Question

A flywheel rotating at 20 rpm undergoes an angular acceleration of +2.0 rad/s^2 for precisely 5.0 rev. What is the angular velocity of the flywheel at the end of the acceleration?

Answer 1

Answer:

final angular velocity = 11.4 rad/sec = 108.86 rpm

Explanation:

To start with, notice that the quantities given are in various units for angle measurement (revolutions, and radians), and also different units of time (minutes and seconds). So in order to have them combined in a single equation, we need to convert them to the same type of angle and time units. If we select to keep radians as the unit of angle, and seconds for the unit of time, we need to convert:

1) the initial angular velocity $\omega_i$  of 20 rpm into radians per second using:

$20\,\frac{rev}{min} = \frac{20\,* 2\, \pi }{60} \,\frac{rad}{sec}=\frac{2\,\pi }{3} \,\frac{rad}{sec}=2.094\,\frac{rad}{sec}$

2) the 5 revolutions during which the flywheel is under acceleration into radians:

$5\,rev= 5\,*\,2\, \pi\, rad= 10\, \pi\,rad=31.41\,rad$

Now  notice as well that the information we are given does not contain the time under which the flywheel is accelerated, but the angle (number of revolutions), so it would be convenient to use a kinematic relationship for accelerated rotational motion where the variable time has been reduced from the formula. We therefore choose the following formula:

$\omega_f^2-\omega_i^2=2\,\alpha\,(\theta_f-\, \theta_i)$

where we know all the variables except for the final angular velocity $\omega_f$.

Therefore, replacing for the known quantities, and solving for the unknown, we get:

$\omega_f^2-\omega_i^2=2\,\alpha\,(\theta_f-\, \theta_i)\\\omega_f^2-(2.094)^2=2\,(2)\,(31.41)\\\omega_f^2=(2.094)^2+2\,(2)\,(31.41)\\\omega_f^2=130.0248\,(\frac{rad}{sec})^2 \\\omega_f=11.40\,\frac{rad}{sec}$

If the answer is requested in revolutions per minute, we can multiply this result by 60 and divide it by $2\,\pi$, to obtain its equivalent : 108.86 rpm