Question

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

Answer 1

Step 1: Given data

The initial angular speed of truck= $900rpm$

The angular speed of the truck after $26seconds$=$2460rpm$

Step 2: Convert the angular speed of the truck in $rad/sec$

The initial angular speed of the truck in $rad/sec$ is,

$\begin{array}{rcl}{\omega }_i& =& 900\times \frac{2\mathrm{\pi }}{60}\\ & =& 30\mathrm{\pi }\mathrm{rad}/\sec \end{array}$

The final speed of the truck in $rad/sec$ is,

$\begin{array}{rcl}{\omega }_f& =& 2460\times \frac{2\mathrm{\pi }}{60}\\ & =& 82\mathrm{\pi }\mathrm{rad}/\sec \end{array}$

Step 3: Obtain the magnitude of angular acceleration

So, the magnitude of angular acceleration is,

$\begin{array}{rcl}\alpha & =& \frac{{\omega }_f-{\omega }_i}{t}\\ & =& \frac{82\mathrm{\pi }-30\mathrm{\pi }}{26}\\ & =& 2\mathrm{\pi }\mathrm{rad}/{\sec }^2\end{array}$

Step4: Compute the number of revolutions

It is understood that $\theta =\frac{{\omega }_f^2-{\omega }_i^2}{2\alpha }$

Suppose the number of revolutions is $N$. So, $\theta =2\mathrm{\pi N}$

$$\begin{gathered} 2\mathrm{\pi N}=\frac{\left({\omega }_f+{\omega }_i\right)\left({\omega }_f-{\omega }_i\right)}{2\alpha } \\ N=\frac{\left(82\mathrm{\pi }+30\mathrm{\pi }\right)\left(82\mathrm{\pi }-30\mathrm{\pi }\right)}{2\times 2\mathrm{\pi }\times 2\mathrm{\pi }} \\ N=728 \end{gathered}$$

Thus, the number of revolutions by the truck wheel during the given time is $728$.