A block of mass M with a semicircular track of radius R, rests on a horizontal frictionless surface.

Q: Why is the vertical displacement of the cylinder $(R-r)$ and not $R$?

The cylinder is not a point mass; it has a radius $r$. Its center of mass starts at a height $R$ relative to the track's center of curvature, but when it reaches the bottom, its center is at a height $r$ from the bottom. Thus, the actual vertical drop of its center of mass is $(R-r)$.

Q: Why does the block move to the left?

Because there is no external force in the horizontal direction, the total horizontal momentum must remain zero. As the cylinder gains velocity to the right by sliding down the track, the block must gain velocity to the left to keep the total momentum zero.

Q: Can we use the work-energy theorem here?

Yes! Conservation of mechanical energy is essentially the work-energy theorem applied to conservative forces. The work done by gravity on the cylinder converts entirely into the kinetic energy of both the block and the cylinder.

Q: Why is the absolute displacement of the cylinder $(R-r) - x$?

Displacement is a vector. Relative to the block, the cylinder moves right by $(R-r)$. However, the block itself moves left by $x$ relative to the ground. So, the net displacement of the cylinder relative to the ground is $(R-r)$ to the right, minus $x$ to the left.

Q: What happens to the normal force between the cylinder and the track?

The normal force is an internal force for the block-cylinder system. It does no net work on the system as a whole because the track is frictionless, and it only serves to transfer momentum between the cylinder and the block.

Question

A block of mass $M$ with a semicircular track of radius $R$, rests on a horizontal frictionless surface. A uniform cylinder of radius $r$ and mass $m$ is released from rest at the top point $A$ (see fig.). The cylinder slips on the semicircular frictionless track.
(a) How far has the block moved when the cylinder reaches the bottom (point $B$) of the track ?
(b) How fast is the block moving when the cylinder reaches the bottom of the track?

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a massive block of mass $M$ featuring a perfectly smooth semicircular track of radius $R$. This block is resting peacefully on a frictionless horizontal floor. Now, we introduce a small, uniform cylinder of mass $m$ and radius $r$, releasing it from rest at the very top edge of the track (point $A$).

As the cylinder begins its descent, gravity pulls it down, but the curvature of the track forces it to move horizontally as well. Because the floor is completely frictionless, there is absolutely no external force acting on the entire block-cylinder system in the horizontal direction. This is the crucial insight: the center of mass of the system is horizontally locked in place. If the cylinder moves to the right, the block must recoil to the left!

### Conservation of Center of Mass

Let's determine exactly how far the block moves. Suppose the block slides to the left by a distance $x$. Meanwhile, the cylinder slides down the track. Relative to the block, the cylinder's center of mass moves horizontally from the edge of the track to the center. Since the track has radius $R$ and the cylinder has radius $r$, this relative horizontal displacement is $(R-r)$.

However, because the block itself has moved left by $x$, the absolute displacement of the cylinder to the right is $(R-r) - x$. Since the center of mass of the system cannot shift horizontally, the mass moments must perfectly balance out:

$$M(x) = m((R-r) - x)$$

Expanding and rearranging this equation to solve for $x$:

$$Mx = m(R-r) - mx$$

$$x(M+m) = m(R-r)$$

$$x = \frac{m(R-r)}{M+m}$$

This elegant expression gives us the exact recoil distance of the block when the cylinder reaches the bottom.

### Conservation of Momentum and Energy

Now, let's find the speed of the block when the cylinder hits the bottom (point $B$). Let $v_1$ be the absolute speed of the cylinder to the right, and $v_2$ be the absolute speed of the block to the left.

Just as the center of mass position is conserved, the total horizontal momentum must remain zero. Therefore, the momentum of the cylinder must equal the momentum of the block:

$$mv_1 = Mv_2$$

Next, we apply the principle of conservation of mechanical energy. As the cylinder drops, its center of mass descends by a vertical distance of $(R-r)$. The loss in gravitational potential energy is entirely converted into the kinetic energy of both the cylinder and the block:

$$mg(R-r) = \frac{1}{2}mv_1^2 + \frac{1}{2}Mv_2^2$$

### Final Calculation

We need to find $v_2$, the speed of the block. From our momentum equation, we can express $v_1$ in terms of $v_2$:

$$v_1 = \frac{M}{m}v_2$$

Substituting this into the energy equation:

$$mg(R-r) = \frac{1}{2}m\left(\frac{M}{m}v_2\right)^2 + \frac{1}{2}Mv_2^2$$

Squaring the term inside the parenthesis and simplifying:

$$mg(R-r) = \frac{1}{2}\frac{M^2}{m}v_2^2 + \frac{1}{2}Mv_2^2$$

Notice that we can factor out $\frac{1}{2}Mv_2^2$ from the right side:

$$mg(R-r) = \frac{1}{2}Mv_2^2 \left(\frac{M}{m} + 1\right)$$

Finding a common denominator for the terms in the bracket:

$$mg(R-r) = \frac{1}{2}Mv_2^2 \left(\frac{M+m}{m}\right)$$

Finally, we isolate $v_2^2$ and take the square root to find the speed of the block:

$$v_2^2 = \frac{2m^2g(R-r)}{M(M+m)}$$

$$v_2 = m\sqrt{\frac{2g(R-r)}{M(M+m)}}$$

And there we have it! A beautiful interplay of momentum and energy conservation yielding the exact dynamics of the system.

Visualized Solution

System Orientation

$Δ x_{C M} = 0$

Relative Displacement

Calculating $x$

Velocity Setup

Conservation of Momentum

Conservation of Energy

Energy Equation

Substituting $v_{1}$

Simplifying the Equation

Final Speed $v_{2}$

The Sigma Insight: Motion of Centre of Mass

Analyzing the Setup

Conservation of Center of Mass

Conservation of Momentum and Energy

Final Calculation

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