91Ó°ÊÓ

The angular velocity of the flywheel obeys the following equation.

${\mathrm{Ó\neg }}_z\left(t\right)=A+B{t}^2$

Where, A = 2.75 and B = 1.50.

The first radius is$r_a=2.5m$ .

$\mathbf{Î\pm }\left(t\right)\mathbf{=}\frac{{\mathbf{dÓ\neg }}_{\mathbf{z}}\left(t\right)}{\mathbf{dt}}$The derivative of the angular velocity is called angular acceleration.

Here,$\mathbf{Î\pm }\left(t\right)$ is angular acceleration, and${\mathbf{Ó\neg }}_{\mathbf{z}}\left(t\right)$ is angular velocity.

The angular acceleration is given by,

${\mathrm{Î\pm }}_z\left(t\right)=\frac{d{\mathrm{Ó\neg }}_z\left(t\right)}{dt}$

Substitute ${\mathrm{Ó\neg }}_z\left(t\right)=\mathrm{γ}-\mathrm{β}{t}^2$in the above equation.

$$\begin{gathered} {\mathrm{Î\pm }}_z\left(t\right)=\frac{d}{dt}\left[\mathrm{γ}-\mathrm{β}{t}^2\right] \\ =-2\mathrm{β}t \\ =-2\left(0.8\right)t \\ =-1.6t \end{gathered}$$ .......................(1)

Therefore, the angular acceleration as a function of time is ${\mathrm{Î\pm }}_z\left(t\right)=-1.6t$.

Substitute in equation .

$$\begin{gathered} {\mathrm{Î\pm }}_{\mathrm{z}}\left(\mathrm{t}\right)=-1.6\left(3\mathrm{s}\right) \\ =-4.8\mathrm{rad}/{\mathrm{s}}^2 \end{gathered}$$

Find the average angular acceleration as follows.

$$\begin{gathered} {\mathrm{Î\pm }}_{\mathrm{av}-\mathrm{z}}=\frac{{\mathrm{Ó\neg }}_{\mathrm{z}}\left({\mathrm{t}}_1\right)-{\mathrm{Ó\neg }}_{\mathrm{z}}\left({\mathrm{t}}_0\right)}{{\mathrm{t}}_1-{\mathrm{t}}_0} \\ =\frac{\left(\mathrm{γ}-{\mathrm{ββ³Ù}}_1^2\right)-\left(\mathrm{γ}-{\mathrm{ββ³Ù}}_0^2\right)}{{\mathrm{t}}_1-{\mathrm{t}}_0} \\ =-\mathrm{β}\left(\frac{{\mathrm{t}}_1^2-{\mathrm{t}}_0^2}{{\mathrm{t}}_1-{\mathrm{t}}_0}\right) \end{gathered}$$

Substitute all the values in the above expression.

$$\begin{gathered} {\mathrm{Î\pm }}_{\mathrm{av}-\mathrm{z}}=-\left(0.8\mathrm{rad}/{\mathrm{s}}^3\right)\left(\frac{3^2-0^2}{3-0}\right) \\ =-2.4\mathrm{rad}/{\mathrm{s}}^2 \end{gathered}$$

The average angular acceleration and the instantaneous angular acceleration are different, and the instantaneous angular acceleration is twice the size of average angular acceleration. They are different because the acceleration is not constant, and it is changing linearly with time.

Therefore, the instantaneous angular acceleration is $-4.8\mathrm{rad}/{\mathrm{s}}^2$, and average angular acceleration is $-2.4\mathrm{rad}/{\mathrm{s}}^2$.