Question

(part 1 of 3) A flywheel in the form of a heavy circular disk of diameter 0.504 m and mass 213 kg is mounted on a frictionless bearing. A motor connected to the flywheel accelerates it from rest to 1010 rev/min. What is the moment of inertia of the flywheel? 6.76318

(part 2 of 3) How much work is done on it during this acceleration? 37828.7 J

(paer 3 of 3) After 1010 rev/min is achieved, the motor is disengaged. A friction brake is used to slow the rotational rate to 616 rev/min. What is the magnitude of the energy dissipated as heat from the friction brake? Answer in units of J

I've found part 1 and 2 but can't get 3. Please help me!

Answer 1

Solution:

The momentum of inertial is

$I = 0.5 \times MR^2$

where R = radius of disk. subustitute the given values

$I = 0.5 \times 213 \times (0.504/2)^2$

$I = 6.76$ kg m^2

b)

work done

the work done is calculated from the work-energy theorem

W$W = \Delta KE = KE_f - KE_i$

KE_f =final KE

KE_i = initial KE

disk starts from rest. so the initial KE is Zero

the final KE is

$KE_f = 0.5 \times I \omega ^2$

$\omega = 1010 \times 2\pi/60$

$\omega = 105.713$

$KE_f = 0.5 \times 6.76 \times( 105.713)^2$

$KE_f = 37772.30 J$

$W = 37772.30 J$

c)

The energy lost:

the initial KE is 37772.30

the final KE is

$\omega = 616 \times 2\pi/60$

$\omega = 64.474$

$KE_f = 0.5 \times 6.76 \times( 64.474)^2$

$KE_f = 14050.310J$

the energy lost is

$E = 14050.310 - 37772.30 J$

$E = -23721.99 J$

the negative sign indicates the enrgy is dissipated