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:

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 ([tex]\tau[/tex]), measured in Newton-meters, is:

[tex]\tau = I\cdot \alpha[/tex] (1)

Where:

[tex]I[/tex] - Moment of inertia, measured in Newton-meter-square seconds.

[tex]\alpha[/tex] - Angular acceleration, measured in radians per square second.

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

[tex]\alpha = \frac{\omega - \omega_{o}}{t}[/tex] (2)

Where:

[tex]\omega_{o}[/tex] - Initial angular speed, measured in radians per second.

[tex]\omega[/tex] - Final angular speed, measured in radians per second.

[tex]t[/tex] - Time, measured in seconds.

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

[tex]\alpha = \frac{145.875\,\frac{rad}{s}-0\,\frac{rad}{s}}{4\,s}[/tex]

[tex]\alpha = 36.469\,\frac{rad}{s^{2}}[/tex]

[tex]I = \frac{\tau}{\alpha}[/tex]

[tex]I = \frac{3\,N\cdot m}{36.469\,\frac{rad}{s^{2}} }[/tex]

[tex]I = 0.0823\,N\cdot m\cdot s^{2}[/tex]

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

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