Question

Question: A rotating wheel requiresrotating through 37.0 revolutions. Its angular speed at the end of theinterval is. What is the constant angular acceleration of the wheel?

Short answer

The solution of the constant angular acceleration is$\alpha =13.7\text{rad}/{\text{s}}^2$.

Step-by-step solution

Converting the given units and deriving angular speed

First converting units:

Multiply the angular speed $\left({\omega }_i\right)$by a conversion factor to convert its units from

(rev/min) to (rad/sec)

$\theta =37\left(rev\right)\left(\frac{2\pi \text{rad}}{1 \text{rev}}\right)=232.47 \text{rad}$

Second, solving the problem:

Model the wheel as a rigid object under constant angular acceleration and use the following equation to find the initial angular speed.

${\omega }_f={\omega }_i+\alpha t$

Solve for ${\omega }_i$

${\omega }_i={\omega }_f-\alpha t$

Substitute the known numerical values.

${\omega }_i=98-3\alpha$

Deriving the equation and finding angular acceleration

Use the following equation to find the angular acceleration of the wheel:

$\theta ={\omega }_{i}t+\frac{1}{2}\alpha t^2$

Solve for $\left(\alpha \right)$.

$\alpha =\frac{2\left(\theta -{\omega }_{i}t\right)}{t^2}$

Substitute for$\left({\omega }_i\right)$ from Equation (1) in Equation (2).

$\begin{array}{rcl}\alpha & =& \frac{2[\theta -(98-3\alpha t\left)\right]}{t^2}\\ \alpha & =& \frac{2\theta -196t+6\alpha t}{t^2}\\ \alpha t^2& =& 2\theta -196t+6\alpha t\\ \alpha \left(t^2-6t\right)& =& 2\theta -196t\end{array}$

Solve for$\left(\alpha \right)$

$\alpha =\frac{2\theta -196t}{t^2-6t^2}$

Substitute numerical values.

$\begin{array}{rcl}\alpha & =& \frac{2(232.47)-196\left(3\right)}{3^2-6\left(3\right)}\\ & =& 13.7 \text{rad}/{\text{s}}^2\end{array}$

Hence, the answer is $13.7 \text{rad}/{\text{s}}^2$.