Question 206: Centre of Mass and Rotational Motion

Question

A sphere of radius 'a' and mass 'm' rolls along a horizontal plane with constant speed v0. It encounters an inclined plane at angle $$	heta$$ and climbs upward. Assuming that it rolls without slipping, how far up the sphere will travel?

Accepted Answer

From energy conservation
mgh = $${1 \over 2}mv_0^2 + {1 \over 2}I{\omega ^2}$$
mgh = $${1 \over 2}mv_0^2 + {1 \over 2} 	imes {1 \over 5}m{a^2} 	imes {{{v_0}^2} \over {{a^2}}}$$
gh = $${1 \over 2}{v_0}^2 + {1 \over 5}{v_0}^2$$
gh = $${7 \over {10}}{v_0}^2$$
h = $${7 \over {10}}{{{v_0}^2} \over g}$$
from triangle, $$\sin 	heta  = {h \over l}$$
then h = l sin$$	heta$$
$$l\sin 	heta  = {7 \over {10}}{{{v_0}^2} \over g}$$
$$l = {7 \over {10}}{{{v_0}^2} \over {g\sin 	heta }}$$

Answer

$${{v_0^2} \over {2g\sin 	heta }}$$

Answer

$${{7v_0^2} \over {10g\sin 	heta }}$$

Answer

$${2 \over 5}{{v_0^2} \over {g\sin 	heta }}$$

Answer

$${{v_0^2} \over {5g\sin 	heta }}$$

Centre of Mass and Rotational Motion

A sphere of radius 'a' and mass 'm' rolls along a horizontal plane with constant speed v0. It encounters an inclined plane at angle $\theta$ and climbs upward. Assuming that it rolls without slipping, how far up the sphere will travel?

Choose the correct answer: