Question

A smooth ring A of mass m can slide on a fixed horizontal rod. A string tied to the ring passes over a fixed pulley B and carries a block C of mass $M$ =$2m$ as shown in figure. At an instant the string between the ring and the pulley makes an angle $\theta$ with the rod. (a) if the ring slides with a speed $v$, then the block descends with what speed? (b) With what acceleration will the ring start moving if system is released from rest with $\theta$ = ${30}^{\circ }$?

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

If ring moves right then block moves down

If ring moves right then length of the string reduces and when C moves down its trying to increase the length of the string so the velocity's component that's trying to reduce the string should be equal to the component that's trying to increase it

So $Vcos\theta$ = $V_c$

$\Rightarrow$$V_{block}$ = $Vcos\theta$ downwards

Alternate solution

$L$ = $x_r+x_b$

Differentiating $0$ = $-vcos\theta +v_b$

$v_b$ = $vcos\theta$

same constraint relation will be there for acceleration.

If ring moves on the rod with acceleration a then block goes down with acceleration $acos\theta$.

Now free body diagram of block

$2mg-T=2macos\theta$ ----------(I)

$Tcos\theta$ = $ma$ -----------(II)

Solving we get a of ring as $6.78m/s^2$