You are analysing the motion of a large flywheel that has radius0.800m. In one test run, the wheel starts from rest and turns with constant angular acceleration. An accelerometer on the rim of the flywheel measures the magnitude of the resultant acceleration of a point on the rim of the flywheel as a function of the angle through which the wheel has turned. You collect these results:

Construct a graph of ${\mathit{a}}^{\mathbf{2}}\left({\text{inm}}^{\text{2}}{\text{/s}}^{\text{4}}\right)$versus in ${\left(\theta -{\theta }_0\right)}^{\mathbf{2}}$.

1. What are the slope and y-intercept of the straight line that gives the best fit to the data?
2. Use the slope from part (a) to find the angular acceleration of the flywheel?
3. What is the linear speed of a point on the rim of the flywheel when the wheel has turned through an angle of${\mathbf{135}}^{\mathbf{0}}$?
4. When the flywheel has turned through an angle of${\mathbf{90}}^{\mathbf{\circ }}$, what is the angle between the linear velocity of a point on its rim and the resultant acceleration of that point?

${\mathit{a}}_{\mathbf{t}}\mathbf{=}\mathit{\alpha }\mathit{r}$ (1)

Where$\mathit{\alpha }$is the body's angular acceleration's magnitude.

${\mathit{a}}_{\mathbf{r}}\mathbf{=}\frac{{\mathbf{v}}^{\mathbf{2}}}{\mathbf{r}}\mathbf{=}{\mathit{\omega }}^{\mathbf{2}}\mathit{r}$ (2)

Where localid="1668082158377" $\mathit{\omega }$is the magnitude of the angular velocity of the body.

$\begin{array}{l}{\mathbf{\omega }}_{\mathbf{f}}\mathbf{=}{\mathbf{\omega }}_{\mathbf{i}}\mathbf{+}\mathbf{\alpha }\mathbf{t}\\ {\mathbf{\theta }}_{\mathbf{f}}\mathbf{=}{\mathbf{\theta }}_{\mathbf{i}}\mathbf{+}{\mathbf{\omega }}_{\mathbf{i}}\mathbf{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\alpha }{\mathbf{t}}^{\mathbf{2}}\end{array}$

${{\mathit{\omega }}_{\mathbf{f}}}^{\mathbf{2}}\mathbf{=}{{\mathit{\omega }}_{\mathbf{i}}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathit{\alpha }\left({\theta }_f-{\theta }_i\right)$ (3)

localid="1668083513765" ${\mathit{\theta }}_{\mathbf{f}}\mathbf{=}{\mathit{\theta }}_{\mathbf{i}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(}{\mathbf{\omega }}_{\mathbf{i}}\mathbf{+}{\mathbf{\omega }}_{\mathbf{f}}\mathbf{)}\mathit{t}$

(d) When a rigid object rotates about a fixed axis, the angular speed is related t the translational speed through the relationship

$\mathit{v}\mathbf{=}\mathit{r}\mathit{\omega }$ (4)

Here we have radius of flywheel is$r=0.800\text{m}$

Wheel starts from rest. So, initial angular speed is ${\omega }_i=0$.

Also we have the table which shows the magnitude of the resultant acceleration of a point on the rim of the flywheel as a function of the angle$\theta \text{-}{\theta }_0$ .

Here we have a table of acceleration and $\theta \text{-}{\theta }_0$ from that table we have,

Now, to find slope of line which is obtained in above graph is given by

$m=\frac{1.14-0.46}{1-0.25}=0.91m^2{\text{/s}}^4$

Now, from the graph we can see that y-intercept is at$0.233{\text{\hspace{0.17em}m}}^2{\text{/s}}^4$

The tangential and radial components make up the linear acceleration of the location on the flywheel's rim.

Since both components are perpendicular in this case, the Pythagorean theorem is used to get the linear acceleration that results:

$a^2={a_t}^2+{a_r}^2$

Now, from equation (1) and (2)

$\begin{array}{l}\Rightarrow a^2={\left(\alpha r\right)}^2+{\left({\omega }^{2}r\right)}^2\\ \Rightarrow a^2={\alpha }^2{r}^2+{\left({\omega }^{2}r\right)}^2\end{array}$

Here we have initial angular speed is zero.

Substitute equation (3) in above equation. We get,

$\begin{array}{l}\Rightarrow a^2={\alpha }^2{r}^2+{\left[2\alpha \left(\theta -{\theta }_0\right)\right]}^2{r}^2\\ \Rightarrow a^2={\alpha }^2{r}^2+4{\alpha }^2{r}^2{\left(\theta -{\theta }_0\right)}^2\end{array}$

Now, above equation is in the form of straight line.

With ${\left(\theta -{\theta }_0\right)}^2$and$a^2$.

Also, it has slope of $4{\alpha }^2{r}^2$and y-intercept at${\alpha }^2{r}^2$.

Now, from step 4 we get slope is$m=0.91m^2{\text{/s}}^4$

So, we have

$\begin{array}{l}4{\alpha }^2{r}^2=0.91m^2{\text{/s}}^4\\ \Rightarrow \alpha =\sqrt{\frac{0.91m^2{\text{/s}}^4}{4{\left(0.800\text{\hspace{0.17em}m}\right)}^2}}=0.600{\text{\hspace{0.17em}\hspace{0.17em}rad/s}}^2\end{array}$

So, angular acceleration is $0.600{\text{\hspace{0.17em}\hspace{0.17em}rad/s}}^2$.

The angular speed of the wheel is found from equation (3)

${\omega }^2={{\omega }_i}^2+2\alpha \left(\theta -{\theta }_0\right)$

Here we have initial angular speed is ${\omega }_i=0$and$\alpha =0.600{\text{\hspace{0.17em}\hspace{0.17em}rad/s}}^2$

$\begin{array}{l}\Rightarrow {\omega }^2=0+2\left(0.600{\text{\hspace{0.17em}\hspace{0.17em}rad/s}}^2\right)135\left(\frac{\pi \text{rad}}{{180}^{\circ }}\right)\\ \Rightarrow \omega =\sqrt{2\left(0.600{\text{\hspace{0.17em}\hspace{0.17em}rad/s}}^2\right)135\left(\frac{\pi \text{rad}}{{180}^{\circ }}\right)}\\ \Rightarrow \omega =1.68\text{\hspace{0.17em}rad/s}\end{array}$

Now, from equation (4) we have,

$\begin{array}{l}v=r\omega \\ \Rightarrow v=\left(0.800\text{\hspace{0.17em}\hspace{0.17em}m}\right)\left(1.68\text{\hspace{0.17em}rad/s}\right)\\ \Rightarrow v=1.35\text{\hspace{0.17em}m/s}\end{array}$

the linear speed of a point on the rim of the flywheel when the wheel has turned through an angle of ${135}^{\circ }$is $1.35\text{\hspace{0.17em}m/s}$.

Let $\delta$be an angle between the linear velocity of a point on its rim and the resultant acceleration

Now, we know that

$\begin{array}{l}\tan \delta =\frac{a_t}{a_r}=\frac{{\omega }^{2}r}{\alpha r}\\ \Rightarrow \tan \delta =\frac{{\omega }^2}{\alpha }\end{array}$

Now, from equation (3),

$\begin{array}{l}\Rightarrow \tan \delta =\frac{2\alpha \left(\theta -{\theta }_0\right)}{\alpha }\\ \Rightarrow \tan \delta =2\left(\theta -{\theta }_0\right)\\ \Rightarrow \delta ={\tan }^{-1}\left[2\left(\theta -{\theta }_0\right)\right]\end{array}$

Here we have $\theta ={90}^{\circ }$

$\begin{array}{l}\delta ={\tan }^{-1}\left[2\times 90\left(\frac{\pi \text{\hspace{0.17em}rad}}{180}\right)\right]\\ \delta ={\tan }^{-1}\left(\pi \right)\\ \delta ={72.3}^{\circ }\end{array}$

Hence, angle between the linear velocity of a point on its rim and the resultant acceleration is$\delta ={72.3}^{\circ }$