The angular speed of a motor wheel is increased from $1200$ rpm to $3120$ rpm in $16$ seconds. $(i)$ What is its angular acceleration, assuming the acceleration to be uniform? $(ii)$ How many revolutions does the engine make during this time?
Example-(6.11)
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$(i)$ We shall use $\omega=\omega_0+\alpha t$
$\omega_0=$ initial angular speed in $rad / s$
$=2 \pi \times \text { angular speed in rev } / s$
$=\frac{2 \pi \times \text { angular speed in rev } / min }{60 s / min }$
$=\frac{2 \pi \times 1200}{60}\ rad / s$
$=40 \pi\ rad / s$
Similarly$\omega=$ final angular speed in $rad / s$
$=\frac{2 \pi \times 3120}{60}\ rad / s$
$=2 \pi \times 52\ rad / s$
$=104 \pi\ rad / s$
$\therefore$ Angular acceleration
$\alpha=\frac{\omega-\omega_0}{t} \quad=4 \pi\ rad / s ^2$
The angular acceleration of the engine $=4 \pi\ rad / s ^2$
$(ii)$ The angular displacement in time $t$ is given by
$\theta=\omega_0 t+\frac{1}{2} \alpha t^2$
$= \left(40 \pi \times 16+\frac{1}{2} \times 4 \pi \times 16^2\right)\ rad$
$= (640 \pi+512 \pi)\ rad$
$= 1152 \pi\ rad$
Number of revolutions $=\frac{1152 \pi}{2 \pi}=576$
$\omega_0=$ initial angular speed in $rad / s$
$=2 \pi \times \text { angular speed in rev } / s$
$=\frac{2 \pi \times \text { angular speed in rev } / min }{60 s / min }$
$=\frac{2 \pi \times 1200}{60}\ rad / s$
$=40 \pi\ rad / s$
Similarly$\omega=$ final angular speed in $rad / s$
$=\frac{2 \pi \times 3120}{60}\ rad / s$
$=2 \pi \times 52\ rad / s$
$=104 \pi\ rad / s$
$\therefore$ Angular acceleration
$\alpha=\frac{\omega-\omega_0}{t} \quad=4 \pi\ rad / s ^2$
The angular acceleration of the engine $=4 \pi\ rad / s ^2$
$(ii)$ The angular displacement in time $t$ is given by
$\theta=\omega_0 t+\frac{1}{2} \alpha t^2$
$= \left(40 \pi \times 16+\frac{1}{2} \times 4 \pi \times 16^2\right)\ rad$
$= (640 \pi+512 \pi)\ rad$
$= 1152 \pi\ rad$
Number of revolutions $=\frac{1152 \pi}{2 \pi}=576$
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