Question

A light, rigid rod is 59.7 cm long. Its top end is pivoted on a frictionless horizontal axle. The rod hangs straight down at rest with a small, massive ball attached to its bottom end. You strike the ball, suddenly giving it a horizontal velocity so that it swings around in a full circle. What minimum speed at the bottom is required to make the ball go over the top of the circle?

Answer 1

Answer:

minimum speed at the bottom, $v_{b(min)}$ = 4.84 m/s

Given:

Length of the rod, l = 59.7 cm =0.597 m

Solution:

According to the conditions given in the question, the gravitation force is the only conservative force acting.

Now by the law of conservation of energy for the rod:

total energy at the bottom end = total energy at the top end

Since, total energy is the sum of Kinetic Energy (KE) and Potential energy (PE)

and we know that PE is maximum at the top and minimum or zero at the bottom whereas KE is min at the top or zero and max at the bottom as:

KE = $\frac{1}{2}m\times v^{2}$

PE = mgl

As the rod goes over ther top of the circle, the length at the top becomes twice and hence

$PE_{top} mg\times 2l$ = 2mgl

Therefore,

$KE_{bottom} + PE_{bottom} = KE_{top} + PE_{top}$

$\frac{1}{2}m\times v_{b(min)}^{2} + 0 = 0 + 2mgl$

⇒ $v^{2} = 4gl$

⇒ $v = \sqrt 4gl$

⇒ $v = \sqrt 4\times 9.8\times 0.597 = 4.84 m/s$

Therefore, minimum speed at the bottom is $v_{b(min)}$ = 4.84 m/s