System Of Particles And Rotational Motion Ques 86
86. A solid cylinder of mass $m$ and radius $R$ rolls down an inclined plane of height $h$ without slipping. The speed of its centre of mass when it reaches the bottom is
[2003, 1989]
(a) $\sqrt{(2 g h)}$
(b) $\sqrt{4 g h / 3}$
(c) $\sqrt{3 g h / 4}$
(d) $\sqrt{4 g / h}$
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Answer:
Correct Answer: 86.(b)
Solution:
- (b) $K . E .=\frac{1}{2} I \omega^{2}+\frac{1}{2} m v^{2}$
$K . E . =\frac{1}{2}(\frac{1}{2} m r^{2}) \omega^{2}+\frac{1}{2} m v^{2}$
$ =\frac{1}{4} m v^{2}+\frac{1}{2} m v^{2}=\frac{3}{4} m v^{2}$
Now, gain in $K.E$. $=$ Loss in $P.E.$
$\frac{3}{4} m v^{2}=m g h \Rightarrow v=\sqrt{(\frac{4}{3}) g h} $
BETA
