A thin circular ring of mass M and radius R is rotating with a constant angular velocity 2 rads -1 in a horizontal plane about an axis vertical to its plane and passing through the center of the ring. If two objects each of mass m be attached gently to the opposite ends of a diameter of ring, the ring will then rotate with an angular velocity (in rads -1 ).

Question

A thin circular ring of mass M and radius R is rotating with a constant angular velocity 2 rads -1 in a horizontal plane about an axis vertical to its plane and passing through the center of the ring. If two objects each of mass m be attached gently to the opposite ends of a diameter of ring, the ring will then rotate with an angular velocity (in rads -1 ). (A) (M/(M+m)) (B) (( M +2 m )/2 M ) (C) (2 M/(M+2 m)) (D) (2( M +2 m )/ M )

Tardigrade Admin · Accepted Answer

Applying conservation of angular momentum MR 2 ω=( MR 2+2 mR 2) ω prime ω prime=(2 M/M+2 m)

$\frac{M}{( M + m )}$

$\frac{( M + 2 m )}{2 M}$

$\frac{2 M}{( M + 2 m )}$

$\frac{2 ( M + 2 m )}{M}$

Applying conservation of angular momentum
$M R^{2} ω = (M R^{2} + 2 m R^{2}) ω^{'}$
$ω^{'} = \frac{2 M}{M + 2 m}$

(M+m)M​

2M(M+2m)​

(M+2m)2M​

M2(M+2m)​

Applying conservation of angular momentum MR2ω=(MR2+2mR2)ω′ ω′=M+2m2M​

$\frac{M}{( M + m )}$

$\frac{( M + 2 m )}{2 M}$

$\frac{2 M}{( M + 2 m )}$

$\frac{2 ( M + 2 m )}{M}$

Applying conservation of angular momentum
$M R^{2} ω = (M R^{2} + 2 m R^{2}) ω^{'}$
$ω^{'} = \frac{2 M}{M + 2 m}$