Question:
The angular speed of truck wheel is increased from $900 \mathrm{rpm}$ to $2460 \mathrm{rpm}$ in 26 seconds. The number of revolutions by the truck engine during this time is
(Assuming the acceleration to be uniform).
Solution:
We know, $\theta=\left(\frac{\omega_{1}+\omega_{2}}{2}\right) \mathrm{t}$
Let number of revolutions be $\mathrm{N}$
$\therefore 2 \pi N=2 \pi\left(\frac{900+2460}{60 \times 2}\right) \times 26$
$\mathrm{N}=728$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.