Question 272: Centre of Mass and Rotational Motion

Question

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 $$	heta $$, where $$	heta $$
is the angle by which it has rotated, is given as
k$$	heta $$2. If its moment of inertia is I then the
angular acceleration of the disc is :

Accepted Answer

Kinetic energy KE = $${1 \over 2}l{\omega ^2} = k{	heta ^2}$$
$$ \Rightarrow {\omega ^2} = {{2k{	heta ^2}} \over l} \Rightarrow \omega  = \sqrt {{{2k} \over l}} 	heta $$ .... (A)
Differentiate (A) wrt time $$ 	o $$
$${{d\omega } \over {dt}} = \alpha  = \sqrt {{{2k} \over l}} \left( {{{d	heta } \over {dt}}} ight)$$
$$ \Rightarrow \alpha  = \sqrt {{{2k} \over l}} .\sqrt {{{2k} \over l}} 	heta \,\{ by\,(1)\} $$
$$ \Rightarrow \alpha  = {{2k} \over l}	heta \,$$

Answer

$${k \over {4I}}	heta $$

Answer

$${k \over {I}}	heta $$

Answer

$${k \over {2I}}	heta $$

Answer

$${2k \over {I}}	heta $$

Centre of Mass and Rotational Motion

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 :

Choose the correct answer: