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

The torque \(\tau\) is given by \(\tau = rF\sin\theta\), where \(r\) is the distance from the pivot, \(F\) is the force, and \(\theta\) is the angle between \(r\) and \(F\). Here, \(\theta = 90^\circ\) so \(\sin\theta = 1\), and \(r=\frac{L}{2}\) (midpoint). The moment of inertia \(I\) for a rod pivoted at one end is \(I=\frac{1}{3}mL^2\). Also, \(\tau = I\alpha\), where \(\alpha\) is the angular acceleration.

Step2: Calculate Torque

Substitute \(r = \frac{L}{2}\), \(F = 10\,\text{N}\), \(L = 1.5\,\text{m}\) into \(\tau = rF\):
\(\tau=\frac{L}{2}\times F=\frac{1.5}{2}\times10 = 7.5\,\text{N·m}\)

Step3: Calculate Moment of Inertia

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

Step4: Solve for Angular Acceleration

From \(\tau = I\alpha\), we get \(\alpha=\frac{\tau}{I}\). Substitute \(\tau = 7.5\,\text{N·m}\) and \(I = 1.5\,\text{kg·m}^2\):
\(\alpha=\frac{7.5}{1.5}=5.0\,\text{rad/s}^2\)

Answer:

\(5.0\,\text{rad/s}^2\)