Question

A block of mass M is moving with a velocity $v_1$ on a frictionless surface as shown in the figure. It passes over a cylinder of radius R and mass M, which has a fixed axis and is initially at rest. When the block first makes contact with the cylinder, it slips on the cylinder, but the friction is large enough so that slipping ceases before it losses contact with the cylinder. Finally, it goes to the dotted position with velocity $v_2 = \frac{2}{n}v_1$. Find the value of n.

Answer 1

n = 3

Answer 2

We analyze the system by conserving the angular momentum about the cylinder’s center. Before contact, the only contribution is from the block. Assuming the block’s line of motion is tangent to the cylinder (at a distance R from its center), the block’s initial angular momentum is

$$L_i = M v_1 R.$$

After rolling without slipping is established, the block and cylinder satisfy the non‐slipping condition

$$v_2 = \omega R.$$

Now, the block (mass M) has translational angular momentum about the center

$$L_{\text{block}} = M v_2 R,$$

and the cylinder (mass M, moment of inertia $I = \tfrac{1}{2}MR^2$ for a uniform cylinder) has rotational angular momentum

$$L_{\text{cyl}} = I \omega = \frac{1}{2} M R^2 \frac{v_2}{R} = \frac{1}{2} M R v_2.$$

Thus, the total final angular momentum is

$$L_f = L_{\text{block}} + L_{\text{cyl}} = M R v_2 + \frac{1}{2} M R v_2 = \frac{3}{2} M R v_2.$$

Assuming no net external torque about the center (since the support forces have zero moment about the fixed center), equate the initial and final angular momenta:

$$M R v_1 = \frac{3}{2} M R v_2 \quad \Longrightarrow \quad v_2=\frac{2}{3}v_1.$$

Since the problem states $v_2=\frac{2}{n}v_1,$ we must have

$$\frac{2}{n}=\frac{2}{3}\quad\Longrightarrow\quad n=3.$$