A thin circular ring of mass M and radius r is rotating about its axis with an angular speed $$\omega$$. Two particles having mass m each are now attached at diametrically opposite points. The angular speed of the ring will become :

A sphere of mass 2 kg and radius 0.5 m is rolling with an initial speed of 1 ms^-1 goes up an inclined plane which makes an angle of 30$$^\circ$$ with the horizontal plane, without slipping. How long will the sphere take to return to the starting point A?

A triangular plate is shown. A force $$\overrightarrow F $$ = 4$$\widehat i$$ $$-$$ 3$$\widehat j$$ is applied at point P. The torque at point P with respect to point 'O' and 'Q' are :

A mass M hangs on a massless rod of length l which rotates at a constant angular frequency. The mass M moves with steady speed in a circular path of constant radius. Assume that the system is in steady circular motion with constant angular velocity $$\omega$$. The angular momentum of M about point A is L_A which lies in the positive z direction and the angular momentum of M about point B is L_B. The correct statement for this system is :