Question

A uniform-density 6 kg disk of radius 0.3 m is mounted on a nearly frictionless axle. Initially it is not spinning. A string is wrapped tightly around the disk, and you pull on the string with a constant force of 25 N through a distance of 0.6 m. Now what is the angular speed?

Short answer

The angular speed is, 10.54 rad/s .

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The mass of disk is, m=6 kg.
• The radius of disk is, r =0.3 m .
• The pulling force is, F =25 N .
• The string distance, s=0.6 m .

Significance of angular speed

The angular speed measures the rate at which the central angle of a rotating body changes with time.

Determination of the angular speed

The relation of final angular speed with respect to initial angular speed and angular acceleration, it is expressed as,

${\omega }_f^2={\omega }_{\mathrm{i}}^2+2\alpha \theta$ ...(i)

Here ${\omega }_f$is the final angular speed, $\omega$is the initial angular speed and $\alpha$is the angular acceleration.

The initial angular speed is zero i.e ${\omega }_i=0$then the equation (i) written as,

${\omega }_f^2=2\alpha \theta$ ...(ii)

The relation of torque is expressed as,

$\tau =I\alpha =Fr$ ...(iii)

Here I is the moment of inertia and $\tau$is the torque.

The moment of inertia for disk is expressed as,

role="math" localid="1657800893463" $I=\frac{1}{2}m{r}^2$

Substitute 6 kg for m and 0.3 m for r in the above equation.

$I=\frac{1}{2}m{r}^2$

The value of Isubstitute in equation (iii)

$$\begin{gathered} \frac{1}{2}m{r}^2\times \alpha =Fr \\ \alpha =\frac{2F}{mr} \end{gathered}$$

Substitute 6 kg for m , 25 N for F and 0.3 m for r in the above equation.

$$\begin{gathered} \alpha =\frac{2\times 25\mathrm{N}}{6\mathrm{kg}\times 0.3\mathrm{m}} \\ =27.78\mathrm{rad}/{\mathrm{s}}^2 \end{gathered}$$

Here the angular displacement is expressed as,

$\theta =\frac{s}{r}$

Substitute 0.6 m for s , and 0.3 m for r in the above equation.

$$\begin{gathered} \mathrm{\theta }=\frac{0.6\mathrm{m}}{0.3\mathrm{m}} \\ =2\mathrm{rad} \end{gathered}$$

Substitute 2 rad for $\theta$, and 27.78 $\mathrm{rad}/{\mathrm{s}}^2$for $\alpha$in the equation (ii)

$$\begin{gathered} {\omega }_f^2=2\times 27.78\mathrm{rad}/{\mathrm{s}}^2\times 2\mathrm{rad} \\ {\mathrm{\omega }}_{\mathrm{f}}=10.54\mathrm{rad}/\mathrm{s} \end{gathered}$$

Hence the angular speed is, $10.54\mathrm{rad}/\mathrm{s}$.