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 \(\tau = rF\sin\theta\), here \(\theta = 90^\circ\), so \(\tau = rF\). The moment of inertia \(I\) for a rod pivoted at one end is \(I=\frac{1}{3}mL^{2}\). Angular acceleration \(\alpha=\frac{\tau}{I}\).

Step2: Calculate Torque

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

Step3: Calculate Moment of Inertia

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

Step4: Calculate Angular Acceleration

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

Answer:

The resulting angular acceleration of the rod is \(5\ rad/s^{2}\)