Question

38. II FIGURE EX4.38 shows the angular-velocity-versus-time graph for a particle moving in a circle. How many revolutions does the object make during the first 4 s?

Short answer

The number of revolution made by a particle moving in a circle during the first 4s is 9.55 rev.

Step-by-step solution

Step 1. Given information

From the graph up to t =2s the particle is having angular acceleration and from time t = 4s the particle is having constant angular velocity.

Then for $t=2\text{s}$

${\alpha }_2=\frac{d\omega }{dt}=\frac{20\text{rad}/\text{s}-0}{2\text{s}}=10\text{rad}/{\text{s}}^2$

Step 2. Explanation

Initial angular velocity is

${\omega }_0=0\text{rad}/\text{s}$

$\text{Then}\Delta {\theta }_2={\omega }_{0}t+\frac{1}{2}{\alpha }_2{t}^2=0+\frac{1}{2}\left(10\text{rad}/{\text{s}}^2\right){(2\text{s})}^2=20\text{rad}$

Now from $t=2\text{s}$to $t=4\text{s},\omega =20\text{rad}/\text{s}$

Then

role="math" localid="1652331307170" $\begin{array}{l}\Delta {\theta }_4=\omega \Delta t\\ =(20\text{rad}/\text{s})2\text{s}\\ =40\text{rad}\end{array}$

Then total angular displacement is

$\begin{array}{l}\Delta \theta =\Delta {\theta }_L+\Delta {\theta }_4\\ =20\text{rad}+40\text{rad}\\ =60\text{rad}\end{array}$

Now we know that$\Delta \theta =2\pi N$

Thus number of revolutions is

$\begin{array}{l}N=\frac{\Delta \theta }{2\pi }\\ =\frac{60\text{rad}}{2\pi \text{rad}/\text{rev}}\\ =9.55\text{rev}\end{array}$

Therefore, the number of revolutions made by a particle moving in a circle during the first 4s is 9.55 rev.