JEE MAIN - Physics (2019 - 9th April Morning Slot - No. 8)
A stationary horizontal disc is free to rotate
about its axis. When a torque is applied on it,
its kinetic energy as a function of $$\theta $$, where $$\theta $$
is the angle by which it has rotated, is given as
k$$\theta $$2. If its moment of inertia is I then the
angular acceleration of the disc is :
$${k \over {4I}}\theta $$
$${k \over {I}}\theta $$
$${k \over {2I}}\theta $$
$${2k \over {I}}\theta $$
Explanation
Kinetic energy KE = $${1 \over 2}l{\omega ^2} = k{\theta ^2}$$
$$ \Rightarrow {\omega ^2} = {{2k{\theta ^2}} \over l} \Rightarrow \omega = \sqrt {{{2k} \over l}} \theta $$ .... (A)
Differentiate (A) wrt time $$ \to $$
$${{d\omega } \over {dt}} = \alpha = \sqrt {{{2k} \over l}} \left( {{{d\theta } \over {dt}}} \right)$$
$$ \Rightarrow \alpha = \sqrt {{{2k} \over l}} .\sqrt {{{2k} \over l}} \theta \,\{ by\,(1)\} $$
$$ \Rightarrow \alpha = {{2k} \over l}\theta \,$$
$$ \Rightarrow {\omega ^2} = {{2k{\theta ^2}} \over l} \Rightarrow \omega = \sqrt {{{2k} \over l}} \theta $$ .... (A)
Differentiate (A) wrt time $$ \to $$
$${{d\omega } \over {dt}} = \alpha = \sqrt {{{2k} \over l}} \left( {{{d\theta } \over {dt}}} \right)$$
$$ \Rightarrow \alpha = \sqrt {{{2k} \over l}} .\sqrt {{{2k} \over l}} \theta \,\{ by\,(1)\} $$
$$ \Rightarrow \alpha = {{2k} \over l}\theta \,$$
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