Question

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

• A. $4\pi {\text{rads}}^{-2}$ and 200
• B. $2\pi {\text{rads}}^{-2}$ and 470
• C. $4\pi {\text{rads}}^{-2}$ and 576
• D. $3\pi {\text{rads}}^{-2}$ and 390

Answer 1

Answer: (C)

We shall use $\omega ={\omega }_0+\alpha t$
${\omega }_0=$ Initial angular speed in ${\text{rads}}^{-1}$
$=2\pi \times$ Angular speed in revs ${}^{-1}$
$=\frac{2\pi \times \text{ Angular speed in }rev/\min }{60s/\min }$
$=\frac{2\pi \times 1200}{60}{\text{rads}}^{-1}=40\pi {\text{rads}}^{-1}$
Similarly, $\omega =$ Final angular speed is ${\text{rads}}^{-1}$
$=\frac{2\pi \times 3120}{60}=2\pi \times 52=104\pi {\text{rads}}^{-1}$
$\therefore$ Angular acceleration,
$\alpha =\frac{\omega -{\omega }_0}{t}=\frac{104\pi -40\pi }{16}=4\pi {\text{rads}}^{-2}$
The angular acceleration of the motor is $4\pi {\text{rads}}^{-2}$
(ii) The angular displacement in time $t$ is given by
$\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2=40\pi \times 16+\frac{1}{2}\times 4\pi \times 16^2$
$=640\pi +512\pi =1152\pi rad$
Number of revolutions $=\frac{1152\pi }{2\pi }=576$