A ball weighing 2 kg is mounted on a rod 81 cm long, which rotates in a vertical plane around

A ball weighing 2 kg is mounted on a rod 81 cm long, which rotates in a vertical plane around the other end at a frequency of 30 rpm. Find the force acting from the rod on the ball at the bottom point of the trajectory.

m = 2 kg.

g = 9.8 m / s ^ 2.

L = 81 cm = 0.81 m.

v = 30 rpm = 0.5 rpm = 0.5 Hz.

N -?

At the lowest point of the trajectory, two forces act on the ball:
– the force of gravity m * g directed vertically downward:

is the elastic force of the rod N directed vertically upward.

When moving along a circle, the acceleration of the body a directed to the center of the circle.

At the bottom point of the circle, the acceleration of the body a, will be directed vertically upward.

We write 2 Newton’s law in projections and express the elastic force of the rod N.
m * a = N – m * g;

N = m * a + m * g;

N = m * (a + g).

The centripetal acceleration a, is determined by the formula: a = V ^ 2 / L, where V is the speed of movement, L is the radius of the circle, in our case the length of the rod.
Acceleration is determined by the formula: a = 4 * P ^ 2 * v ^ 2 * L, where P is the number of Pi, v is the rotational speed.
a = 4 * 3.14 ^ 2 * 0.5 Hz ^ 2 * 0.81 m = 7.98 m / s ^ 2.

The elastic force N of the rod will be determined by the formula: N = 2 kg * (7.98 m / s ^ 2 + 9.8 m / s ^ 2) = 35.56 N.
Answer: a force N = 35.56 N. acts on the ball from the side of the rod.