Question

A nail is fixed at a point P vertically below the point of suspension 'O' of a simple pendulum of length 1 m. The bob is released when the string of pendulum makes an angle $30^\circ$ with horizontal. The bob reaches lowest point and then describes vertical circle whose centre coincides with P. The least distance of P from O is

• A. 0.4 m
• B. 0.5 m
• C. 0.6 m
• D. 0.8 m

Answer 1

Answer: (D)

0.8 m

Answer 2

The problem involves two main parts: the swing of a pendulum and the subsequent circular motion around a nail.

1. Calculate the velocity of the bob at the lowest point (just before hitting the nail): Let the length of the pendulum be $L = 1$ m. The bob is released when the string makes an angle of $30^\circ$ with the horizontal. This means the angle the string makes with the vertical is $\theta = 90^\circ - 30^\circ = 60^\circ$. The vertical height dropped by the bob from its initial position to the lowest point is $h_{drop} = L - L\cos\theta = L(1 - \cos60^\circ)$. $h_{drop} = 1 \text{ m} (1 - 0.5) = 0.5 \text{ m}$.

Using the principle of conservation of mechanical energy, the potential energy at the initial position is converted into kinetic energy at the lowest point: $mgh_{drop} = \frac{1}{2}mv_B^2$ where $v_B$ is the velocity of the bob at the lowest point. $g(0.5) = \frac{1}{2}v_B^2$ $v_B^2 = g$ (since $L=1$, this is $gL$)

2. Apply the condition for completing a vertical circle: After hitting the nail at point P, the bob starts moving in a vertical circle with P as its center. Let the distance of the nail P from the suspension point O be $h$. The radius of this new circular path is $r = L - h$. For the bob to complete a vertical circle, the minimum velocity at the lowest point of the circle (which is $v_B$) must satisfy the condition: $v_B^2 = 5gr$

3. Equate the expressions for $v_B^2$ to find $r$: From step 1, $v_B^2 = gL$. From step 2, $v_B^2 = 5gr$. Equating these two expressions: $gL = 5gr$ $L = 5r$ $r = \frac{L}{5}$

Substitute the given value of $L = 1$ m: $r = \frac{1 \text{ m}}{5} = 0.2 \text{ m}$.

4. Calculate the distance of P from O: We know that $r = L - h$. Therefore, $h = L - r$. $h = 1 \text{ m} - 0.2 \text{ m} = 0.8 \text{ m}$.

The least distance of P from O is 0.8 m.