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 \(\theta = 90^\circ\), so \(\sin\theta = 1\). The moment of inertia \(I\) 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 \(F = 10\space N\), \(L = 1.5\space m\), \(\tau=rF=\frac{L}{2}\times F=\frac{1.5}{2}\times10 = 7.5\space N\cdot m\).

Step3: Calculate Moment of Inertia

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

Step4: Calculate Angular Acceleration

Using \(\alpha=\frac{\tau}{I}\), substitute \(\tau = 7.5\space N\cdot m\) and \(I = 1.5\space kg\cdot m^2\), \(\alpha=\frac{7.5}{1.5}=5.0\space rad/s^2\).

Answer:

The resulting angular acceleration of the rod is \(5.0\space rad/s^2\) (or \(5\space rad/s^2\)).