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 7.00 Nm is applied to the tire for 2.00 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 125 s. Assume the direction the wheel is rotating is the positive direction.

Part A

Compute the moment of inertia of the wheel about the rotation axis.

Part B

Compute the friction torque.

Part C

Compute the total number of revolutions made by the wheel in the 125-s time interval.

Explanation / Answer

= 100rev/min * 2rad/rev * 1min/60s = 10.5 rad/s
= / t = 10.5rad/s / 2.00s = 5.24 rad/s²

A)

Torque T = I*alpha

I is the moment of inertia and alpha is the angular accelaration

I = ?

alpha = (wf-wi)/t

wf = 100 rpm = 100*2*3.142/60 = 10.5 rad/s
wi = 0 rad/s
t = 2 S

alpha = (10.5)/2 = 5.25 rad/s^2

T = I*alpha

moment of inertia is I = T/alpha = (7)/5.25 = 1.34 kg m^2

B) Now = 1.34kg·m² * 10.5rad/s / 125s = 0.112 N·m

C) = *t - ½t² = 10.5rad/s*125s - ½(10.5rad/s/125s)(125s)²
= 656 rads = 104 revs

Hope this helps!!

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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