Question

At $\mathbf{t}\mathbf{=}\mathbf{}\mathbf{0}$, a flywheel has an angular velocity of$\mathbf{4.7}\mathbf{\text{\hspace{0.17em}rad}}\mathbf{/}\mathbf{\text{s}}$, a constant angular acceleration of$\mathbf{-}\mathbf{0.25}\mathbf{\text{\hspace{0.17em}rad}}\mathbf{/}{\mathbf{\text{s}}}^{\mathbf{\text{2}}}$, and a reference line at${\mathbf{\theta }}_{\mathbf{0}}\mathbf{=}\mathbf{0}$.

(a) Through what maximum angle${\mathit{\theta }}_{\mathbf{max}}$will the reference line turn in the positive direction? What are the

(b) first and

(c) second times the reference line will be$\mathbf{\theta }\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{\theta }}_{\mathbf{max}}$?

At what(d) negative time and

(e) positive times will the reference line be at$\mathbf{\theta }\mathbf{=}\mathbf{10.5}\mathbf{\text{\hspace{0.17em}rad}}$?

(f) Graph$\mathbf{\theta }$versus$\mathbf{t}$, and indicate your answers.

Short answer

1. The maximum angle of rotation in positive direction is,$44\text{\hspace{0.17em}rad}$
2. The first time reference line is at$\mathrm{\theta }=\frac{1}{2}{\mathrm{\theta }}_{\max }$is$5.5\text{\hspace{0.17em}s}$
3. The second time reference line is at$\mathrm{\theta }=\frac{1}{2}{\mathrm{\theta }}_{\max }$is$32\text{\hspace{0.17em}s}$
4. The negative time at which reference line is at$\mathrm{\theta }=10.5\text{\hspace{0.17em}rad}$is$\mathrm{t}=-2.1\text{\hspace{0.17em}s}$
5. The positive time at which reference line is at$\theta =10.5\text{\hspace{0.17em}rad}$is$\mathrm{t}=40\text{\hspace{0.17em}s}$
6. Graph of $\mathrm{\theta }$vs role="math" localid="1660900831861" $\mathbf{t}$is plotted.

Step-by-step solution

Listing the given quantities

The initial angular speed of the flywheel ${\mathrm{\omega }}_0=4.7\text{rad}/\text{s}$

At$\mathrm{t}=0$, the position of flywheel is ${\theta }_0=0$

The constant angular acceleration is,$\mathrm{\alpha }=-0.25\text{\hspace{0.17em}rad}/{\text{s}}^{\text{2}}$

Understanding the kinematic equations

The flywheel undergoes rotational motion about an axis passing through its axel. Hence, we determine its angles of rotation at instants of time using rotational kinematic equations.

Formula:

${\mathbf{\omega }}^{\mathbf{2}}\mathbf{=}\mathbf{}{\mathbf{\omega }}_{\mathbf{0}}^{\mathbf{2}}\mathbf{+}\mathbf{}\mathbf{2}\mathbf{}\mathbf{\alpha }\mathbf{}\mathbf{(}\mathbf{\theta }\mathbf{-}{\mathbf{\theta }}_{\mathbf{0}}\mathbf{)}$

$\mathbf{\theta }\mathbf{-}{\mathbf{\theta }}_{\mathbf{0}}\mathbf{=}\mathbf{}{\mathbf{\omega }}_{\mathbf{0}}\mathbf{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{}{\mathbf{\alpha t}}^{\mathbf{2}}$

(a) Maximum angle of rotation in positive direction

The constant angular acceleration is negative. Thus, the flywheel will turn through maximum angle when it stops. i.e. the final angular velocity is zero. Hence, we use the kinematic equation to determine this angle.

$\begin{array}{c}{\mathrm{\omega }}^2={{\mathrm{\omega }}_0}^2+2\mathrm{\alpha }(\mathrm{\theta }-{\mathrm{\theta }}_0)\\ 0^2={(4.7\text{\hspace{0.17em}rad}/\text{s})}^2+2\times (-0.25\text{\hspace{0.17em}rad}/{\text{s}}^{\text{2}})(\mathrm{\theta }-0)\\ \mathrm{\theta }=\frac{{(4.7\text{\hspace{0.17em}rad}/\text{s})}^2}{2\times (0.25\text{\hspace{0.17em}rad}/{\text{s}}^{\text{2}})}\\ =44\text{\hspace{0.17em}rad}\end{array}$

$\mathrm{The}\mathrm{maximum}\mathrm{angle}\mathrm{of}\mathrm{rotation}\mathrm{in}\mathrm{positive}\mathrm{direction}\mathrm{is},44\text{\hspace{0.17em}rad}.$

(b) the first time reference line is at θ= 12θmax

It is given that$\mathrm{\theta }=\frac{1}{2}{\mathrm{\theta }}_{\max }$so we get,

$\begin{array}{c}\mathrm{\theta }=\frac{1}{2}{\mathrm{\theta }}_{\max }\\ =\frac{44\text{\hspace{0.17em}rad}}{2}\\ \mathrm{\theta }=22\text{\hspace{0.17em}rad}\end{array}$

To determine the time to reach this angle, we will use another kinematic equation as

$\begin{array}{c}22\text{\hspace{0.17em}rad}=(4.7\text{\hspace{0.17em}rad}/\text{s})\mathrm{t}+\frac{1}{2}(-0.25\text{\hspace{0.17em}rad}/{\text{s}}^{\text{2}}){\mathrm{t}}^2\\ 22\text{\hspace{0.17em}rad}=(4.7\text{\hspace{0.17em}rad}/\text{s})\mathrm{t}-(0.13\text{\hspace{0.17em}rad}/{\text{s}}^{\text{2}}){\mathrm{t}}^2\\ (0.13\text{\hspace{0.17em}rad}/{\text{s}}^{\text{2}}){\mathrm{t}}^2-(4.7\text{\hspace{0.17em}rad}/\text{s})\mathrm{t}+22\text{\hspace{0.17em}rad}=0\end{array}$

We solve this quadratic equation using the formula

$\mathrm{x}=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}}$

For our equation,$\mathrm{a}=(0.13\text{\hspace{0.17em}rad}/{\text{s}}^{\text{2}}),\mathrm{b}=-(4.7\text{\hspace{0.17em}rad}/\text{s}),\mathrm{c}=22\text{\hspace{0.17em}rad}$

Using these values in the above equation, we have

$\begin{array}{c}\mathrm{t}=32.12\text{\hspace{0.17em}s}\\ \approx 32\text{\hspace{0.17em}s}\end{array}$

And

$\begin{array}{c}\mathrm{t}=5.48\text{\hspace{0.17em}s}\\ \approx 5.5\text{\hspace{0.17em}s}\end{array}$

Thus, the flywheel will reach the angle $\mathrm{\theta }=\frac{1}{2}{\mathrm{\theta }}_{\max }$for the first time at$\mathrm{t}=5.5\text{\hspace{0.17em}s}$

(c) the second time reference line is at θ= 12θmax

The other root of the equation gives us the second time at which the flywheel will reach the same angle. So $\mathrm{t}=32\text{\hspace{0.17em}s}$.

(d) the negative time at which reference line is at θ= 10.5 rad 

Using the same kinematic equation as above to determine the time at which$\mathrm{\theta }=10.5\text{rad}$

As we need to determine the negative time, we consider$\mathrm{\theta }=-10.5\text{rad}$

$\begin{array}{c}\mathrm{\theta }-{\mathrm{\theta }}_0={\mathrm{\omega }}_0\mathrm{t}+\frac{1}{2}{\mathrm{\alpha t}}^2\\ -10.5\text{\hspace{0.17em}rad}=(4.7\text{\hspace{0.17em}rad}/\text{s})\left(\mathrm{t}\right)+\frac{1}{2}(-0.25\text{\hspace{0.17em}rad}/{\text{s}}^{\text{2}}){\left(\mathrm{t}\right)}^2\\ -10.5\text{\hspace{0.17em}rad}=(4.7\text{\hspace{0.17em}rad}/\text{s})\mathrm{t}-(0.125\text{\hspace{0.17em}rad}/{\text{s}}^{\text{2}}){\mathrm{t}}^2\\ (0.125\text{\hspace{0.17em}rad}/{\text{s}}^{\text{2}}){\mathrm{t}}^2-(4.7\text{\hspace{0.17em}rad}/\text{s})\mathrm{t}-10.5\text{\hspace{0.17em}rad}=0\end{array}$

We solve this quadratic equation using the formula

$\mathrm{x}=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}}$

For our equation,$\mathrm{a}=(0.125\text{\hspace{0.17em}rad}/{\text{s}}^{\text{2}}),\mathrm{b}=-(4.7\text{\hspace{0.17em}rad}/\text{s}),\mathrm{c}=-10.5\text{\hspace{0.17em}rad}$

Using these values in above equation, we get

$\begin{array}{c}\mathrm{t}=-2.115\text{\hspace{0.17em}s}\\ \approx -2.1\text{\hspace{0.17em}s}\end{array}$

And

$\begin{array}{c}t=39.71\text{\hspace{0.17em}s}\\ \approx 40\text{\hspace{0.17em}s}\end{array}$

Thus, the flywheel will reach the angle $\mathrm{\theta }=10.5\text{\hspace{0.17em}rad}$ for the negative time at$t=-2.1\text{\hspace{0.17em}s}$ .

(e) the positive time at which reference line is at θ= 10.5 rad

The other root of the equation gives us the second (positive) time at which the flywheel will reach the same angle $\mathrm{\theta }=10.5\text{\hspace{0.17em}rad}$so$t=40\text{\hspace{0.17em}s}$.

(f) Graph of  θ vs. t

The graph of $\mathrm{\theta }\text{vst}$appears as

The flywheel rotates about its axel. It is decelerating with constant magnitude. Thus, we need to use the rotational kinematic equations to determine the time and the corresponding angular positions of the wheel.