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 N * m is applied to the tire for 2.00 s, the angular speed of the tire increases from zero to 120 rev/min. The external torque is then removed, and the wheel is brought to rest in 145 s by friction in its bearings.

Compute the moment of inertia of the wheel about the axis of rotation.
I = ? Units= kg * m^2

Compute the friction torque.
Units = N * m

Compute the total number of revolutions made by the wheel in the 145 s time interval.
N= ?

Answer 1

Answer:

$I=\frac{3}{\pi}\ kg.m^2$

$\tau_f=\frac{1}{725} \ N.m$

$n=17255\ rev$

Explanation:

Given:

Torque applied on the wheel, $\tau=6\ N.m$

time for which the torque is applied, $t=2\ s$

initial angular speed, $\omega_i=0\ rad.s^{-1}$

final angular speed, $\omega_f=120\ rev.min^{-1}=120\times \frac{2\pi}{60}=4\pi\ rad.s^{-1}$

time after which the wheel comes to rest due to friction, $t_r=145\ s$

As most mass of the tyre is on its periphery so we consider it as a ring:

Moment of inertia of the ring is given as,

$I=\frac{\tau}{\alpha}$ ....................(1)

where:

$\alpha=$ angular acceleration

given as:

$\alpha=\frac{\omega_f-\omega_i}{t}$

$\alpha=\frac{4\pi-0}{2}$

$\alpha=2\pi\ rad.s^{-2}$

Now

$I=\frac{6}{2\pi}$

$I=\frac{3}{\pi}\ kg.m^2$

The torque when the wheel comes to rest form the final velocity to zero:

The angular acceleration during the deceleration will be:

$\alpha_r=\frac{0-2\pi}{145}$

$\alpha_r=-\frac{2\pi}{145} \ rad.s^{-2}$ (negative sign denotes deceleration)

From eq. (1)

$\tau_f=I.\alpha_r$

$\tau_f=0.1\times \frac{2\pi}{145}$

$\tau_f=\frac{1}{725} \ N.m$

Now the no. of revolutions made in 145 seconds:

Using equation of motion:

$n=\omega_f.t+\alpha_r.t^2$ (keep values in revolutions)

$n=120\times 145-\frac{1}{145} \times 145^2$            (∵ $rad.s^{-2}\div (2\pi)=rev.s^{-2}$)

$n=17255\ rev$