Question

A sphere of radius r0 = 23.0 cm and mass m = 1.20kg starts from rest and rolls without slipping down a 35.0 ∘ incline that is 13.0 m 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?

Answer 1

Given,

r = 23 cm ; m = 1.2 kg ; theta = 35 deg ; L = 13 m

A)H = L sin(theta)

H = 13 x sin35 = 7.46 m

PE = m g h = 1.2 x 9.8 x 7.46 = 87.73

from conservation of energy

PE = KEtrans + KErot

PE = 1/2 m v^2 + 1/2 I w^2

m g h = 1/2 m v^2 + 1/2 x 2/5 m r^2 v^2/r^2

since, I = 2/5 m r^2 and v = r w => w = v/r

87.73 = 7/10 m v^2

v = sqrt (10 x 87.73/7 x 1.2) = 11.2 m/s

Hence, v = 10.22 m/s

B)w = v/r = 10.22/0.23 = 44.43 rad/s

Hence, w = 44.43 rad/s

C)KEtrans = 1/2 m v^2

KEtrans = 0.5 x 1.2 x 11.2^2 = 62.67 J

KErot = 87.73 - 62.67 = 25.06 J

KEtran/KErot = 62.67/25.06 = 2.5

Hence, KEtrans/KErot = 2.5