An experimental bicycle wheel is placed on a test stand so that it is free to tu

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 4.00 N m is applied to the tire for 4.00 s, the angular speed of the tire increases from 0 to 100 rev/min. The external torque is then removed, and the wheel is brought to rest by friction in its bearings in 100 s. (a) Compute the moment of inertia of the wheel about the rotation axis.
kg m2

(b) Compute the friction torque.

N m

(c) Compute the total number of revolutions made by the wheel in the 100-s time interval.

rev 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 4.00 N m is applied to the tire for 4.00 s, the angular speed of the tire increases from 0 to 100 rev/min. The external torque is then removed, and the wheel is brought to rest by friction in its bearings in 100 s. (a) Compute the moment of inertia of the wheel about the rotation axis.
kg m2

(b) Compute the friction torque.

N m

(c) Compute the total number of revolutions made by the wheel in the 100-s time interval.

rev 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 4.00 N m is applied to the tire for 4.00 s, the angular speed of the tire increases from 0 to 100 rev/min. The external torque is then removed, and the wheel is brought to rest by friction in its bearings in 100 s. (a) Compute the moment of inertia of the wheel about the rotation axis.
kg m2

(b) Compute the friction torque.

N m

(c) Compute the total number of revolutions made by the wheel in the 100-s time interval.

rev (a) Compute the moment of inertia of the wheel about the rotation axis.
kg m2

(b) Compute the friction torque.

N m

(c) Compute the total number of revolutions made by the wheel in the 100-s time interval.

rev

Explanation / Answer

a) w _ final = 100 rev/min = 100 x 2 x pi / 60 rad/s =10.47 rad/s
w_final = w_initial + alpha x t
10.47 = 0 + alpha x 4
alpha = 2.62 rad/s2
after that
alpha due to frcitional torque
0= 10.47 + alpha1 x 100
alpha1 = -0.1047 rad/s2
so frictional torque   = I x alpha1
net trque =   external torque + frictional torque = I x alpha
4 + I x -0.1047 = I x 2.62
I = 1.468 kg.m^2

b) frictional torque = 1.468 x 0.1047 = - 0.154 N-m

c)   theta = wf^2 - wi^2 / 2 xalpha1   = 0^2 - 10.47^2 / (2 x -0.1047) = 523.5 rad
rev = 523.5 / 2pi = 83.31 rev

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