A disc of mement of inertia `I` is rotating due to external torque. Its kinetic energy is equal to `Ktheta^(2)`. Where K is the positive constant. Its angular acceleration at an angle `theta` will be:

A disc of moment of inertia I_1 , is rotating in horizontal plane about an axis passing through its centre and perpendicular to its plane with constant angular speed omega_1 . Another disc of moment of inertia I_2 having angular speed omega_2 The energy lost by the initial rotating disc is

If the rotational kinetic energy and moment of inertia for a body be E and I respetively then its relation with the angular momentum will be

A disc of the moment of inertia 'l_(1)' is rotating in horizontal plane about an axis passing through a centre and perpendicular to its plane with constant angular speed 'omega_(1)' . Another disc of moment of inertia 'I_(2)' . having zero angular speed is placed discs are rotating disc. Now, both the discs are rotating with constant angular speed 'omega_(2)' . The energy lost by the initial rotating disc is

A disc of the moment of inertia 'l_(1)' is rotating in horizontal plane about an axis passing through a centre and perpendicular to its plane with constant angular speed 'omega_(1)' . Another disc of moment of inertia 'I_(2)' . having zero angular speed is placed discs are rotating disc. Now, both the discs are rotating with constant angular speed 'omega_(2)' . The energy lost by the initial rotating disc is

For a body rotating with constant angular velcoity, its kinetic energy is directly proportional to the square of its

A disc of moment of inertia la is rotating in a horizontal plane about its symmetry axis with constant angular speed co. Another discinitially at rest of moment of inertia I, is dropped coaxially on to the rotating disc. Then, both the discs rotate with same constant angular speed The loss of kinetic energy due to friction in this process is,

If angular velocity of a disc depends an angle rotated theta as omega=theta^(2)+2theta , then its angular acceleration alpha at theta=2 rad is :

If angular velocity of a disc depends an angle rotated theta as omega=theta^(2)+2theta , then its angular acceleration alpha at theta=1 rad is :