JEE MAIN - Physics (2021 - 25th February Evening Shift - No. 3)
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?
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$${{v_0^2} \over {2g\sin \theta }}$$
$${{7v_0^2} \over {10g\sin \theta }}$$
$${2 \over 5}{{v_0^2} \over {g\sin \theta }}$$
$${{v_0^2} \over {5g\sin \theta }}$$
Explanation
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From energy conservation
mgh = $${1 \over 2}mv_0^2 + {1 \over 2}I{\omega ^2}$$
mgh = $${1 \over 2}mv_0^2 + {1 \over 2} \times {1 \over 5}m{a^2} \times {{{v_0}^2} \over {{a^2}}}$$
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, $$\sin \theta = {h \over l}$$
then h = l sin$$\theta$$
$$l\sin \theta = {7 \over {10}}{{{v_0}^2} \over g}$$
$$l = {7 \over {10}}{{{v_0}^2} \over {g\sin \theta }}$$
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