Question

The angular speed of a motor wheel is increased from $1200rpm$ to $3120rpm$ in $16$ seconds. $\left(i\right)$ What is its angular acceleration to be uniform? $\left(ii\right)$ How many revolutions does the engine make during this time?

Answer 1

$\left(i\right)$ We shall use $\omega ={\omega }_0+\alpha t$
${\omega }_0=$ initial anugular speed in rad/s
$=2\pi \times angularspeedinrev/s$
$=\frac{2\pi \times angularspeedinrev/min}{60s/min}$
$=\frac{2\pi \times 1200}{60}rad/s$
$=40\pi rad/s$
Similarly $\omega =$ final angular speed in rad/s
$=\frac{2\pi \times 3120}{60}rad/s$
$=2\pi \times 52rad/s$
$=104\pi rad/s$
$\therefore$ Angular acceleration
$\alpha =\frac{\omega -{\omega }_0}{t}=4\pi rad/s^2$
The angular acceleration of the engine $=4\pi rad/s^2$
$\left(ii\right)$ The angular displacement in time $t$ is given by
$\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$
$=(40\pi \times 16+\frac{1}{2}\times 4\pi \times {16}^2)$ rad
$=(640\pi +512\pi )rad$
$=1152\pi rad$
Number of revolutions $=\frac{1152\pi }{2\pi }=576$