The moment of inertia of a flywheel is `4 kg m^(2)`. What angular acceleration will be produced in it by applying a torque of 10 Nm on it?

The moment of inertia of a flywheel is 4 kg- m^(2) . What angular acceleration will be produced in it by applying a torque of 10 N-m on it ?

The moment of inertia of body about an axis of rotation is 150 kgm^2 .calculate the angular acceleration produced in it when a torque of magnitude 30Nm acts on it .

The moment of inertia of flywheel is 4 kg m^2 . If a torque of 10 Nm is applied on it, then the angular acceleration produced will be

The moment of inertia of a body is 2.5 kg m^(2) . Calculate the torque required to produce an angular acceleration of 18 rad s^(-2) in the body.

The moment of inertia of a solid flywheel about its axis is 0.1 kg-m^(2) .A tangential force of 2 kg-wt. is applied round the circumference of the flywheel with the help of a string and mass arrangement as shown in the figure. If the radius of the wheel is 0.1 m, find the acceleration ("in rad/sec"^(2)) of the mass :

A flywheel having moment of inertia 20kgm^(2) rotates about its axis with an angular velocity of 100rads^(-1) . If the moment of inertia of the flywheel is reduced to 10kgm^(2) without applying external torque, calculate the new angular velocity of the Flywheel.

A flywheel having moment of inertia 20kgm^(2) rotates about its axis with an angular velocity of 100rads^(-1) . If the moment of inertia of the flywheel is reduced to 10kgm^(2) without applying external torque, calculate the new angular velocity of the Flywheel.

A flywheel having moment of inertia 20kgm^(2) rotates about its axis with an angular velocity of 100rads^(-1) . If the moment of inertia of the flywheel is reduced to 10kgm^(2) without applying external torque, calculate the new angular velocity of the Flywheel.

A wheel of radius 0.4m can rotate freely about its axis as shown in the figure. A string is wrapped over its rim and a mass of 4 kg is hung. An angular acceleration of 8 rad//s^(2) is produced in it due to the torque. Then, the moment of inertia of the wheel is ( g=10m//s^(2) )