The problem states:

Two wheels, at a certain instant, rotate at 120 r.p.m. and 240 r.p.m., with radii of 20 cm and 40 cm, respectively. A brake is applied to each wheel, with the smaller one stopping in 16 seconds and the larger one in 8 seconds, both decelerating uniformly. We are asked:

1. At what instant do both wheels have the same angular velocity?
2. At that moment, how many rotations has each wheel completed?

Solution Steps:

1. Convert initial speeds to radians per second:

   • Smaller wheel: 120 r.p.m. = $\frac{120 \times 2\pi}{60}$ rad/s = $4\pi$ rad/s.
   • Larger wheel: 240 r.p.m. = $\frac{240 \times 2\pi}{60}$ rad/s = $8\pi$ rad/s.

2. Calculate the angular deceleration:

   • For the smaller wheel: Initial angular velocity ($\omega_0$) = $4\pi$ rad/s, final velocity = 0 rad/s, time = 16 s. $\alpha_{\text{small}} = \frac{\omega_0 - \omega}{t} = \frac{4\pi}{16} = \frac{\pi}{4} \, \text{rad/s}^2$
   • For the larger wheel: Initial angular velocity ($\omega_0$) = $8\pi$ rad/s, final velocity = 0 rad/s, time = 8 s. $\alpha_{\text{large}} = \frac{8\pi}{8} = \pi \, \text{rad/s}^2$

3. Determine when the angular velocities are equal: Let $t$ be the time when both angular velocities are the same.

   • Angular velocity of the smaller wheel at time $t$: $\omega_{\text{small}} = 4\pi - \frac{\pi}{4} \times t$.
   • Angular velocity of the larger wheel at time $t$: $\omega_{\text{large}} = 8\pi - \pi \times t$.

   Set the angular velocities equal: $4\pi - \frac{\pi}{4}t = 8\pi - \pi t$ Solve for $t$ to find the time when both velocities match.

4. Calculate the rotations completed by each wheel up to that time