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 (1/3)ML². 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) immediately after it starts falling off would be :

• A. (2/3) gcos$\theta$
• B. (2/3) g
• C. g
• D. (3/2) gcos$\theta$

Answer 1

Answer: (D)

(3/2) gcos$\theta$

Answer 2

The problem involves the rotational motion of a uniform flag pole pivoted at one end. We need to find the linear acceleration of the free end immediately after it starts falling.

1. Identify the forces and torque: The flag pole has mass $M$ and length $L$. It is pivoted at one end (let's call it the origin). The only force producing torque about the pivot is gravity, acting at the center of mass of the pole. For a uniform pole, the center of mass is at $L/2$ from the pivot. The flag pole makes an angle $\theta$ with the horizontal. The vertical component of gravity acts downwards. The perpendicular distance from the pivot to the line of action of gravity is $(L/2) \cos\theta$.

   The torque ($\tau$) due to gravity about the pivot is: $\tau = Mg \times (L/2) \cos\theta$

2. Identify the moment of inertia: The problem states that the moment of inertia of the flag pole about one end is $I = (1/3)ML^2$. This is the moment of inertia about the axis of rotation (the pivot).

3. Apply Newton's second law for rotation: The rotational equivalent of Newton's second law is $\tau = I \alpha$, where $\alpha$ is the angular acceleration. Substituting the expressions for $\tau$ and $I$: $Mg (L/2) \cos\theta = (1/3)ML^2 \alpha$

4. Calculate the angular acceleration ($\alpha$): We can cancel $M$ and one $L$ from both sides: $g (1/2) \cos\theta = (1/3)L \alpha$ Now, solve for $\alpha$: $\alpha = \frac{g (1/2) \cos\theta}{(1/3)L}$ $\alpha = \frac{3g \cos\theta}{2L}$

5. Calculate the linear acceleration of the free end (P): The linear acceleration ($a$) of a point at a distance $r$ from the axis of rotation is given by $a = r \alpha$. The free end P is at a distance $L$ from the pivot. So, the linear acceleration of point P ($a_P$) is: $a_P = L \alpha$ Substitute the value of $\alpha$: $a_P = L \left( \frac{3g \cos\theta}{2L} \right)$ $a_P = \frac{3}{2} g \cos\theta$

The linear acceleration of the free end of the flag pole immediately after it starts falling off is $(3/2) g \cos\theta$.

Explanation of the solution:

1. Calculate torque ($\tau$) due to gravity about the pivot: $\tau = Mg(L/2)\cos\theta$.
2. Use the given moment of inertia ($I = (1/3)ML^2$) about the pivot.
3. Apply rotational second law: $\tau = I\alpha$.
4. Solve for angular acceleration ($\alpha$): $\alpha = \frac{Mg(L/2)\cos\theta}{(1/3)ML^2} = \frac{3g\cos\theta}{2L}$.
5. Calculate linear acceleration of the free end P (at distance L from pivot): $a_P = L\alpha = L \left(\frac{3g\cos\theta}{2L}\right) = \frac{3}{2}g\cos\theta$.