An acrobat on a motorcycle describes a “loop” with a radius of 9.8 m. At what slowest speed should

An acrobat on a motorcycle describes a “loop” with a radius of 9.8 m. At what slowest speed should the acrobat pass the top of the loop to avoid falling off?

R = 9.8 m.

g = 9.8 m / s2.

Vmin -?

Let’s write Newton’s 2 law when the acrobat passes the top point of the “dead loop”: m * a = m * g + N, where m is the mass of the acrobat with a motorcycle, a is the centripetal acceleration, m * g is the force of gravity, N is the force with which the surface acts on the acrobat.

At the moment of separation, the surface reaction force will be equal to 0: N = 0.

m * a = m * g – 2 Newton’s law at the separation of the acrobat at the top point.

Centripetal acceleration a is determined by the formula: a = Vmin2 / R.

Vmin2 / R = g.

Vmin = √ (g * R).

Vmin = √ (9.8 m / s2 * 9.8 m) = 9.8 m / s.

Answer: the minimum travel speed of the acrobat’s top point is Vmin = 9.8 m / s.