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 (\tau = rF). The moment of inertia (I) for a rod pivoted at one end is (I=\frac{1}{3}mL^{2}). Angular acceleration (\alpha=\frac{\tau}{I}).

Step2: Calculate Torque

The force is applied at the midpoint, so (r = \frac{L}{2}). Given (F = 10\ N), (L = 1.5\ m), (\tau=\frac{L}{2}\times F=\frac{1.5}{2}\times10 = 7.5\ N\cdot m).

Step3: Calculate Moment of Inertia

Given (m = 2.0\ kg), (L = 1.5\ m), (I=\frac{1}{3}\times2.0\times(1.5)^{2}=\frac{1}{3}\times2.0\times2.25 = 1.5\ kg\cdot m^{2}).

Step4: Calculate Angular Acceleration

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

Answer:

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