Question

The angular speed of a motor wheel is increased from $1200\, rpm$ to $3120\, rpm$ in $16\, s$. (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 ${ }^{-1}$
$=\frac{2 \pi \times \text { Angular speed in } rev / \min }{60\, s / \min }$
$=\frac{2 \pi \times 1200}{60} \text{rads} ^{-1}=40\, \pi\, \text{rads} ^{-1}$
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 $=\frac{1152 \pi}{2 \pi}=576$