A 2.0 kg bike wheel, which can be taken as a hoop witha radius of 0.40 m and a 0

Question

A 2.0 kg bike wheel, which can be taken as a hoop witha radius of 0.40 m and a 0.50 kg disk with a radius of 0.15 m areattached to a common axle and are initially rotating together at aconstant angular velocity of 2.0 rev/s. A brake is applied tothe outer edge of the disk such that the system comes to rest in2.5s. (a) If it decelerates uniformly, what is the angularacceleration of the system during this 2.5 s? (b) What is themoment of inertia of the system? (c) What is the torque on thesystem? (d) How much frictional force is applied in order to stopthe wheels? (Assume the axle and the bike tire spokes havenegligible mass.) A 2.0 kg bike wheel, which can be taken as a hoop witha radius of 0.40 m and a 0.50 kg disk with a radius of 0.15 m areattached to a common axle and are initially rotating together at aconstant angular velocity of 2.0 rev/s. A brake is applied tothe outer edge of the disk such that the system comes to rest in2.5s. (a) If it decelerates uniformly, what is the angularacceleration of the system during this 2.5 s? (b) What is themoment of inertia of the system? (c) What is the torque on thesystem? (d) How much frictional force is applied in order to stopthe wheels? (Assume the axle and the bike tire spokes havenegligible mass.)

Explanation / Answer

Mass of wheel m = 2 kg radius of wheel r = 0.4 m Moment of inertia of wheel I = m r ^ 2                                            = 0.32 kg m ^ 2 mass of disk m ' = 0.5 kg radius of disk r ' = 0.15 m moment of inetia of disk I ' = ( 1/ 2) m ' r ' ^ 2                                         = 0.005625 kg m ^ 2 (a). Initial angular speed w = 2 rev / s= 2.5 * 2rad / s                                         = 15.707 rad / s final angular speed w ' = 0 time t= 2.5 s angular accleration = ( w ' - w ) / t                                = -6.283 rad / s ^ 2 (b). Moment of inertia of the system I " = I + I '                                                            = 0.325625 kg m ^2 (c). torque of the system = I "                                         = 2.045 N m (d). frictional force F = / r                                 = 5.114 N

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