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: Find the moment of inertia of the rod

For a rod pivoted at one end, the moment of inertia \( I \) is given by the formula \( I=\frac{1}{3}mL^{2} \).
Substituting \( m = 2.0\space kg \) and \( L=1.5\space m \):
\( I=\frac{1}{3}\times2.0\times(1.5)^{2}=\frac{1}{3}\times2.0\times2.25 = 1.5\space kg\cdot m^{2} \)

Step2: Calculate the torque

The torque \( \tau \) is given by \( \tau = rF\sin\theta \). Here, the force is applied at the midpoint, so \( r=\frac{L}{2}=\frac{1.5}{2} = 0.75\space m \), \( F = 10\space N \) and \( \theta = 90^{\circ} \) (since the force is perpendicular to the rod, \( \sin\theta=1 \)).
So, \( \tau=0.75\times10\times1 = 7.5\space N\cdot m \)

Step3: Find the angular acceleration

Using the formula \( \tau=I\alpha \), we can solve for angular acceleration \( \alpha \):
\( \alpha=\frac{\tau}{I} \)
Substituting \( \tau = 7.5\space N\cdot m \) 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 \( \boldsymbol{5.0\space rad/s^{2}} \)