Question

An electric motor’s rotor is spinning at 1500 rpm when its load and power are cut off. The rotor, which has a mass of 50 kg and a radius of gyration of 200 mm, then coasts down to rest. Due to kinetic friction, a constant torque of 10 Nm acts on the rotor as it coasts down.
The number of revolutions executed by the rotor before it comes to rest is .......... (in integer).

Hint:

For problems involving deceleration due to a constant torque, use the kinematic equation for angular motion: \( \theta = \frac{\omega_0^2}{2\alpha} \), and remember to convert radians to revolutions using \( 2\pi \).

Answer 1

Given: 
- Initial angular velocity, \( \omega_0 = 1500 \) rpm = \( \frac{1500 \times 2\pi}{60} \) rad/s = 157.08 rad/s 
- Mass of rotor, \( m = 50 \, {kg} \) 
- Radius of gyration, \( k = 0.2 \, {m} \) 
- Torque due to friction, \( \tau = 10 \, {Nm} \) 
Step 1: Moment of Inertia 
The moment of inertia \( I \) of the rotor is given by: 
\[ I = m k^2 = 50 \times (0.2)^2 = 50 \times 0.04 = 2 \, {kg} \cdot {m}^2 \] 
Step 2: Angular Deceleration 
Using the equation for torque, we can find the angular acceleration: 
\[ \tau = I \alpha \Rightarrow \alpha = \frac{\tau}{I} = \frac{10}{2} = 5 \, {rad/s}^2 \] This is the angular deceleration as the rotor is slowing down.
Step 3: Number of Revolutions 
The equation for angular displacement \( \theta \) when the object is under constant angular acceleration is: 
\[ \theta = \frac{\omega_0^2}{2\alpha} \] Substituting the values: 
\[ \theta = \frac{(157.08)^2}{2 \times 5} = \frac{24670.22}{10} = 2467.02 \, {rad} \] To convert from radians to revolutions, we divide by \( 2\pi \): 
\[ {Number of revolutions} = \frac{2467.02}{2\pi} \approx 393 \, {revolutions} \] Thus, the number of revolutions executed by the rotor before it comes to rest is approximately \( 393 \) revolutions.