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A stationary horizon 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 accerlation of the disc is :

Option 1)
$\frac{k}{4I}\theta$

Option 2)
$\frac{k}{I}\theta$

Option 3)
$\frac{k}{2I}\theta$

Option 4)
$\frac{2k}{I}\theta$

Answers (1)

best_answer

$kE=k\theta ^{2}$

$\frac{1}{2}I\omega ^{2}=k\theta ^{2}$

$\omega ^{2}=\frac{2\; k\theta ^{2}}{I}$

$a=\frac{d\omega }{d\theta }$

$2\omega \frac{d\omega }{d\theta }=\frac{2k2\theta }{I}$

$\frac{d\omega }{d\theta }=\frac{2k\theta }{I\omega }$

$a=\frac{2k\theta }{I\omega }$

angular acceleration

$\alpha =a\omega$

$=\left ( \frac{2k\theta }{I\omega } \right )\left ( \omega \right )$

$=\frac{2k\theta }{I}$

Option 1)

$\frac{k}{4I}\theta$

Option 2)

$\frac{k}{I}\theta$

Option 3)

$\frac{k}{2I}\theta$

Option 4)

$\frac{2k}{I}\theta$

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