Question

A uniform flag pole of length $L$ and mass $M$ is pivoted on the ground with a frictionless hinge. The flag pole makes an angle $\theta$ with the horizontal. The moment of inertia of the flag pole about one end is $\left(1/3\right)M{L}^2$. If it starts falling from the position shown in the accompanying figure, the linear acceleration of the free end of the flag pole - labeled P - would be:

• A. $\left(2/3\right)g\cos \theta$
• B. $\left(2/3\right)g$
• C. $g$
• D. $\left(3/2\right)g\cos \theta$

Answer 1

The correct option is D $\left(3/2\right)g\cos \theta$

Torque at point O,

$mg\cos \theta .\frac{L}{2}=I\alpha =\frac{1}{3}m{L}^2\alpha$

$\alpha =\frac{3g\cos \theta }{2L}$

Acceleration at point P, whose distance from O is L

$a_p=L\alpha =\frac{3g\cos \theta }{2}$

Hence, acceleration at point P is $\frac{3g\cos \theta }{2}$