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:

Step 1: Recall Torque and Moment of Inertia Formulas

Torque \(\tau = rF\sin\theta\), where \(r\) is the distance from the pivot to the force application point, \(F\) is the force, and \(\theta = 90^\circ\) (so \(\sin\theta = 1\)). For a rod pivoted at one end, the moment of inertia \(I=\frac{1}{3}mL^2\). Angular acceleration \(\alpha=\frac{\tau}{I}\).

Step 2: Calculate Torque

The force is applied at the midpoint, so \(r = \frac{L}{2}\). Given \(F = 10\,\text{N}\), \(L = 1.5\,\text{m}\), so \(\tau=\frac{L}{2}\times F\times1=\frac{1.5}{2}\times10 = 7.5\,\text{N·m}\).

Step 3: Calculate Moment of Inertia

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

Step 4: Calculate Angular Acceleration

Using \(\alpha=\frac{\tau}{I}\), 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:

The resulting angular acceleration of the rod is \(\boldsymbol{5.0\,\text{rad/s}^2}\).