JEE MAIN - Physics (2020 - 7th January Morning Slot - No. 9)
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As shown in the figure, a bob of mass m is tied by a massless string whose other end portion is wound on a fly wheel (disc) of radius r and mass m. When released from rest the bob starts falling vertically. When it has covered a distance of h, the angular speed of the wheel will be:$$r\sqrt {{3 \over {2gh}}} $$
$$r\sqrt {{3 \over {4gh}}} $$
$${1 \over r}\sqrt {{{4gh} \over 3}} $$
$${1 \over r}\sqrt {{{2gh} \over 3}} $$
Explanation
mgh = $$\Delta $$KE
= $${1 \over 2}m{v^2} + {1 \over 2}I{\omega ^2}$$
For no slipping, v = $$\omega $$R
$$ \therefore $$ mgh = $${1 \over 2}m{\omega ^2}{R^2} + {1 \over 2}{{m{R^2}} \over 2}{\omega ^2}$$
$$ \Rightarrow $$ mgh = $${3 \over 4}m{\omega ^2}{R^2}$$
$$ \Rightarrow $$ $$\omega $$ = $${1 \over r}\sqrt {{{4gh} \over 3}} $$
= $${1 \over 2}m{v^2} + {1 \over 2}I{\omega ^2}$$
For no slipping, v = $$\omega $$R
$$ \therefore $$ mgh = $${1 \over 2}m{\omega ^2}{R^2} + {1 \over 2}{{m{R^2}} \over 2}{\omega ^2}$$
$$ \Rightarrow $$ mgh = $${3 \over 4}m{\omega ^2}{R^2}$$
$$ \Rightarrow $$ $$\omega $$ = $${1 \over r}\sqrt {{{4gh} \over 3}} $$
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