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)

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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