Question

A constant torque of 3 Nm is applied to an unloaded motor at rest at time t = 0. The motor reaches a speed of 1,393 rpm in 4 s. Assuming the damping to be negligible, calculate the motor inertia in Nm·s2.

Answer 1

Answer:

The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.

Explanation:

From Newton's Laws of Motion and Principle of Motion of D'Alembert, the net torque of a system ($\tau$), measured in Newton-meters, is:

$\tau = I\cdot \alpha$ (1)

Where:

$I$ - Moment of inertia, measured in Newton-meter-square seconds.

$\alpha$ - Angular acceleration, measured in radians per square second.

If motor have an uniform acceleration, then we can calculate acceleration by this formula:

$\alpha = \frac{\omega - \omega_{o}}{t}$ (2)

Where:

$\omega_{o}$ - Initial angular speed, measured in radians per second.

$\omega$ - Final angular speed, measured in radians per second.

$t$ - Time, measured in seconds.

If we know that $\tau = 3\,N\cdot m$, $\omega_{o} = 0\,\frac{rad}{s }$, $\omega = 145.875\,\frac{rad}{s}$ and $t = 4\,s$, then the moment of inertia of the motor is:

$\alpha = \frac{145.875\,\frac{rad}{s}-0\,\frac{rad}{s}}{4\,s}$

$\alpha = 36.469\,\frac{rad}{s^{2}}$

$I = \frac{\tau}{\alpha}$

$I = \frac{3\,N\cdot m}{36.469\,\frac{rad}{s^{2}} }$

$I = 0.0823\,N\cdot m\cdot s^{2}$

The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.