Question

A 250 g, 25-cm-diameter plastic disk is spun on an axle through its center by an electric motor. 

What torque must the motor supply to take the disk from 0to 1600 rpm in 4.8 s?

Answer 1

Concepts and reason

The concepts used to solve this problem are angular kinematics and moment of inertia. Initially, the angular acceleration can be calculated by using the angular kinematics formula. Then, the moment of inertia can be calculated for the disk by using its formula. Finally, the torque acting on the system can be calculated by multiplying the moment of inertia and angular acceleration.

Fundamentals

The expression for the angular acceleration from angular kinematics is, \(\omega_{f}=\omega_{i}+\alpha t\)

Here, \(\omega_{f}\) is the final angular speed, \(\omega_{i}\) is the initial angular speed, \(\alpha\) is the angular acceleration, and \(t\) is the time period. The moment of inertia for the disk is, \(I=\frac{1}{2} M R^{2}\)

Here, \(I\) is the moment of inertia, \(M\) is the mass of the disk, and \(R\) is the radius of the disk. The formula for the torque is, \(\tau=I \alpha\)

Here, \(\tau\) is the torque acting on the system. Rpm is revolution per min which is expressed as \(\mathrm{rpm}=\frac{\mathrm{rev}}{\mathrm{min}}\)

 

The expression for the angular acceleration is, \(\omega_{f}=\omega_{i}+\alpha t\)

Substitute 1600 rpm for \(\omega_{f}, 0\) for \(\omega_{i}, 4.8 s\) for \(t\)

$$ \begin{array}{c} 0+4.8 \alpha=\left(1600 \frac{\mathrm{rev}}{\mathrm{min}}\right)\left(\frac{1 \mathrm{~min}}{60 \mathrm{~s}}\right)(2 \pi \mathrm{rad}) \\ 167.47=4.8 \alpha \\ \alpha=34.9 \mathrm{rad} / \mathrm{s}^{2} \end{array} $$

The angular acceleration is calculated by using the angular kinematics formula. The initial angular speed is converted from rev/min to rad/s.

 

The radius of the plastics disk is, \(R=\frac{d}{2}\)

Here, \(\mathrm{d}\) is the diameter of the disk. Substitute \(25 \mathrm{~cm}\) for \(\mathrm{d}\) to find \(\mathrm{R}\).

$$ \begin{aligned} R=& \frac{(25 \mathrm{~cm})\left(\frac{10^{-2} \mathrm{~m}}{1 \mathrm{~cm}}\right)}{2} \\ &=0.125 \mathrm{~m} \end{aligned} $$

The expression for the moment of inertia is, \(I=\frac{1}{2} M R^{2}\)

Substitute \(250 \mathrm{~g}\) for \(M\) and \(0.125 \mathrm{~m}\) for \(R\)

$$ \begin{array}{c} I=\frac{1}{2}(250 \mathrm{~g})\left(\frac{1 \times 10^{-3} \mathrm{~kg}}{1 \mathrm{~g}}\right)\left[(12.5 \mathrm{~cm})\left(\frac{1 \times 10^{-2} \mathrm{~m}}{1 \mathrm{~cm}}\right)\right]^{2} \\ =1.95 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2} \end{array} $$

The moment of inertia for the disk is calculated by substituting the mass of the disk with the radius of the disk. Here, the moment of inertia of the disk is calculated to determine the torque acting on the system.

 

The expression for the torque is, \(\tau=I \alpha\)

Substitute \(1.95 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\) for \(I\) and \(34.9 \mathrm{rad} / \mathrm{s}^{2}\) for \(\alpha\)

$$ \begin{array}{l} \tau=\left(1.95 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\right)\left(34.9 \mathrm{rad} / \mathrm{s}^{2}\right) \\ \quad=0.0681 \mathrm{~N} \cdot \mathrm{m} \\ \quad \approx 0.068 \mathrm{~N} \cdot \mathrm{m} \end{array} $$

The torque acting on the system is \(0.068 \mathrm{~N} \cdot \mathrm{m}\).

The torque acting on the system is calculated by using the angular kinematics equation and moment of inertia.

 

The torque acting on the system is \(0.068 \mathrm{~N} \cdot \mathrm{m}\).