Question

A steel rod 1 M long and having a mass of 2 kg is pinned at one of its ends. A force is applied perpendicularly to the rod at a distance of 75 cm from the pinned end. What is the magnitude of the resulting angular acceleration of the rod? Moment of inertia is 2 kg m^2

Answer 1

$$F=W=mg$$

Total moment is

$$M=-W(0.5L)+Fx=-0.5mgL+mgx$$

$$M=(2)(10)(0.75-0.5(1))=5\ Nm$$

The resulting angular acceleration of the rod:

$$\alpha=\frac{M}{I}=\frac{5}{2}=2.5\frac{rad}{s^2}$$