91Ó°ÊÓ

The angular position of a point on the rim of a rotating wheel is

$\mathrm{θ}=4.0\mathrm{t}\left(\frac{\text{rad}}{\text{s}}\right)−3.0{\mathrm{t}}^2\left(\frac{\text{rad}}{{\text{s}}^{\text{2}}}\right)+{\mathrm{t}}^3\left(\frac{\text{rad}}{{\text{s}}^{\text{3}}}\right)$

To find the angular velocity and angular acceleration, we can take the first and second derivative respectively. We can use the expression of average angular acceleration and find its value.

Formulae:

$\mathrm{Ó\neg }=\frac{\text{d}\mathrm{θ}}{\text{dt}}$

${\mathrm{Î\pm }}_{\mathrm{avg}}=\frac{{\mathrm{Ó\neg }}_{\mathrm{f}}−{\mathrm{Ó\neg }}_{\mathrm{i}}}{\text{dt}}$

$\mathrm{Î\pm }=\frac{\text{d}\mathrm{Ó\neg }}{\text{dt}}$

The angular position of a point on the rim of a rotating wheel is

$\mathrm{θ}=4.0\mathrm{t}\left(\frac{\text{rad}}{\text{s}}\right)−3.0{\mathrm{t}}^2\left(\frac{\text{rad}}{{\text{s}}^{\text{2}}}\right)+{\mathrm{t}}^3\left(\frac{\text{rad}}{{\text{s}}^{\text{3}}}\right)$

The angular velocity of a point on the rim of a rotating wheel is

$\begin{array}{c}\mathrm{Ó\neg }=\frac{\mathrm{»åθ}}{\mathrm{dt}}\\ =\frac{\mathrm{d}\left[4.0\mathrm{t}\left(\frac{\mathrm{rad}}{\mathrm{s}}\right)−3.0{\mathrm{t}}^2\left(\frac{\mathrm{rad}}{{\mathrm{s}}^2}\right)+{\mathrm{t}}^3\left(\frac{\mathrm{rad}}{{\mathrm{s}}^3}\right)\right]}{\text{dt}}\\ =4.0−6.0\mathrm{t}+3{\mathrm{t}}^2\end{array}$

The angular velocity of the point:

$\begin{array}{c}\mathrm{Ó\neg }=4.0−6.0\mathrm{t}+3{\mathrm{t}}^2\\ {\mathrm{Ó\neg }}_2=4.0−6.0\left(2.0\text{s}\right)+3{(2.0\text{s})}^2\\ =4.0\text{ }\frac{\text{rad}}{\text{s}}\end{array}$

The angular velocity at $\text{t}=2.0\text{s}$ is ${\mathrm{Ó\neg }}_2=4.0\text{ }\frac{\text{rad}}{\text{s}}$

The angular velocity of the point is

$\begin{array}{l}\mathrm{Ó\neg }=4.0−6.0\mathrm{t}+3{\mathrm{t}}^2\\ {\mathrm{Ó\neg }}_4=4.0−6.0(4.0\text{s})+3{(4.0\text{s})}^2\\ {\mathrm{Ó\neg }}_4=28\text{ }\frac{\text{rad}}{\text{s}}\end{array}$

The angular velocity at $\text{t}=4.0\text{s}$ is ${\mathrm{Ó\neg }}_4=28\text{ }\frac{\text{rad}}{\text{s}}$

$\begin{array}{c}{\mathrm{Î\pm }}_{\mathrm{avg}}=\frac{{\mathrm{Ó\neg }}_{\mathrm{f}}−{\mathrm{Ó\neg }}_{\mathrm{i}}}{\text{dt}}\\ =\frac{{\mathrm{Ó\neg }}_4−{\mathrm{Ó\neg }}_2}{{\text{t}}_4−{\text{t}}_2}\end{array}$

Substitute the values and solve as:

$\begin{array}{c}{\mathrm{Î\pm }}_{\mathrm{avg}}=\frac{28\text{ }\frac{\text{rad}}{\text{s}}−4.0\text{ }\frac{\text{rad}}{\text{s}}}{(4.0\text{s})−(2.0\text{s})}\\ =12\text{ }\frac{\text{rad}}{{\text{s}}^2}\end{array}$

The average angular acceleration for the time interval from $\mathrm{t}=2.0\text{s}$ to the $\mathrm{t}=4.0\text{s}$ is ${\mathrm{Î\pm }}_{\mathrm{avg}}=12\text{ }\frac{\text{rad}}{{\text{s}}^2}$

The instantaneous angular acceleration of the point is

The instantaneous angular acceleration of the point is

$\begin{array}{c}\mathrm{Î\pm }=−6.0+6.0\mathrm{t}\\ {\mathrm{Î\pm }}_2=−6.0+6.0(2.0\text{s})\\ {\mathrm{Î\pm }}_2=6.0\text{ }\frac{\text{rad}}{{\text{s}}^2}\end{array}$

The instantaneous angular acceleration at the beginning is ${\mathrm{Î\pm }}_2=6.0\text{ }\frac{\text{rad}}{{\text{s}}^2}$

The instantaneous angular acceleration at the end of the time interval:

$\begin{array}{l}\mathrm{Î\pm }=−6.0+6.0\mathrm{t}\\ {\mathrm{Î\pm }}_4=−6.0+6.0(4.0\text{s})\\ {\mathrm{Î\pm }}_4=18\text{ }\frac{\text{rad}}{{\text{s}}^2}\end{array}$

The instantaneous angular acceleration at the end of the time interval ${\mathrm{Î\pm }}_4=18\text{ }\frac{\text{rad}}{{\text{s}}^2}$