Question

a uniform rod of mass ( m = 2.0 , \text{kg} ) and length ( l = 1.5 , \text{m} ) is pivoted at one end. a force of ( f = 10 , \text{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 \(\sin\theta = 1\), and \(r=\frac{L}{2}\) (mid - point). 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

Given \(F = 10\ N\), \(r=\frac{L}{2}=\frac{1.5}{2}=0.75\ m\). Then \(\tau=rF=(0.75\ m)\times(10\ N) = 7.5\ N\cdot m\).

Step3: Calculate Moment of Inertia

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

Step4: Calculate Angular Acceleration

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

Answer:

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