Question 691: Centre of Mass and Rotational Motion

Question

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 ?

Accepted Answer

K.E. = $${1 \over 2}I{\omega ^2} + {1 \over 2}m{v^2}$$
K.E. = $${1 \over 2}\left( {{1 \over 2}m{r^2}} \right){\omega ^2} + {1 \over 2}m{v^2}$$
$$ = {1 \over 4}m{v^2} + {1 \over 2}m{v^2} = {3 \over 4}m{v^2}$$
Now, gain in K.E. = Loss in P.E.
$${3 \over 4}m{v^2} = mgh \Rightarrow v = \sqrt {\left( {{4 \over 3}} \right)gh} $$

Answer

$$\sqrt {2gh} $$

Answer

$$\sqrt {{3 \over 4}gh} $$

Answer

$$\sqrt {{4 \over 3}gh} $$

Answer

$$\sqrt {4gh} $$

Centre of Mass and Rotational Motion

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 ?

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