Rotational Motion Question 26
Question: A solid sphere is rolling on a surface , with a translational velocity $ vm{s^{-1}}. $ If it is to climb the inclined surface continuing to roll without slipping, then minimum velocity for this to happen is
[JEE ONLINE 12-05-2012]
Options:
A) $ \sqrt{2gh} $
B) $ \sqrt{\frac{7}{5}gh} $
C) $ \sqrt{\frac{7}{2}gh} $
D) $ \sqrt{\frac{10}{7}gh} $
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Answer:
Correct Answer: D
Solution:
- [d] Minimum velocity for a body rolling without slipping For solid sphere,
$\frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 = mgh$
$\frac{1}{2} m v^2 + \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v^2}{r^2}\right) = mgh$
$\frac{v^2}{2} + \frac{2v^2}{10} = g h$
$\frac{5v^2}{2} + \frac{2v^2}{10} = g h \Rightarrow v^2 = \frac{10}{7} gh$
$v \geq \sqrt{\frac{10}{7} gh}$
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