A body weighing 400 g is tied to a rod 0.5 m long, which turns in a vertical plane.

A body weighing 400 g is tied to a rod 0.5 m long, which turns in a vertical plane. The rod breaks at a pulling force of 9N. At what angular velocity will the rod break?

l = 0.5 m.

m = 400 g = 0.4 kg.

g = 10 m / s2.

F = 9 N.

w -?

When the load rotates in the vertical plane, the maximum load on the bar will be at the lowest point of the trajectory. The maximum force with which the load acts on the rod is expressed by the formula: F = m * (g + a), where m is the mass of the load, g is the acceleration of gravity, a is the acceleration with which the rod is rotated.

We express the centripetal acceleration a by the formula: a = V ^ 2 / l, where V is the rotation speed, l is the radius of the circle, in our case the length of the rod.

The linear velocity V is related to the angular w by the formula: V = w * l.

a = V ^ 2 / l = (w * l) 2 / l = w2 * l.

F = m * (g + w ^ 2 * l).

w ^ 2 * l = F / m – g.

w = √ (F / m * l – g / l).

w = √ (9 N / 0.4 kg * 0.5 m – 10 m / s2 / 0.5 m) = 5 rad / s.

Answer: the rod will break at an angular velocity w = 5 rad / s.