A uniform-density wheel of mass and radius rotates on a low-friction axle. Starting from rest, a string wrapped around the edge exerts a constant force of $\mathbf{15}\mathit{N}$for$\mathbf{0}\mathbf{.}\mathbf{6}\mathit{s}$ (a) what is the final angular speed? (b) what is the average angular speed? (c) Through how big an angle did the wheel turn? (d) How much string come off the wheel?

Angular speed is the rate at which the central angle of a spinning body varies over time.

The rate of change of angular displacement is known as angular speed.

Mass of wheel,$m=6kg$

Radius of the wheel,$r=0.3m$

As the wheel starting from rest, initial angular speed of the wheel,${\omega }_i=0rad/s$

Applied force at the edge of the wheel,$F=15N$

Total time that force applied at the edge of the wheel,$t=0.6s$

Torque acting on the system related to initial angular momentum and final angular momentum as

${\overrightarrow{\tau }}_{net}=\frac{L_f-L_i}{\Delta t}$

Here,

$L_f=I{\omega }_f\to$Final angular momentum of the wheel

$L_i=I{\omega }_i\to$Initial angular momentum of the wheel

$\Delta t\to$Time taken

${\overrightarrow{\tau }}_{net}\to$Net torque

Net torque acting on the wheel${\overrightarrow{\tau }}_{net}\to$related to force acting at the edge$\left(F\right)$ and perpendicular distance between the axis of rotation and force as

${\overrightarrow{\tau }}_{net}=Fr$

From the above two relations we get

$Fr=\frac{I{\omega }_f-I{\omega }_i}{\Delta t}$

Isolating the final angular velocity of the wheel from above equation we get

${\omega }_f=\frac{Fr-\Delta t}{\Delta t}+{\omega }_i$

(a)Moment of inertia of the uniform density wheel is

$I=\frac{m{r}^2}{2}$

Now the final angular velocity of the wheel is

$\begin{array}{rcl}{\omega }_f& =& \frac{2Fr\Delta t}{m{r}^2}+0\\ & =& \frac{2Fr\Delta t}{m{r}^2}\\ & =& \frac{2F\Delta t}{mr}\end{array}$ (Since${\omega }_i=0rad/s$ )

By substituting the given values in above equation, we get

$\begin{array}{rcl}{\omega }_i& =& \frac{2\left(15N\right)\left(0.6s\right)}{\left(6kg\right)\left(0.3m\right)}\\ & =& 10rad/s\end{array}$

Final angular speed of the wheel is$10rad/s$ .

(b)Initial angular speed,${\omega }_i=0$

Final angular speed,${\omega }_f=10rad/s$

Now the average angular speed of the wheel is

$\begin{array}{rcl}{\omega }_{avg}& =& \frac{{\omega }_i+{\omega }_f}{2}\\ & =& \frac{0+10rad/s}{2}\\ & =& 5rad/s\end{array}$

The average angular speed of the wheel is$5rad/s$ .

(c)Angular displacement of the wheel is

$\begin{array}{rcl}\Delta \theta & =& \left(\frac{{\omega }_i+{\omega }_f}{2}\right)\Delta t\\ & =& \left(5rad/s\right)\left(0.6s\right)\\ & =& 3.0rad\end{array}$

Angle turn by the wheel in $0.6s$is $3.0rad$

(d) The length of the string that comes off the wheel related to angular displacement of the wheel as

$\begin{array}{rcl}\Delta x& =& r\Delta \theta \\ & =& \left(0.3m\right)\left(3.0rad/s\right)\\ & =& 0.9m\end{array}$

The total length of the string that comes off the wheel is$0.9m$ .