Q-30) A ring of radius R is rolling without slipping over rough horizontal surface with velocity v0 Two points - Physics - System Of Particles And Rotational Motion

Question

Q-30) A ring of radius R is rolling without slipping over rough
horizontal surface with velocity v0 Two points are
located at A and B on rim of ring Find angular velocity of A w r t
B

1   v 0 R 2   2 v 0 R 3   2 v 0 R 4   v 0 2 R

Aarushi Mishra · Accepted Answer

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vP→=vQ→+ω→×r→PQ, where P and Q are general point on rigid body and r→PQ is position vector of P with respect to QChoose P=A and Q=CvA→=vC→+ω→×r→ACGivenvC→=v0 i^r→AC=Rj^ω→=ωk^By condition of no slipping, ω=v0R ω→=-v0R k^ body is rolling clockwise so direction of angular velocity will be along -z axisThereforevA→=v0i^-v0Rk^×Rj^vA→=v0i^-v0RRk^×j^vA→=v0i^+v0i^vA→=2v0i^SimilarlyvB→=vC→+ω→×r→BCr→BC=Ri^vB→=v0i^-v0Rk^×Ri^vB→=v0i^-v0RRk^×i^vB→=v0i^-v0j^Now we know, let angular velocity of A with respect to B be ω→'ω→'=ω'k^vA→=vB→+ω→'×r→ABr→AB=-Ri^+Rj^2v0i^=v0i^-v0j^+ω'k^×-Ri^+Rj^2v0i^=v0i^-v0j^+ω'Rk^×-i^+ω'Rk^×j^2v0i^=v0i^-v0j^-ω'Rj^-ω'Ri^v0i^+v0j^=-ω'Rj^-ω'Ri^Clearly we have-ω'Rj^=v0ω'=-v0Rω '→=-v0Rk^
You can take is as a general result that angular velocity
of any point of the body with respect to any other point is same
and equal to angular velocity of the body

Q-30) A ring of radius R is rolling without slipping over rough horizontal surface with velocity v0. Two points are located at A and B on rim of ring. Find angular velocity of A w.r.t. B. 1 v 0 R 2 2 v 0 R 3 2 v 0 R 4 v 0 2 R

Q-30) A ring of radius R is rolling without slipping over rough horizontal surface with velocity v₀. Two points are located at A and B on rim of ring. Find angular velocity of A w.r.t. B.

$(1) \frac{v_{0}}{R} (2) \frac{\sqrt{2} v_{0}}{R} (3) \frac{2 v_{0}}{R} (4) \frac{v_{0}}{\sqrt{2} R}$