Rotational Motion Question 148
Question: A solid cylinder of mass m & radius R rolls down inclined plane without slipping. The speed of C.M. when it reaches the bottom is
Options:
A) $\sqrt{2gh}$
B) $\sqrt{4gh/3}$
C) $\sqrt{3/4gh}$
D) $\sqrt{4gh}$
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Answer:
Correct Answer: B
Solution:
[b] By energy conservation
$ (K .E)_i+(P .E)_i=(K .E)_f+(P .E)_f $
$ (K .E)_i=0,(P .E)_i=m g h,(P .E)_f=0 $
$ (K .E)_f={ } _{1 / 2} I \omega^2+{ }_1{ }_2 m v _{c m}^2 $
$\text { Where I } $ moment of inertia=$ 1 / 2 m R^2 \text { (for solid cylinder) so } $
$ m g h={ }^1 / 2\left(1 / 2 m R^2\right)\left(\frac{v^2 c m}{R^2}\right)+{ }_1{ }_2 m v _{c m}^2 \text {.} $
$ \Rightarrow \quad v _{c m}=\sqrt{4 g h / 3} $
