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…

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}). Moment of inertia for a rod pivoted at one end is (I=\frac{1}{3}mL^2). Also, (\tau = I\alpha) (rotational analog of (F = ma)), so (\alpha=\frac{\tau}{I}).

Step2: Calculate Torque

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

Step3: Calculate Moment of Inertia

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

Step4: Calculate Angular Acceleration

(\alpha=\frac{\tau}{I}=\frac{7.5}{1.5}=5\space rad/s^2).

Answer:

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