Sliding on a block with a circular cut

JEE Advanced 2017 Paper 1, Question 2

A block of mass $M$ has a circular cut with a frictionless surface as shown. The block rests on the horizontal frictionless surface of a fixed table. Initially the right edge of the block is at $x=0$ , in a co-ordinate system fixed to the table. A point mass $m$ is released from rest at the topmost point of the path as shown and it slides down. When the mass loses contact with the block, its position is $x$ and the velocity is $v$ . At that instant, which of the following options is/are correct?

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The position of the point mass $m$ is: $x=-\sqrt{2} \frac{m R}{M+m}$
The velocity of the point mass $m$ is: $v=\sqrt{\frac{2 g R}{1+\frac{m}{M}}}$
The $x$ component of displacement of the center of mass of the block $M$ is: $-\frac{m R}{M+m}$
The velocity of the block $M$ is: $V=-\frac{m}{M} \sqrt{2 g R}$

Related problem: Mass on a semicircular block.

Solution

Suppose the mass $m$ has a velocity $v \hat{\bf x}$ , and the block has a velocity $V \hat{\bf x}$ at the instant $m$ loses contact with the block, as shown in the figure below.

Since the surfaces are frictionless, energy is conserved. Furthermore there is no external force in the $\hat{\bf x}$ direction, which means momentum is conserved along the horizontal axis. Then,

(1) $\begin{align*} M V + m v &= 0 \\ \frac{1}{2} M V^2 + \frac{1}{2} m v^2 &= m g R . \end{align*}$

In writing these equations we have used the fact that the initial momentum of the system is zero, and the total potential energy decreases by $m g R$ . Denoting the final momentum of the mass $m$ as $p \equiv mv = -MV$ , we may write the energy conservation equation gives

(2) $\begin{equation*} \frac{p^2}{2 M} + \frac{p^2}{2 m} = m g R \implies p = \sqrt{\frac{2 m g R}{\frac{1}{m}+\frac{1}{M}}} . \end{equation*}$

We have chosen the positive square root because the mass $m$ ends up moving in the $+x$ direction. Thus,

(3) $\begin{equation*} \begin{aligned} v = \frac{p}{m} = \sqrt{\frac{2 g R}{1+\frac{m}{M}}} &\implies \text{\bf option (B) is correct} , \\ V = -\frac{p}{M} = \sqrt{\frac{2 m g R/M}{1+\frac{M}{m}}} &\implies \text{\bf option (D) is incorrect} . \end{aligned} \end{equation*}$

Calculating the diplacement of the masses is slightly more involved because we need to integrate the velocity at each instant of time from start to finish. Furthermore, when the mass $m$ is rolling down the circular cut there are two parts to its motion: the circular motion as it rolls down the cut, and the horizontal displacement as it remains in contact with the moving block during transit. In such situations the fool proof method to find the correct velocity is the write down the position of the mass in question and take its time derivative.

Tip: To find the velocity of a mass undergoing “compound” motion, write down an expression for its position and differentiate it with respect to time.

Here is how it works. We choose a co-ordinate system as shown in figure. The top right corner of the block is $(X,h)$ , and the mass $m$ makes an angle $\theta$ with respect to the horizontal, as indicated.

The mass $m$ is at a position

(4) $\begin{equation*} {\bf r} = (X-R\cos\theta,h-R\sin\theta) . \end{equation*}$

The velocity of $m$ is

(5) $\begin{equation*} {\bf v} = \dot{\bf r} = (\dot{X} + R\sin\theta \dot{\theta} ,-R\cos\theta \dot{\theta}) \end{equation*}$

where the overhead dots represent derivatives w.r.t. time, $\dot{f} = \frac{{\rm d}f}{{\rm d}t}$ . The velocity of the block is ${\bf V} = (\dot{X},0)$ . Again, by conservation of momentum in the horizontal direction

(6) $\begin{equation*} M \dot{X} + m (\dot{X} + R\sin\theta \dot{\theta}) = 0 , \end{equation*}$

which we can invert to obtain

(7) $\begin{equation*} \dot{X} = - \frac{m R \sin \theta \dot{\theta}}{(M+m)} . \end{equation*}$

At the moment it loses contact with the block, the total horizontal displacement of the mass $m$ is the same as the $x-$ co-ordinate of the corner,

(8) $\begin{equation*} X = \int {\rm d}X = \int \frac{{\rm d}X}{{\rm d}t} {\rm d}t \equiv \int \dot{X} {\rm d}t . \end{equation*}$

(9) $\begin{equation*} X = - \int_{0}^{\pi/2} \frac{m R \sin \theta}{(M+m)} {\rm d}\theta = - \frac{m R }{(M+m)} . \end{equation*}$

This $X$ is also the displacement of the center of mass of the block. Thus, we conclude that option (A) is incorrect but option (C) is correct.

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