Energy Absorbed by a Brake

Energy Absorbed by a Brake

The energy absorbed by a brake depends upon the type of motion of the moving body. The motion of a body may be either pure translation or pure rotation or a combination of both translation and rotation. The energy corresponding to these motions is kinetic energy. Let us consider these motions as follows :

1.     When the motion of the body is pure translation. Consider a body of mass (m) moving with a velocity v1 m / s. Let its velocity is reduced to v2 m / s by applying the brake. Therefore, the change in kinetic energy of the translating body or kinetic energy of translation,

E1 = (1/2) m [(v1)2 - (v2)2]

This energy must be absorbed by the brake. If the moving body is stopped after applying the brakes, then v2 = 0, and

E1 = (1/2) m (v1)2

2. When the motion of the body is pure rotation. Consider a body of mass moment of inertia I (about a given axis) is rotating about that axis with an angular velocity ω1 rad / s. Let its angular velocity is reduced to ω2 rad / s after applying the brake. Therefore, the change in

kinetic energy of the rotating body or kinetic energy of rotation,

E2 = (1/2) I [(ω1)2 - (ω2)2]

This energy must be absorbed by the brake. If the rotating body is stopped after applying the brakes, then ω2 = 0, and

E1 = (1/2) I (ω1)2

3. When the motion of the body is a combination of translation and rotation. Consider a body having both linear and angular motions, e.g. in the locomotive driving wheels and wheels of a moving car. In such cases, the total kinetic energy of the body is equal to the sum ∴of the kinetic energies of translation and rotation.

Total kinetic energy to be absorbed by the brake,

E = E1 + E2

Sometimes, the brake has to absorb the potential energy given up by objects being lowered by hoists, elevators etc. Consider a body of mass m is being lowered from a height h1 to h2 by applying the brake. Therefore the change in potential energy,

E3 = m.g (h1 – h2)