Question

A torque of 35.6 N · m is applied to an initially motionless wheel which rotates around a fixed axis. This torque is the result of a directed force combined with a friction force. As a result of the applied torque the angular speed of the wheel increases from 0 to 10.1 rad/s. After 5.90 s the directed force is removed, and the wheel comes to rest 59.0 s later. (a) What is the wheel's moment of inertia (in kg · m2)? kg · m2 (b) What is the magnitude of the torque caused by friction (in N · m)? N · m (c) From the time the directed force is initially applied, how many revolutions does the wheel go through? revolutions

Answer 1

(a) $20.8 kg m^2$

First of all, we can find the angular acceleration of the wheel when both the directed force and the friction force are acting on it:

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

where

$\omega_f = 10.1 rad/s$ is the final angular velocity

$\omega_i=0$ is the initial angular velocity

t = 5.90 s is the time taken

Substituting,

$\alpha=\frac{10.1 rad/s-0}{5.90 s}=1.71 rad/s^2$

Now we can find the moment of inertia of the wheel by using the equivalent of Newton's second law for rotational motions:

$\tau = I\alpha$ (1)

where

$\tau=35.6 Nm$ is the torque applied

I is the moment of inertia

$\alpha=1.71 rad/s^2$ is the angular acceleration

Solving the equation for I,

$I=\frac{\tau}{\alpha}=\frac{35.6 Nm}{1.71 rad/s^2}=20.8 kg m^2$

(b) -3.5 Nm

In the second part, the directed force is removed, and only the friction force acts on the wheel. The wheel comes to rest after t=59.0 s, so the angular acceleration in this part is

$\alpha=\frac{\omega_f-\omega_i}{t}=\frac{0-(10.1 rad/s)}{59.0 s}=-0.17 rad/s^2$

And it is negative since it is a deceleration. The moment of inertia of the wheel has not changed, so we can still use eq.(1) to find the torque caused by the friction only:

$\tau=I \alpha=(20.8 kg m^2)(-0.17 rad/s^2)=-3.5 Nm$

(c) 52.5 revolutions

The total angular displacement covered by the wheel in the first part of the motion is given by:

$\omega_f^2 - \omega_i^2 = 2 \alpha_1 \theta_1$

where

$\omega_f = 10.1 rad/s\\\omega_i = 0\\\alpha_1 = 1.71 rad/s^2$

Solving for $\theta_1$,

$\theta_1 = \frac{\omega_f^2-\omega_i^2}{2\alpha_1}=\frac{(10.1 rad/s)^2-0}{2(1.71 rad/s^2)}=29.8 rad$

The total angular displacement covered by the wheel in the second part of the motion is given by:

$\omega_f^2 - \omega_i^2 = 2 \alpha_2 \theta_2$

where

$\omega_f = 0 rad/s\\\omega_i = 10.1 rad/s\\\alpha_2 = -0.17 rad/s^2$

Solving for $\theta_2$,

$\theta_2 = \frac{\omega_f^2-\omega_i^2}{2\alpha_2}=\frac{0-(10.1 rad/s)^2}{2(-0.17 rad/s^2)}=300.0 rad$

So the total angular displacement in radians is

$\theta=\theta_1+\theta_2=29.8 rad+300.0 rad=329.8 rad$

And since $1 rev = 2\pi rad$

the angle convered in revolutions is

$\theta=\frac{329.8 rad}{2\pi rad/rev}=52.5 rev$