Question

problem 8:
a grinding wheel of radius 8 cm and mass 600 g may be thought of as a
uniform disk. a force perpendicular to the radius vector is applied at the
edge of the wheel to give it an angular acceleration of 9.2 rad/s². find the
magnitude of the force applied on the wheel.

Explanation:

Step1: Find the moment of inertia of the disk

The moment of inertia \(I\) of a uniform disk about an axis perpendicular to its plane and passing through its center is \(I=\frac{1}{2}mr^{2}\).
Given \(m = 600\ g=0.6\ kg\) and \(r = 8\ cm = 0.08\ m\).

$$I=\frac{1}{2}\times0.6\times(0.08)^{2}$$

$$I = 0.3\times0.0064= 0.00192\ kg\cdot m^{2}$$

Step2: Use the torque - angular acceleration relation

The torque \(\tau\) is related to the angular acceleration \(\alpha\) by \(\tau=I\alpha\). Also, for a force \(F\) applied at a distance \(r\) from the axis of rotation (and \(F\perp r\)), \(\tau = rF\).
Since \(\tau=I\alpha\) and \(\tau = rF\), we have \(rF=I\alpha\).
We know \(I = 0.00192\ kg\cdot m^{2}\), \(\alpha=9.2\ rad/s^{2}\) and \(r = 0.08\ m\).

$$F=\frac{I\alpha}{r}$$

$$F=\frac{0.00192\times9.2}{0.08}$$

$$F=\frac{0.017664}{0.08}$$

$$F = 0.2208\ N$$

Answer:

\(0.22\ N\) (rounded to two decimal places)