91Ó°ÊÓ

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

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

The constant angular acceleration is,$\mathrm{Î\pm }=−0.25\text{ r²¹»å}/{\text{s}}^{\text{2}}$

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:

${\mathbf{Ó\neg }}^{\mathbf{2}}\mathbf{=}\mathbf{}{\mathbf{Ó\neg }}_{\mathbf{0}}^{\mathbf{2}}\mathbf{+}\mathbf{}\mathbf{2}\mathbf{}\mathbf{Î\pm }\mathbf{}\mathbf{(}\mathbf{θ}\mathbf{−}{\mathbf{θ}}_{\mathbf{0}}\mathbf{)}$

$\mathbf{θ}\mathbf{−}{\mathbf{θ}}_{\mathbf{0}}\mathbf{=}\mathbf{}{\mathbf{Ó\neg }}_{\mathbf{0}}\mathbf{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{}{\mathbf{Î\pm ³Ù}}^{\mathbf{2}}$

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.

$\begin{array}{c}{\mathrm{Ó\neg }}^2={{\mathrm{Ó\neg }}_0}^2+2\mathrm{Î\pm }(\mathrm{θ}−{\mathrm{θ}}_0)\\ 0^2={(4.7\text{ r²¹»å}/\text{s})}^2+2×(−0.25\text{ r²¹»å}/{\text{s}}^{\text{2}})(\mathrm{θ}−0)\\ \mathrm{θ}=\frac{{(4.7\text{ r²¹»å}/\text{s})}^2}{2×(0.25\text{ r²¹»å}/{\text{s}}^{\text{2}})}\\ =44\text{ r²¹»å}\end{array}$

$\mathrm{The}\mathrm{maximum}\mathrm{angle}\mathrm{of}\mathrm{rotation}\mathrm{in}\mathrm{positive}\mathrm{direction}\mathrm{is},44\text{ r²¹»å}.$

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

$\begin{array}{c}\mathrm{θ}=\frac{1}{2}{\mathrm{θ}}_{\max }\\ =\frac{44\text{ r²¹»å}}{2}\\ \mathrm{θ}=22\text{ r²¹»å}\end{array}$

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

$\begin{array}{c}22\text{ r²¹»å}=(4.7\text{ r²¹»å}/\text{s})\mathrm{t}+\frac{1}{2}(−0.25\text{ r²¹»å}/{\text{s}}^{\text{2}}){\mathrm{t}}^2\\ 22\text{ r²¹»å}=(4.7\text{ r²¹»å}/\text{s})\mathrm{t}−(0.13\text{ r²¹»å}/{\text{s}}^{\text{2}}){\mathrm{t}}^2\\ (0.13\text{ r²¹»å}/{\text{s}}^{\text{2}}){\mathrm{t}}^2−(4.7\text{ r²¹»å}/\text{s})\mathrm{t}+22\text{ r²¹»å}=0\end{array}$

We solve this quadratic equation using the formula

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

For our equation,$\mathrm{a}=(0.13\text{ r²¹»å}/{\text{s}}^{\text{2}}),\mathrm{b}=−(4.7\text{ r²¹»å}/\text{s}),\mathrm{c}=22\text{ r²¹»å}$

Using these values in the above equation, we have

$\begin{array}{c}\mathrm{t}=32.12\text{ s}\\ ≈32\text{ s}\end{array}$

And

$\begin{array}{c}\mathrm{t}=5.48\text{ s}\\ ≈5.5\text{ s}\end{array}$

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

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

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

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

$\begin{array}{c}\mathrm{θ}−{\mathrm{θ}}_0={\mathrm{Ó\neg }}_0\mathrm{t}+\frac{1}{2}{\mathrm{Î\pm ³Ù}}^2\\ −10.5\text{ r²¹»å}=(4.7\text{ r²¹»å}/\text{s})\left(\mathrm{t}\right)+\frac{1}{2}(−0.25\text{ r²¹»å}/{\text{s}}^{\text{2}}){\left(\mathrm{t}\right)}^2\\ −10.5\text{ r²¹»å}=(4.7\text{ r²¹»å}/\text{s})\mathrm{t}−(0.125\text{ r²¹»å}/{\text{s}}^{\text{2}}){\mathrm{t}}^2\\ (0.125\text{ r²¹»å}/{\text{s}}^{\text{2}}){\mathrm{t}}^2−(4.7\text{ r²¹»å}/\text{s})\mathrm{t}−10.5\text{ r²¹»å}=0\end{array}$

We solve this quadratic equation using the formula

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

For our equation,$\mathrm{a}=(0.125\text{ r²¹»å}/{\text{s}}^{\text{2}}),\mathrm{b}=−(4.7\text{ r²¹»å}/\text{s}),\mathrm{c}=−10.5\text{ r²¹»å}$

Using these values in above equation, we get

$\begin{array}{c}\mathrm{t}=−2.115\text{ s}\\ ≈−2.1\text{ s}\end{array}$

And

$\begin{array}{c}t=39.71\text{ s}\\ ≈40\text{ s}\end{array}$

Thus, the flywheel will reach the angle $\mathrm{θ}=10.5\text{ r²¹»å}$ for the negative time at$t=−2.1\text{ s}$ .

The other root of the equation gives us the second (positive) time at which the flywheel will reach the same angle $\mathrm{θ}=10.5\text{ r²¹»å}$so$t=40\text{ s}$.

The graph of $\mathrm{θ}\text{vst}$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.