Question

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

Answer: (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 rotting 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
$\alpha =\frac{C h a n g e \, i n \, a n g u l a r \, v e l o c i t y}{T i m e \, i n t e r v a l}$
$\alpha =\frac{\omega _{2} - \omega _{1}}{\Delta t}$
$=\frac{100 \pi - 60 \pi }{20}$
$=\frac{40 \pi }{20}$
$=2\pi \, r a d / s^{2}$