Question

A flywheel with a diameter of 0.679 m is rotating at an angular speed of 221 rev/min. (a) What is the angular speed of the flywheel in radians per second? (b) What is the linear speed of a point on the rim of the flywheel? (c) What constant angular acceleration (in revolutions per minute-squared) will increase the wheel's angular speed to 730 rev/min in 68.5 s? (d) How many revolutions does the wheel make during that 68.5 s? (a is in rad/s, b is in m/s, c is in rev/min^2, d is rev)

Answer 1

PART a)

Angular speed is given by

$\omega = 2\pi f$

$\omega = 2\pi*\frac{221}{60}$

$\omega = 23.14 rad/s$

PART B)

Linear speed is given by

$v = R\omega$

$v = \frac{0.679}{2}*23.14$

$v =7.86 m/s$

PART C)

Constant angular acceleration is given by

$\alpha = \frac{2\pi(f_2 - f_1)}{\Delta t}$

$\alpha = \frac{2\pi(\frac{730}{60} -\frac{221}{60})}{68.5}$

$\alpha = 0.778 rad/s^2$

PART d)

Number of revolutions in the given time will be

$N = \frac{f_1+ f_2}{2}*t$

$N = \frac{(\frac{730}{60} +\frac{221}{60})}{2}*68.5$

$N = 542.8rev$