Question

A rotating wheel changes angular speed from $1800\,rpm$ to $3000\, rpm$ in $20\,s$. What is the angular acceleration assuming to be uniform ?

• A. $60\,\pi\, rad\,s^{-2}$
• B. $90\,\pi\, rad\,s^{-2}$
• C. $2\,\pi\, rad\,s^{-2}$
• D. $40\,\pi\, rad\,s^{-2}$

Answer 1

The Correct Option is C

We know that
$\omega =2 \pi n$
$\therefore \omega_{1}=2 \pi n_{1}$
where $n_{1}=1800\, rpm$
$n_{2}=3000\, rpm$
$\Delta t=20\, s$
$\omega_{1}=2 \pi \times \frac{1800}{60}=2 \pi \times 30=60 \pi$
Similarly, $\omega_{2}=2 \pi n_{2}=2 \pi \times \frac{3000}{60}$
$=2 \pi \times 50$
$=100 \pi$
If the angular velocity of a rotating wheel about on axis changes by change in angular velocity in a time interval $\Delta t$, then the angular acceleration of rotating wheel about that axis is
$\alpha =\frac{\text { Change in angular velocity }}{\text { Time interval }}$
$\alpha =\frac{\omega_{2}-\omega_{1}}{\Delta t}$
$=\frac{100 \pi-60 \pi}{20}$
$=\frac{40 \pi}{20}$
$=2 \pi\, rad / s ^{2}$