A ballet dancer spins about a vertical axis at 90 rpm with arms outstretched. With the arms folded, the moment of inertia about the same axis of rotation changes to 75 %. Calculate the new speed of rotation.

A ballet dancer spins about a vertical axis at 2.5pi rad s^-1 with his arms outstretched. With the arms folded, the M.I. about the same axis of rotation changes by 25%.Calculate the new speed of rotation in r.p.m.

A ballet dancer spins about a vertical axis at 90 rpm with arms outstretched. With her arms folded, the M.I. about the axis of rotation decreases to 75%.Calculate the new rate of revolutions

Aballet dancer spins about a vertical axis at 120rpm with arms outstretched. With her arms folded the moment of inertia about the axis of rotation decreases by 40 %. What isnew rate of revolution ?

A flywheel rotating about a fixed axis has a kinetic energy of 360 J when its angularspeed is 30 rad/s, then the moment of inertia of fly wheel about the axis of rotation will be

A dancer spins about himself with an angular speed omega with his arms extended. When he draws his hands in, his moment of inertia reduces by 40%. Then his new angular velocity would be

State an expression for moment of inertia of a uniform solid sphere about on axis of rotation coinciding with its diameter.

A circular disc of radius R and thickness R/6 has moment of inertia I about an axis passing through. Its is melted and recast into a solid sphere. What is the moment of inertia of the solid sphere about its diameter as the axis of rotation?

A man standing on a frictionless rotating platform rotates at 1 rps, his arms are outstretched and he holds a weight in his hands. In this position, the total moment of inertia of the system is 6 kg m^2 , he leaves the weights, the moment of inertia decreases to 2kg m^2 . The resulting angular speed of the platform is

Moment of inertia of a disc about the axis of rotation is 10 kgm^2 . A constant torque of 50 Nm acts on it. Find :Angular acceleration .

A person stands on a uniformly rotating turn table with outstretched arms holding two identical weights. The moment of inertia of the system is 60 kgm^2 . When he brings the arms close to his body, the M.I. of inertia reduces to 58 kgm^2 . The angular speed of the system now becomes 3 rad s^-1 . Find the initial angular speed and the final K.E.