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 the formula for torque and rotational inertia

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 \(r=\frac{L}{2}\) (since the force is applied at the midpoint). The rotational inertia \(I\) of a rod pivoted at one end is \(I=\frac{1}{3}mL^2\). The rotational version of Newton's second law is \(\tau = I\alpha\), where \(\alpha\) is the angular acceleration.

First, calculate the torque:
\(r = \frac{L}{2}=\frac{1.5\ m}{2}=0.75\ m\)
\(\tau = rF=(0.75\ m)(10\ N)=7.5\ N\cdot m\)

Step2: Calculate the rotational inertia

\(I=\frac{1}{3}mL^2=\frac{1}{3}(2.0\ kg)(1.5\ m)^2\)
\(I=\frac{1}{3}(2.0\ kg)(2.25\ m^2)=1.5\ kg\cdot m^2\)

Step3: Solve for angular acceleration

From \(\tau = I\alpha\), we get \(\alpha=\frac{\tau}{I}\)
Substitute \(\tau = 7.5\ N\cdot m\) and \(I = 1.5\ kg\cdot m^2\):
\(\alpha=\frac{7.5\ N\cdot m}{1.5\ kg\cdot m^2}=5\ rad/s^2\)

Answer:

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