^2 + 0.5 * m * v^2 = m * g * (s * sin(theta))

0.5 * (0.5 * m * r^2) * (v/r)^2 + 0.5 * m * v^2 = m * g * (s * sin(theta))

0.7 * v^2 = 9.81 * 0.011 * sin(38)

solving for v

v = 0.31 m/s

translational speed when it reaches the bottom is 0.31 m/s

b)

the angular velocity , w = v /r

w = 0.31 /0.22 rad/s = 1.4 rad/s

c)

the ratio of translational to rotational kinetic energy at the bottom , R = (0.5 * m * v^2) /(0.5 * I * w^2)

R = (0.5 * m * v^2) /(0.5 * (0.4 * m * r^2) * (v/r)^2)

R = 2.5

d)

No , the answer will not depend on the radius and mass of ball

e)

Yes , the answer will not depend on the radius of ball

d)

No , the answer will not depend on the radius and mass of ball