Question 202: Centre of Mass and Rotational Motion

Question

A sphere of radius 'a' and mass 'm' rolls along a horizontal plane with constant speed v
0
. It encounters an inclined plane at angle $	heta$ and climbs upward. Assuming that it rolls without slipping, how far up the sphere will travel?
![](https://imagex.cdn.examgoal.net/1kltdtp4h/ffd17cac-4699-46af-9815-82654a8e1ebb/2aa8ce10-7c17-11eb-b0eb-7d97e0807144/file-1kltdtp4i-263w.jpg)

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$
$l\sin 	heta  = {7 \over {10}}{{{v_0}^2} \over g}$
$l = {7 \over {10}}{{{v_0}^2} \over {g\sin 	heta }}$
![](https://imagex.cdn.examgoal.net/1km7octud/9ba9a263-d064-4ff2-ba4a-17895a3cabc2/530fad40-83f3-11eb-bd5c-29fe96a48eb3/file-1km7octue-680w.jpeg)

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 v 0 . It encounters an inclined plane at angle \theta and climbs upward. Assuming that it rolls without slipping, how far up the sphere will travel?

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