Question

a uniform rod of mass m = 2.0 kg and length l = 1.5 m is pivoted at one end. a force of f = 10 n is applied perpendicularly to the rod at its midpoint. what is the resulting angular acceleration of the rod?

Explanation:

Step1: Recall Torque and Moment of Inertia Formulas

Torque \(\tau = rF\sin\theta\), for perpendicular force \(\theta = 90^\circ\), \(\sin\theta = 1\), so \(\tau = rF\). Moment of inertia for a rod pivoted at one end is \(I=\frac{1}{3}mL^{2}\). Also, \(\tau = I\alpha\), so \(\alpha=\frac{\tau}{I}\).

Step2: Calculate Torque

The force is applied at the midpoint, so \(r = \frac{L}{2}\). Given \(F = 10\,\text{N}\), \(L = 1.5\,\text{m}\), \(\tau = \frac{L}{2}\times F=\frac{1.5}{2}\times10 = 7.5\,\text{N·m}\).

Step3: Calculate Moment of Inertia

\(m = 2.0\,\text{kg}\), \(L = 1.5\,\text{m}\), \(I=\frac{1}{3}mL^{2}=\frac{1}{3}\times2.0\times(1.5)^{2}=\frac{1}{3}\times2.0\times2.25 = 1.5\,\text{kg·m}^2\).

Step4: Calculate Angular Acceleration

Using \(\alpha=\frac{\tau}{I}\), substitute \(\tau = 7.5\) and \(I = 1.5\): \(\alpha=\frac{7.5}{1.5}=5\,\text{rad/s}^2\).

Answer:

\(5\,\text{rad/s}^2\)