Question

A pulley one meter in diameter rotating at 600 revolutionS a minute is brought to rest in 20 s by a constant force of friction on its shift. How many revolutions does it make before coming rest?

Answer 1

In 20 seconds, a pulley with an initial speed of 600 revolutions per minute slows down to rest under constant friction. By using kinematic equations adapted for rotational motion, it is determined that the pulley makes 100 revolutions before coming to a stop.

Step-by-step explanation:

The question revolves around rotational kinematics, specifically the process of a rotating object slowing down under friction.

At the start, the pulley is spinning at 600 revolutions per minute (rpm). Since it is being slowed down by a constant force of friction, this implies a uniform deceleration. To find the total number of revolutions before coming to rest, we need to calculate the angular distance (number of revolutions) covered during deceleration. We can use kinematic equations adapted for rotational motion, considering the initial angular velocity (ω0), final angular velocity (ω), angular acceleration (α), and time (t).

Here's a step-by-step solution:

1. Convert the initial angular velocity from rpm to revolutions per second (rps) by dividing by 60:
2. ω0 = 600 rpm / 60 = 10 rps.
3. Since the pulley comes to rest, the final angular velocity ω = 0 rps.
4. The time taken to come to rest, t = 20 s.
5. Use the kinematic equation for angular displacement (θ), which is θ = ω0t + 0.5αt2. We need to find the angular acceleration first.
6. The angular acceleration can be found using α = (ω - ω0) / t. Plugging in the values gives us α = (0 - 10) / 20 = -0.5 rps2.
7. Now we can calculate the number of revolutions: θ = 10 * 20 + 0.5 * (-0.5) * 202 = 200 - 100 = 100 revolutions.

So, the pulley makes 100 revolutions before it comes to rest.