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:
The moment of inertia of a rigid body in terms of its angular momentum L and kinetic energy K 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
The moment of inertia and rotational kinetic energy of a fly wheel are 20kg-m^(2) and 1000 joule respectively. Its angular frequency per minute would be –
If a body starts rotating from rest due to a torque of 1.5 Nm, then its kinetic energy after it complete 20 revolution will be
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 :
A circular disc of moment of inertia I_(t) is rotating in a horizontal plane about its symmetry axis with a constant angular velocity omega_(i) . Another disc of moment of inertia I_(b) is dropped co-axially onto the rotating disc. Initially, the second disc has zero angular speed. Eventually, both the discs rotate with a constant angular speed omega_(f) . Calculate the energy lost by the initially rotating disc due to friction.
A round disc of moment of inertia I_2 about its axis perpendicular to its plane and passing through its centre is placed over another disc of moment of inertia I_1 rotating with an angular velocity omega about the same axis. The final angular velocity of the combination of discs is.
The moment of inertia of a disc rotating about an axis passing through its centre and perpendicular to its axis is 20 kgm^(2) . If its rotational K.E. is 10 J, then its angular momentum will be
A disc is rolling without slipping. The ratio of its rotational kinetic energy and translational kinetic energy would be -
