A uniform ring of mass m and radius R is released from top of an inclined plane. The plane makes an angle theta with horizontal. The cofficent of friction between the ring and plane is mu. Initially, the point of contact of ring and plane is P. Angular momentum of ring about an axis passing from point P and perpendicular to plane of motion as a function of time t is

A uniform ring of mass m and radius R is released from top of an incli

1.	A uniform ring of mass m and radius R is released from top of an inclined plane. The plane makes an angle theta with horizontal. The cofficent of friction between the ring and plane is mu. Initially, the point of contact of ring and plane is P. Angular momentum of ring about an axis passing from point P and perpendicular to plane of motion as a function of time t is
Answer» `mgR(SINTHETA)t-mumgR(COSTHETA)t` `mgR(sintheta)t` `mgR(sintheta)t+mumgR(costheta)t` `mgR(1-mu^(2))(sintheta)t` Answer :B