Question

A car is traveling along a road, and its engine is turning over with an angular velocity of +200 rad/s. The driver steps on the accelerator, and in a time of 13.0 s the angular velocity increases to +300 rad/s. (a) What would have been the angular displacement of the engine if its angular velocity had remained constant at the initial value of +200 rad/s during the entire 13.0-s interval? (b) What would have been the angular displacement if the angular velocity had been equal to its final value of +300 rad/s during the entire 13.0-s interval? (c) Determine the actual value of the angular displacement during the 13.0-s interval.

Answer 1

a) +2600 rad

b) +3900 rad

c) +3250 rad

Explanation:

a)

The angular velocity of an object in rotation is given by

$\omega=\frac{\theta}{t}$

where

$\theta$ is the angular dispalcement

t is the time elapsed

In this problem:

$\omega=+200 rad/s$ is the constant angular velocity of the engine during the time interval

t = 13.0 s is the time elapsed

Therefore, re-arranging the equation, we can find the angular displacement:

$\theta=\omega t=(+200)(13.0)=+2600 rad$

b)

In this case, instead, we have:

$\omega=+300 rad/s$ is the constant angular velocity during the time interval

t = 13.0 s is the time elapsed

We can use again the same equation as before:

$\omega=\frac{\theta}{t}$

where

$\theta$ is the angular dispalcement

t is the time elapsed

And solving for the angular displacement, we find:

$\theta=\omega t=(+300)(13.0)=+3900 rad$

c)

In this case, the motion of the engine is accelerated, so we can use the equivalent of the suvat equations for rotational motion; we can use:

$\theta = (\frac{\omega_i+\omega_f}{2})t$

where:

$\omega_i$ is the initial angular velocity

$\omega_f$ is the final angular velocity

t is the time elapsed

For the engine in this problem we have:

$\omega_i = +200 rad/s$

$\omega_f=+300 rad/s$

t = 13.0 s

Therefore, the angular displacement of the engine is:

$\theta=(\frac{200+300}{2})(13.0)=+3250 rad$