With what angular velocity `omega` (in rad/s) should we rotate the disc so that a mass hanging on to the periphery by a thread of length l =35/24m is deviated from the vertical by an angle `alpha=37^(@)` in steady state(fig).? Radius of the disc R=1 m

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 :

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 :

If angular velocity of a disc depends an angle rotated theta as omega=theta^(3)+2theta^2 , then its angular acceleration alpha at theta=1 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 :

A uniform rod of mass m and length L lies radialy on a disc rotating with angular speed omega in a horizontal plane about vertical axis passing thorugh centre of disc. The rod does not slip on the disc and the centre of the rod is at a distance 2L from the centre of the disc. them the kinetic energy of the rod is

A uniform rod of mass m and length L lies radialy on a disc rotating with angular speed omega in a horizontal plane about vertical axis passing thorugh centre of disc. The rod does not slip on the disc and the centre of the rod is at a distance 2L from the centre of the disc. them the kinetic energy of the rod is

Initial angular velocity of a circualr disc of mass M is omega_1. The tow sphere of mass ma re attached gently to diametrically opposite points ont eh edge of the disc. What is the final angular velocity on the disc?

A uniform disc of mass m and radius R rotates about a fixed vertical axis passing through its centre with angular velocity omega . A particle of same angular velocity mass m and moving horizontally with velocity 2omegaR towards centre of the disc collides and sticks to its rim. The impulse on the particle due to the disc is