A small ball weighing 200g attached to the end of a 20cm long thread rotates in a vertical

A small ball weighing 200g attached to the end of a 20cm long thread rotates in a vertical plane at the lowest point of the circle, the speed of the ball is 3ms, which is equal to the tension force of the thread at this moment.

m = 200 g = 0.2 kg.

g = 10 m / s2.

R = 20 cm = 0.2 m.

V = 3 m / s.

N -?

When passing the lower point of the trajectory, 2 forces act on the ball: gravity Ft directed vertically downward, thread tension force N directed vertically upward.

m * a = m * g + N – 2 Newton’s law in vector form.

For projections onto the vertical axis 2, Newton’s law will take the form: m * a = – Ft + N.

N = m * a + Fт.

The force of gravity Ft is determined by the formula: Ft = m * g.

N = m * a + m * g.

The centripetal acceleration a is expressed by the formula: a = V ^ 2 / R.

N = m * V ^ 2 / R + m * g = m * (V ^ 2 / R + g).

N = 0.2 kg * ((3 m / s) ^ 2 / 0.2 m + 10 m / s2) = 11 N.

Answer: at the lower point of the trajectory, the tension force of the thread is N = 11 N.