The ball slides along an inclined groove that turns into a vertical loop with a radius of 1m.

The ball slides along an inclined groove that turns into a vertical loop with a radius of 1m. From what height should the ball begin to move in order not to come off at the top of the loop?

R = 1 m.

g = 9.8 m / s2.

h -?

At the top of the loop, when the ball is torn off, only the gravity force m * g will act on it. 2 Newton’s law will have the form: m * a = m * g.

a = g.

We express the centripetal acceleration a by the formula: a = V ^ 2 / R, where V is the speed of the ball at the top point, R is the radius of the loop.

V ^ 2 / R = g.

V ^ 2 = R * g.

When the ball moves, the law of conservation of energy is valid: m * g * h = m * V ^ 2/2 + m * g * R.

h = m * V ^ 2/2 * m * g + m * g * R / m * g = V ^ 2/2 * g + R = R / 2 + R.

h = 1 m / 2 + 1 m = 1.5 m.

Answer: the ball should start moving from a height of h = 1.5 m.