(D) The rotational kinetic energy of the disc is given by $KE = \frac{1}{2}I\omega^2 = K\theta^2.$From this,we can express the angular velocity $\omeg

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(D) The rotational kinetic energy of the disc is given by $KE = \frac{1}{2}I\omega^2 = K\theta^2.$From this,we can express the angular velocity $\omega$ as:$\omega^2 = \frac{2K\theta^2}{I} \implies \omega = \sqrt{\frac{2K}{I}}\theta \quad \dots(1)$Angular acceleration $\alpha$ is the rate of change of angular velocity with respect to time:$\alpha = \frac{d\omega}{dt} = \frac{d}{dt}\left( \sqrt{\frac{2K}{I}}\theta \right) = \sqrt{\frac{2K}{I}} \frac{d\theta}{dt}$Since $\frac{d\theta}{dt} = \omega,$ we substitute the expression for $\omega$ from equation $(1)$:$\alpha = \sqrt{\frac{2K}{I}} \left( \sqrt{\frac{2K}{I}}\theta \right)$$\alpha = \frac{2K}{I}\theta.$

Class 11 Physics System of Particles and Rotational Motion Relation between Torque and Angular acceleration and it's Application

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$A$ 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

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A
$\frac{K}{I}\theta$
B
$\frac{K}{2I}\theta$
C
$\frac{K}{4I}\theta$
D
$\frac{2K}{I}\theta$

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System of Particles and Rotational Motion

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Relation between Torque and Angular acceleration and it's Application

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$A$ thin uniform rod of mass $M$ and length $L$ is pivoted at a height $\frac{L}{3}$ from its lower end as shown in the figure. The rod is allowed to fall from a vertical position and lie horizontally on the table. The angular velocity of this rod when it hits the table top is . . . . . . . ($g$ = gravitational acceleration)

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$A$ 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

Similar Questions

$A$ disc of mass $M \ kg$ and radius $R \ m$ is rotating at an angular speed of $\omega \ rad/s$ when the motor is switched off. Neglecting the friction at the axle,the force that must be applied tangentially to the wheel to bring it to rest in time $t$ is ...............

$A$ rod of mass $M$ and length $L$ is placed in a horizontal plane with one end hinged about a vertical axis. $A$ horizontal force of $F = \frac{Mg}{2}$ is applied at a distance $\frac{5L}{6}$ from the hinged end. The angular acceleration of the rod will be:

$A$ uniform rod of mass $m$ and length $2l$ is balanced on a triangular prism. Now,a length of $l/2$ is cut from one end of the rod and placed over the shortened part such that the ends meet. The initial angular acceleration is:

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