Question 764: Centre of Mass and Rotational Motion

Question

A block of mass $$M$$ has a circular cut with a frictionless surface as shown. The block resets 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?

Accepted Answer

We discuss the options as follows:
$$\bullet$$ For option (B) : From law of conservation of momentum, we have
mv = MV ...... (1)
where v is the velocity of point mass m and V is velocity of block of mass M.
Now, by using law of conservation of energy, after the point mass is released, we have the following:
Loss in P.E. of mass m = Gain in K.E. of both mass M and the mass m.
$$ \Rightarrow mgh = {1 \over 2}m{v^2} + {1 \over 2}M{V^2}$$
Substituting $$V = {{mu} \over M}$$ [from Eq. (1)], we get
$$mgh = {1 \over 2}m{v^2} + {1 \over 2}M{\left( {{{mv} \over M}} \right)^2}$$
$$mgh = {1 \over 2}m{v^2} + {1 \over 2}M{{{m^2}{v^2}} \over {{M^2}}}$$
$$mgR = {1 \over 2}m{v^2} + {1 \over 2}{{{m^2}{v^2}} \over M}$$
$$mgR = {1 \over 2}m{v^2}\left( {1 + {m \over M}} \right)$$
$$gR = {1 \over 2}{v^2}\left( {1 + {m \over M}} \right)$$
Rearranging this equation, we get
$$2gR = {v^2}\left( {1 + {m \over M}} \right)$$
$${v^2} = {{2gR} \over {\left( {1 + {m \over M}} \right)}}$$
$$ \Rightarrow v = \sqrt {{{2gR} \over {\left( {1 + {m \over M}} \right)}}} $$
Hence, option (B) is correct.
$$\bullet$$ For option (C) : Now, from Eq. (1), we have
$$V = {m \over M}v = {m \over M}\sqrt {{{2gR} \over {\left( {1 + {m \over M}} \right)}}} $$
$$V = \sqrt {{{{m^2}2gR} \over {{M^2}\left( {1 + {m \over M}} \right)}}}  = \sqrt {{{{m^2}2gR} \over {{M^2}\left( {{{M + m} \over M}} \right)}}} $$
$$V = \sqrt {{{{m^2}2gR} \over {M(m + M)}}}  \Rightarrow V = m\sqrt {{{2gR} \over {M(m + M)}}} $$
Since, there is no external force acting on the system, the centre of mass does not change.
Now, if the change in position of mass m is $$\Delta$$x and the change in position of mass M is $$\Delta$$X, then
m$$\Delta$$x + M$$\Delta$$x = 0 (since centre of mass will not change)
Now, we know that the change in position of point mass m is
$$\Delta$$x = R $$-$$ x
and the change in position of block of mass M is
$$\Delta$$X = $$-$$x
Therefore,
m(R $$-$$ x) $$-$$ Mx = 0
m(R $$-$$ x) = Mx
mR $$-$$ mx = Mx
mR = Mx + mx
mR = (M + m)x
$$\Rightarrow$$ x = $${{mR} \over {M + m}}$$
The change in position of block of mass M is
$$\Delta$$x = $$-$$x = $${{ - mR} \over {M + m}}$$
Thus, x-component of the displacement of the centre of mass of the block is
$$M = {{ - mR} \over {M + m}}$$
Hence, option (C) is also correct.

Answer

The position of the point mass $$m$$ is :
$$x =  - \sqrt 2 {{mR} \over {M + m}}$$

Answer

The velocity of the point mass $$m$$ is :
$$v = \sqrt {{{2gR} \over {1 + {m \over M}}}} $$

Answer

The $$x$$ component of displacement of the center
of mass of the  block $$M$$ is: $$ - {{mR} \over {M + m}}$$

Answer

The velocity of the block $$M$$ is:
$$V =  - {m \over M}\sqrt {2gR} $$

Centre of Mass and Rotational Motion

Choose the correct answer: