A flywheel in the form of a heavy circular disk of diameter 0.74 mand mass 103 k

Question

A flywheel in the form of a heavy circular disk of diameter 0.74 mand mass 103 kg is mounted on a frictionless bearing. A motorconnected to the flywheel accelerates it from rest to 897rev/min.
a) What is the moment of inertia of the flywheel?
b) How much work is done on it during this acceleration?
c) After 897 rev/min is achieved, the motor is disengaged. Afriction brake is used to slow the rotational rate to 470 rev/min.What is the magnitude of the energy dissipated as heat from thefriction brake?

Explanation / Answer

                Given that the mass of wheel is M = 103 kg                 Diameter of wheel is D = 0.74 m                 Radius of wheel si R = D/2 = 0.37 m                  Intiialangular velocity is 1 = 0                 final angular velocity is 2 = 897 rev/min =897*2 / 60 rad/s     -----------------------------------------------------------------------------------------           Themoment of inertia of disk shape wheel is I = MR2/2                    The work doneW = change in rotational kinetic energy                                      = (1/2)I22 -(1/2)I12                                            = (1/2)I22 - 0                                      = ------------ J        if the wheel slows to theangular velocity 3 = 470 rev/min = 470*2/60 rad/s           The loss of energy due to friction is                   K = (1/2)I32 -(1/2)I22                              = --------- J        

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.