Question

A pendulum of mass m and length l is released from rest in a horizontal position. A nail at a distance d below the pivot causes the mass to move along the path indicated by the doted line. Find the minimum distance d in terms of l such that the mass will swing completely round in the circle shown in figure –

• A. l / 5
• B. 3l / 5
• C. 2l / 5
• D. l / 2

Answer 1

Answer: (B)

3l / 5

Answer 2

Take the mass m as a point mass. At the instant when the pendulum collides with the nail, m has a velocity v = $\sqrt{\mathbf{2g}\mathcal{l}}$. The angular momentum of the mass with respect to the point at which the nail locates is conserved during the collision. Then the velocity of the mass is still v at the instant after the collision and the motion thereafter is such that the mass is constrained to rotate around the nail. Under the critical condition that the mass can just swing completely round in a circle, the gravitational force when the mass is at the top of the circle. Let the velocity of the mass at this instant be v₁, and we have

$\frac{mv_{1}^{2}}{\mathcal{l -}d}$ = mg,

or v₁² = (l – d)g

The energy equation

$\frac{mv^{2}}{2}$ = $\frac{mv_{1}^{2}}{2} + 2mg(\mathcal{l -}d)$,

or 2gl = (l – d)g + 4(l – d)g

then gives the minimum distance as

d = $\frac{\mathbf{3}\mathcal{l}}{\mathbf{5}}$