A fly wheel of moment of inertia I is rotating at n revolutions per sec. The work needed to double the frequency would be -

A fly wheel of moment of inertia 4xx10^(-3)kgm^(2) is makign 10 revolution per second. The torque required to stop it in 5 second is:

A wheel of moment of inertia 30 kg m^(2) is rotating at 10 rotations per minute. The work done in increasing its speed to 5 times its initial value will be

A wheel of moment of inertia 30 kg m^(2) is rotating at 10 rotations per minute. The work done in increasing its speed to 5 times its initial value will be

A wheel of moment of inertia 10 kgm^2 is rotating at 10 rotations per minute. The work done in increasing its speed to 5 times its initial value, will be

A flywheel of moment of inertia I is rotating with uniform angular speed omega . If a torque tau retards the wheel, then the time in which the wheel comes to rest is

A flywheel of moment of inertia 10^(7) g cm^(2) is rotating at a speed of 120 rotations per minute. Find the constant breaking torque required to stop the wheel in 5 rotations.

A wheel of moment of inertia 5xx10^(-3) kg m^(2) is making 20 revolutions per second. If it is stopped in 20 s, then its angular retardation would be