A flywheel in the form of a heavy circular disk of diameter 0.762 m and mass 126

Question

A flywheel in the form of a heavy circular
disk of diameter 0.762 m and mass 126 kg is
mounted on a frictionless bearing. A motor
connected to the flywheel accelerates it from
rest to 1160 rev/min.
What is the moment of inertia of the fly-
wheel?
Answer in units of kg · m2.
IVE ALREADY GOT 9.145 FOR THE 1ST ANSWER BUT I NEED HELP WITH THE LAST TWO

B) How much work is done on it during this
acceleration?
Answer in units of J.
C) After 1160 rev/min is achieved, the motor is
disengaged. A friction brake is used to slow
the rotational rate to 629 rev/min.
What is the magnitude of the energy dissi-
pated as heat from the friction brake?
Answer in units of J.

Explanation / Answer

a) moment of inertia = 9.145 kgm^2 b) Final angular velocity = wf = 1160 * 2pi/60 = 121.475 rad/sec final rotational energy = work done during acceleration = 1/2Iwf^2 =1/2*9.145*(121.475^2) = 67473.6 J c) Initial KE before applying breaks (answer b) = 67473.6 J After applying break, final KE = 1/2Iwf^2 final wf = 629 * 2pi/60 = 65.87 Solve final KE = 1/2 * 9.145*(65.87)^2 = 19839 J Magnitude of energy dissipated = 67473.6-19839 = 47634.6 J

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