6.8 Circular motion with constant acceleration  (Page 4/4)

The slope of the straight line is a negative constant. Therefore, options (a) and (b) are correct. The line crosses time axis, when angular velocity is zero. Thus option (c) is correct. The angular velocities have opposite sign across time axis. It means that the disk reverses its direction. Thus option (d) is correct.

Hence, options (a),(b),(c) and (d) are correct.

We can find the distance covered, if we know the angular displacement. On the other hand, we can find angular displacement if we know the average angular speed as time is given.

$$\begin{array}{l}{\omega }_{\text{avg}}=\frac{{\omega }_i+{\omega }_f}{2}=\frac{40+0}{2}=20\text{rad}/s^2\end{array}$$

We can consider this accelerated motion as uniform motion with average speed as calculated above. The angular displacement is :

$$\begin{array}{l}\Rightarrow \theta ={\omega }_{\text{avg}}Xt=20X05=100\text{rad}\end{array}$$

The distance covered is :

$$\begin{array}{l}\Rightarrow s=\theta Xr=100X0.1=10\text{m}\end{array}$$

We can use the equation of motion for constant acceleration. Here,

$$\begin{array}{l}\theta =2\text{revolution}=2X\mathrm{2\pi }=\mathrm{4\pi }\end{array}$$

$$\begin{array}{l}{\omega }_i=0\end{array}$$

$$\begin{array}{l}{\omega }_f=8\text{rad/s}\end{array}$$

Looking at the data, it is easy to find that the following equation will serve the purpose,

$$\begin{array}{l}{{\omega }_f}^2={{\omega }_i}^2=2\mathrm{\alpha \theta }\end{array}$$

Solving for "α" and putting values,

$$\begin{array}{l}\Rightarrow \alpha =\frac{{{\omega }_f}^2-{{\omega }_i}^2}{2\theta }=\frac{8^2-0^2}{2X\mathrm{4\pi }}=\frac{8}{\mathrm{\pi }}\text{rad}/s_2\end{array}$$

Hence, option (c) is correct.

Here, final angular velocity, angular acceleration and time of motion are given. We can find the angular displacement using equation of motion for angular displacement that involves final angular velocity :

$$\begin{array}{l}\theta ={\omega }_{f}t-\frac{1}{2}\alpha \frac{t}{2}\end{array}$$

Putting values, we have :

$$\begin{array}{l}\Rightarrow \theta =10X2-\frac{1}{2}1\frac{2}{2}=20-2=18\text{rad}\end{array}$$

Hence, option (a) is correct.