Question

A light, rigid rod is 77.0cm 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

A minimum speed of 5.49 m/s of is required to make the ball go over the top of the circle.

Length of the rod = 77 cm = 0.77 m

Let the maximum height to which the ball can reach be h.

The kinetic energy of the ball is

$= \frac{1}{2} mv ^{2}$

The potential energy of the ball is

$= mgh$

The velocity of the ball when it is at the top end is 0.

So, the ball's energy from kinetic energy to potential energy is transferred.

Kinetic energy → Potential energy

$\frac{1}{2} mv ^{2} = mgh$

$h = \frac{1 \:mv ^{2}}{2 \: mgh}$

The maximum height to which the ball can reach is equal to twice the length of the rod.

h = 2 L

$h =2 \:L \frac{1 \:mv ^{2}}{2 \: mgh}$

The minimum speed at the bottom required to make the ball go over the top of the circle is,

$v = \sqrt{2g(2L)}$

$v = 4 \times 9.8 \times 0.77$

$v = 5.49 \: m/s$

Therefore, the minimum speed of 5.49 m/s at the bottom is required to make the ball go over the top of the circle.

To know more about potential energy, refer to the below link:

#SPJ4