Question

The angular speed of truck wheel is increased from $900\text{rpm}$ to $2460\text{rpm}$ in $26$ seconds. The number of revolutions by the truck wheel during this time is ______.

$($Assuming the acceleration to be uniform$)$

Answer 1

Given :
${\omega }_f=2460\times \frac{2\pi }{60}=82\pi \text{rad/s}$

${\omega }_i=900\times \frac{2\pi }{60}=30\pi \text{rad/s}$

$t=26\text{s}$

We know that,

$\alpha =\frac{{\omega }_f-{\omega }_i}{t}=\frac{82\pi -30\pi }{26}=2\pi {\text{rad/s}}^2$

Angular displacement of the wheel is given by,
$\theta ={\omega }_{i}t+\frac{1}{2}\alpha t^2$

$\Rightarrow \theta =30\pi \times 26+\frac{1}{2}\times 2\pi \times {26}^2=1456\pi$

Number of revolution $=\frac{\theta }{2\pi }=\frac{1456\pi }{2\pi }=728$