18.10 Work - kinetic energy theorem for rotational motion  (Page 3/3)

Problem : A uniform disk of mass 5 kg and radius 0.2 m is mounted on a horizontal axle to rotate freely. One end of a rope, attached to the disk, is wrapped around the rim of the disk. If a force of 10 N is applied vertically as shown, then find the work done by the force in rotating the disk for 2 s, starting from rest at t = 0 s.

Rotation

Solution : Here, torque due to vertical force is constant. Thus, in order to determine the work on the disk in rotation, we need to use the expression of work done by a constant torque as :

$$\begin{array}{l}W=\tau \Delta \theta =\tau ({\theta }_2-{\theta }_1)\end{array}$$

It means that we need to calculate torque and angular displacement. Here, torque (positive as rotation is anti-clockwise) is given by :

$$\begin{array}{l}\tau =F_{t}r=10x0.2=2\mathrm{N-m}\end{array}$$

This is a rotational motion with constant acceleration (due to constant torque). Thus, we can use the equation of motion for constant angular acceleration to find the displacement :

$$\begin{array}{l}{\theta }_f={\theta }_{i}t+\frac{1}{2}\alpha t^2\end{array}$$

Here, disk is at rest in the beginning, θi = 0. Hence,

$$\begin{array}{l}\Rightarrow {\theta }_f=\frac{1}{2}\alpha t^2\end{array}$$

But, we do not know the angular acceleration. We can, however, use corresponding Newton's second law for rotation. Since, disk rotates freely, we can assume that there is no other torque involved :

$$\begin{array}{l}\Rightarrow \alpha =\frac{\tau }{I}\end{array}$$

On the other hand, the moment of inertia for the disk about an axis passing through the center of mass and perpendicular to its surface is given by :

$$\begin{array}{l}I=\frac{M{R}^2}{2}=\frac{5x{0.2}^2}{2}=0.1\mathrm{kg}-m^2\end{array}$$

Putting this value of moment of inertia, we get the value of angular acceleration :

$$\begin{array}{l}\Rightarrow \alpha =\frac{\tau }{I}=\frac{2}{0.1}=20\frac{\mathrm{rad}}{s^2}\end{array}$$

Now final angular position is :

$$\begin{array}{l}{\theta }_f=\frac{1}{2}x20x{2}^2=40\mathrm{rad}\end{array}$$

Hence, work done is :

$$\begin{array}{l}W=\tau ({\theta }_2-{\theta }_1)=\tau {\theta }_2=2x40=80J\end{array}$$

Problem : A uniform disk of mass 5 kg and radius 0.2 m is mounted on a horizontal axle to rotate freely. One end of a rope, attached to the disk, is wrapped around the rim of the disk. If a force of 10 N is applied vertically as shown, then find the work done by the force in rotating the disk for 2 s, starting from rest at t = 0 s. Use work - kinetic energy theorem to find the result.

Rotation

Solution : This is the same question as the previous one except that we are required to solve the question, using work-kinetic energy theorem :

$$\begin{array}{l}W=\Delta K=\frac{1}{2}I{{\omega }_f}^2-\frac{1}{2}I{{\omega }_i}^2\end{array}$$

From the previous example, we have :

$$\begin{array}{l}I=\frac{M{R}^2}{2}=\frac{5x{0.2}^2}{2}=0.1\mathrm{kg}-m^2\end{array}$$

and

$$\begin{array}{l}\alpha =\frac{\tau }{I}=20\frac{\mathrm{rad}}{s^2}\end{array}$$

Now, as the disk starts from rest, ${\omega }_i=0$ . In order to find the final angular speed, we use equation of motion for constant angular acceleration :

$$\begin{array}{l}{\omega }_f={\omega }_i+\alpha t=0=2x20=40\mathrm{rad}/s\end{array}$$

Thus,

$$\begin{array}{l}W=\Delta K=\frac{1}{2}I{{\omega }_f}^2-0=\frac{1}{2}I{{\omega }_f}^2\\ \Rightarrow W=\frac{1}{2}x0.1x{40}^2=80J\end{array}$$

As expected, the result is same as calculated in earlier example.