Question

A small ring of mass m is connected with a particle of same mass by an ideal string and the whole system is released as shown in figure. Coefficient of friction between ring A and wire is $3/5$. If ring A starts sliding when connecting string makes an angle $\theta$ with the vertical. This angle $\theta$ is (particle is free to move and ring can slide only):

• A. $30$
• B. $45$
• C. $60$
• D. None of these

Answer 1

The correct option is B $45$

Using energy balance we get:
$mglcos\theta =\frac{m{v}^2}{2}$

Also force along the thread is:
$T-mgcos\theta =2mgcos\theta$
$T=3mgcos\theta$

Now the force along the horizontal direction along the ring would be:
$3mgcos\theta sin\theta =\mu [mg+3mgco{s}^2\theta ]$

Thus we get:
$\theta ={45}^0$