# A ball weighing 200 g rolls at a speed of 1.8 km / h along a chute in the shape of an arc of a circle with

**A ball weighing 200 g rolls at a speed of 1.8 km / h along a chute in the shape of an arc of a circle with a radius of 10 cm. Determine the weight of the ball at the lowest point of the chute and the speed required for the ball to experience triple overload at this point (g = 9.8m / s ^ 2)**

m = 200 g = 0.2 kg.

g = 9.8 m / s ^ 2.

V1 = 1.8 km / h = 0.5 m / s.

R = 10 cm = 0.1 m.

P2 = 3 * m * g.

P1 -?

V2 -?

Let’s write Newton’s 2 law for a ball in vector form: m * a = m * g + N.

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

The reaction force of the support on the ball will look like: N = m * a + m * g = m * (a + g).

According to Newton’s 3 law, with what force the ball acts on the support P, with this very force the support acts on the ball N: P = N.

P1 = m * (a1 + g).

We express the centripetal acceleration by the formula: a1 = V1 ^ 2 / R.

P1 = m * (V1 ^ 2 / R + g).

P1 = 0.2 kg * ((0.5 m / s) ^ 2 / 0.1 m + 9.8 m / s ^ 2) = 2.46 N.

P2 = m * (V2 ^ 2 / R + g).

3 * m * g = m * (V2 ^ 2 / R + g).

3 * g = V2 ^ 2 / R + g.

V2 ^ 2 / R = 3 * g – g.

V2 = √ (2 * g * R).

V2 = √ (2 * 9.8 m / s ^ 2 * 0.1 m) = 1.4 m / s.

Answer: P1 = 2.46 N, V2 = 1.4 m / s.