An experimental bicycle wheel is placed on a test stand so that it is free to tu
ID: 2009866 • Letter: A
Question
An experimental bicycle wheel is placed on a test stand so that it is free to turn on its axle. If a constant net torque of 6.00 m*N is applied to the tire for 2.00 s, the angular speed of the tire increases from zero to 115 rev/min. The external torque is then removed, and the wheel is brought to rest in 125 s by friction in its bearings.
Compute the moment of inertia of the wheel about the axis of rotation.
I =
Compute the friction torque.
=
Compute the total number of revolutions made by the wheel in the 125 s time interval.
N =
Explanation / Answer
(a) Data: Friction torque, = ?. Time of application of friction torque, t = 125 s Initial angular speed, i = 0 rad/s Final angular speed, f = 115 rev/min = 115 * ( 2 / 60 ) rad/s = 12.04 rad/s Solution: Torque, = I = I * [ ( f - i) / t ] 6 = I * [ ( 12.04 - 0 ) / 2 ] I = 0.997 kg.m^2 Ans: Moment of inertia about the axis of rotation: I = 0.997 kg.m^2 (a) Data: Friction torque, = ?. Time of application of friction torque, t = 125 s Initial angular speed, i = 0 rad/s Final angular speed, f = 115 rev/min = 115 * ( 2 / 60 ) rad/s = 12.04 rad/s Solution: Torque, = I = I * [ ( f - i) / t ] 6 = I * [ ( 12.04 - 0 ) / 2 ] I = 0.997 kg.m^2 Ans: Moment of inertia about the axis of rotation: I = 0.997 kg.m^2 Data: Friction torque, = ?. Time of application of friction torque, t = 125 s Initial angular speed, i = 0 rad/s Final angular speed, f = 115 rev/min = 115 * ( 2 / 60 ) rad/s = 12.04 rad/s Solution: Torque, = I = I * [ ( f - i) / t ] 6 = I * [ ( 12.04 - 0 ) / 2 ] I = 0.997 kg.m^2 Ans: Moment of inertia about the axis of rotation: I = 0.997 kg.m^2 (b) Data: Torque, = ? Time of application of torque, t = 125 s Initial angular speed, i = 115 rev/min = 115 * ( 2 / 60 ) rad/s = 12.04 rad/s Final angular speed, f = 0 rad/s = 115 * ( 2 / 60 ) rad/s = 12.04 rad/s Solution: Friction torque, = I = I * [ ( f - i) / t ] = 0.997 * [ ( 0 -12.04 ) / 125 ] = - 0.096 N.m Ans: Friction torque, = - 0.096 N.m Note: Here '-' sign indicates that it produces deceleration (c) Angular displacement, = [ ( i +f ) / 2 ] * t = [ ( 12.04 +0 ) / 2 ] * 125 = 752.5 rad = (752.5 / 2) rev = 119.8 rev Ans: No. of revolutions, N = 119.8 revRelated Questions
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