Question

A rotating wheel changes angular speed from $1800rpm$ to $3000rpm$ in $20s$. 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

Answer: (C)

We know that
$\omega =2\pi n$
$\therefore {\omega }_1=2\pi n_1$
where $n_1=1800rpm$
$n_2=3000rpm$
$\Delta t=20s$
${\omega }_1=2\pi \times \frac{1800}{60}=2\pi \times 30=60\pi$
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
$\alpha =\frac{\text{ Change in angular velocity }}{\text{ Time interval }}$
$\alpha =\frac{{\omega }_2-{\omega }_1}{\Delta t}$
$=\frac{100\pi -60\pi }{20}$
$=\frac{40\pi }{20}$
$=2\pi rad/s^2$