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: Calculate torque from force

Torque $\tau = F \cdot r$, where $r = L/2 = 0.75\,\text{m}$.
$\tau = 10\,\text{N} \cdot 0.75\,\text{m} = 7.5\,\text{N·m}$.

Step2: Find moment of inertia of rod

Moment of inertia for rod pivoted at end: $I = \frac{1}{3}mL^2$.
$I = \frac{1}{3} \cdot 2.0\,\text{kg} \cdot (1.5\,\text{m})^2 = \frac{1}{3} \cdot 2 \cdot 2.25 = 1.5\,\text{kg·m}^2$.

Step3: Compute angular acceleration

Using $\tau = I\alpha$, so $\alpha = \tau/I$.
$\alpha = \frac{7.5}{1.5} = 5\,\text{rad/s}^2$.

Answer:

$5\,\text{rad/s}^2$