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Rotational Motion question

2003 · Q161
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Rotational Motion question

2003 · Q161

NEETPhysicsRotational MotionMCQ+4 / −1
A solid cylinder of mass M and radius R rolls without slipping down an inclined plane of length L and height h. What is the speed of its centre of mass when the cylinder reaches its bottom ?
  1. A
    2gh\sqrt {2gh}2gh​
  2. B
    34gh\sqrt {{3 \over 4}gh}43​gh​
  3. C
    43gh\sqrt {{4 \over 3}gh}34​gh​
  4. D
    4gh\sqrt {4gh}4gh​
View written solutionFree

Correct answer: C

K.E. = 12Iω2+12mv2{1 \over 2}I{\omega ^2} + {1 \over 2}m{v^2}21​Iω2+21​mv2

K.E. = 12(12mr2)ω2+12mv2{1 \over 2}\left( {{1 \over 2}m{r^2}} \right){\omega ^2} + {1 \over 2}m{v^2}21​(21​mr2)ω2+21​mv2

=14mv2+12mv2=34mv2 = {1 \over 4}m{v^2} + {1 \over 2}m{v^2} = {3 \over 4}m{v^2}=41​mv2+21​mv2=43​mv2

Now, gain in K.E. = Loss in P.E.

34mv2=mgh⇒v=(43)gh{3 \over 4}m{v^2} = mgh \Rightarrow v = \sqrt {\left( {{4 \over 3}} \right)gh} 43​mv2=mgh⇒v=(34​)gh​

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