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).
Show Hint
Use energy method:
\[
\frac{1}{2} I \omega_0^2 = T \cdot \theta
\]
Then convert total radians to revolutions.
To determine the number of revolutions executed by the rotor before it comes to rest, we need to follow these steps:
1. **Convert initial rotational speed to radians per second:** The initial speed is given as 1500 revolutions per minute (rpm). Since 1 revolution = 2π radians and 1 minute = 60 seconds, we convert rpm to radians per second: \( \omega_i = 1500 \times \frac{2\pi}{60} = 1500 \times \frac{\pi}{30} \) rad/s.
2. **Calculate initial angular kinetic energy:** The mass of the rotor \( m = 50 \) kg and radius of gyration \( k = 200 \) mm = 0.2 m. The moment of inertia \( I = mk^2 = 50 \times (0.2)^2 = 2 \) kg·m². The initial kinetic energy \( KE_i = \frac{1}{2} I \omega_i^2 \). Substituting the values, we have: \( KE_i = \frac{1}{2} \times 2 \times \left(\frac{1500\pi}{30}\right)^2 \).
3. **Calculate work done by friction:** The constant torque \( \tau = 10 \) Nm acts as the rotor spins down. The work done by friction \( W = \tau \times \theta \) (where \( \theta \) is the angle in radians). Since the rotor comes to rest, initial kinetic energy is entirely dissipated by work done through friction: \( W = KE_i \). \( 10 \times \theta = \frac{1}{2} \times 2 \times \left(\frac{1500\pi}{30}\right)^2 \). Solving for \( \theta \): \( \theta = \left(\frac{1500\pi}{30}\right)^2 \).
4. **Convert total radians to revolutions:** The number of revolutions \( N = \frac{\theta}{2\pi} \). \( N = \frac{\left(\frac{1500\pi}{30}\right)^2}{2\pi} \). Calculating this value gives: \( N = 392 \) revolutions. The number of revolutions before the rotor stops is **392**, which is within the given range of 392,392.
