Q134 / 180 · KCET 2014 · Physics
A rotating wheel changes angular speed from \( 1800 \mathrm{rpm} \) to \( 3000 \mathrm{rpm} \) in \( 20 \mathrm{~s} \). What is the
angular acceleration assuming to be uniform ?
angular acceleration assuming to be uniform ?
- A\( 60 \) п rad s \( ^{-2} \)
- B90п rad s \( ^{-2} \)
- Correct answerC\( 2 \Pi \mathrm{rad} s^{-2} \)
- D40п rad s \( ^{-2} \)
Solution
For angular speed of \( 1800 \mathrm{rpm} \)
\( \omega_{1}=2 \Pi f_{1}=2 \Pi\left(\frac{1800}{60}\right)=60 \Pi \mathrm{rad} s^{-1} \)
For angular speed of \( 3000 \mathrm{rpm} \)
\( \omega_{2}=2 \Pi f_{2}=2 \Pi\left(\frac{3000}{60}\right)=100 \Pi \mathrm{rad} s^{-1} \)
Therefore, angular acceleration \( =\frac{\omega_{2}-\omega_{1}}{t}=\frac{100 \Pi-60 \Pi}{20} \)
\( =\frac{40 \Pi}{20}=2 \Pi \)
Thus, angular acceleration of rotating wheel \( =2 \pi \mathrm{rads}^{-2} \)
\( \omega_{1}=2 \Pi f_{1}=2 \Pi\left(\frac{1800}{60}\right)=60 \Pi \mathrm{rad} s^{-1} \)
For angular speed of \( 3000 \mathrm{rpm} \)
\( \omega_{2}=2 \Pi f_{2}=2 \Pi\left(\frac{3000}{60}\right)=100 \Pi \mathrm{rad} s^{-1} \)
Therefore, angular acceleration \( =\frac{\omega_{2}-\omega_{1}}{t}=\frac{100 \Pi-60 \Pi}{20} \)
\( =\frac{40 \Pi}{20}=2 \Pi \)
Thus, angular acceleration of rotating wheel \( =2 \pi \mathrm{rads}^{-2} \)
Hence, the correct answer is option: 'C'.