A ?ywheel in the form of a heavy circular disk of diameter 0.264 m and mass 302

Question

A ?ywheel in the form of a heavy circular disk of diameter 0.264 m and mass 302 kg is mounted on a frictionless bearing. A motor

connected to the ?ywheel accelerates it from rest to 965 rev/min.

1. What is the moment of inertia of the flywheel? Answer in units of kg-m^2.

2. How much work is done on it during this acceleration? Answer in units of J

3. After 965 rev/min is achieved, the motor is disengaged. A friction brake is used to slow the rotational rate to 611 rev/min. What is the magnitude of the energy dissipated as heat from the friction brake? Answer in units of J.

Explanation / Answer

mass of the wheel m = 302 kg

radius =0.264/2 = 0.132 m

a)moment of inertai of the fly wheel I = mr^2 /2

I = 2.631 kg-m^2

b) work done W = k2-k1

W = 0.5 Iw^2 - 0

w = 965 rev/min = 965*2pi/60 rad/sec = 101 rad/s

W = 0.5 *2.631 x (101)^2 = 13420.30 J

c)

kinetic energy at w = 611 rev/min = 611*2pi/60 rad/sec = 63.98 rad/s

KE2 = 0.5 x 2.631 x 63.98^2 = 5384.92 J

the magnitude of the energy dissipated as heat from the friction brake = KE1-KE2 = 13420.30 - 5384.92 = 8035.37 J

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