The shaft of a motor rotating at $50$ rev s$^{-1}$ is uniformly retarded to $20$ rev s$^{-1}$ in $15$ s. The number of complete rotations the shaft makes in the given time is . (Answer in integer)

Question

The primary purpose of mine ventilation is to:

Answer 1

Here we solve the same problem using the standard equations of rotational motion instead of the average speed shortcut, first finding the angular deceleration and then the angle turned.

Convert the given speeds from revolutions per second to angular velocity in radians per second, using $\omega = 2\pi N$.

\[ \omega_i = 2\pi(50) = 100\pi \text{ rad/s}, \quad \omega_f = 2\pi(20) = 40\pi \text{ rad/s} \]

Since the retardation is uniform, the angular velocity falls at a constant rate $\alpha$ over the time $t = 15$ s. Using $\omega_f = \omega_i - \alpha t$:

\[ 40\pi = 100\pi - \alpha(15) \]\[ \alpha = \frac{100\pi - 40\pi}{15} = \frac{60\pi}{15} = 4\pi \text{ rad/s}^2 \]

Now find the total angle turned $\theta$ using the equation $\theta = \omega_i t - \frac{1}{2}\alpha t^2$.

\[ \theta = 100\pi(15) - \frac{1}{2}(4\pi)(15)^2 \]\[ \theta = 1500\pi - \frac{1}{2}(4\pi)(225) = 1500\pi - 450\pi = 1050\pi \text{ rad} \]

Each complete rotation corresponds to an angle of $2\pi$ radians, so divide the total angle by $2\pi$ to get the number of rotations.

\[ n = \frac{1050\pi}{2\pi} = 525 \]

Let's summarize:

Uniform retardation gives a constant angular deceleration $\alpha = 4\pi$ rad/s^2.
The total angle swept in 15 seconds works out to $1050\pi$ radians.
Dividing by $2\pi$ per revolution gives 525 complete rotations, matching the average-speed shortcut.

So the shaft completes 525 rotations in the given 15 seconds.

Answer 2