A child is pushing a merry-go-round. The angle through which the merry-go-round has turned varies with time according to $\mathit{\theta }\left(t\right)\mathbf{=}\mathit{\gamma }\mathit{t}\mathbf{+}\mathit{\beta }{\mathit{t}}^{\mathbf{3}}$, where $\mathit{\gamma }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{400}\mathbf{}\mathbf{rad}\mathbf{/}\mathbf{s}$and $\mathbf{\beta }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0120}\mathbf{rad}\mathbf{/}{\mathbf{s}}^{\mathbf{3}}$. (a) Calculate the angular velocity of the merry-go-round as a function of time. (b) What is the initial value of the angular velocity ${\mathbf{\omega }}_{\mathbf{z}}\mathbf{}\mathbf{at}\mathbf{}\mathbf{t}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{s}$? (c) Calculate the instantaneous value of the angular velocity at and the average angular velocity for the time interval t = 0 to t = 5.00 s . Show that ${\mathbf{\omega }}_{\mathbf{av}\mathbf{-}\mathbf{z}}$is not equal to the average of the instantaneous angular velocities at t = 0 to t = 5.00 s , and explain.

The equation of angular displacement is as follows.

$\theta \left(t\right)=\gamma t+\beta t^3$

Where, $\mathrm{\gamma }=0.400\mathrm{rad}/\mathrm{s}$, and $\mathrm{\beta }=0.0120\mathrm{rad}/{\mathrm{s}}^3$.

The average angular velocity of the body in the time interval $\mathbf{∆}\mathbf{t}\mathbf{=}{\mathbf{t}}_{\mathbf{2}}\mathbf{-}{\mathbf{t}}_{\mathbf{1}}$is the ratio of the angular displacement to role="math" localid="1667969695670" $\mathbf{∆}\mathit{\theta }\mathbf{=}{\mathit{\theta }}_{\mathbf{2}}\mathbf{-}{\mathit{\theta }}_{\mathbf{1}}\mathbf{∆}\mathbf{t}$

${\mathbf{\omega }}_{\mathbf{av}\mathbf{-}\mathbf{z}}\mathbf{=}\frac{{\mathbf{\theta }}_{\mathbf{2}}\mathbf{-}{\mathbf{\theta }}_{\mathbf{1}}}{{\mathbf{t}}_{\mathbf{2}}\mathbf{-}{\mathbf{t}}_{\mathbf{1}}}$

The angular velocity is given by,

$\omega \left(t\right)=\frac{d\left[\theta \left(t\right)\right]}{dt}$

Substitute$\theta \left(t\right)=\gamma t+\beta t^3$ in the above equation.

role="math" localid="1667969866879" $$\begin{gathered} \omega \left(t\right)=\frac{d}{dt}\left[\gamma t+\beta t^3\right] \\ =\gamma +3\gamma t^2...........\left(1\right) \end{gathered}$$

Therefore, the angular velocity as a function of time is $\omega \left(t\right)=\gamma +3\beta t^2$.

Substitute t = 0 s in equation (1) to find initial value of angular velocity.

$$\begin{gathered} \omega \left(0\right)=\gamma +3\beta {\left(0\right)}^2 \\ =\gamma \\ =0.400\mathrm{rad}/\mathrm{s} \end{gathered}$$

Therefore, the initial value of angular velocity is 0.400 rad/s.

Substitute t = 5 s ,$\gamma =0.400\mathrm{rad}/\mathrm{s}$, and$\mathrm{\beta }=0.0120\mathrm{rad}/{\mathrm{s}}^3$in equation (1) to find instantaneous angular velocity.

role="math" localid="1667970286412" $$\begin{gathered} {\mathrm{\omega }}_{\mathrm{z}}{|}_{\mathrm{t}=5\mathrm{s}}=0.400\mathrm{rad}/\mathrm{s}+3(0.0120\mathrm{rad}/{\mathrm{s}}^3){(5\mathrm{s})}^2 \\ =1.3\mathrm{rad}/\mathrm{s} \end{gathered}$$

Find the angular displacement at t = 0 .

$$\begin{gathered} {\theta }_1=\gamma \times 0\times \beta \times 0 \\ =0 \end{gathered}$$

Find the angular displacement at t = 5 s .

$$\begin{gathered} {\mathrm{\theta }}_2=(0.400\mathrm{rad}/\mathrm{s})\times \left(5\mathrm{s}\right)+\left(0.0120\mathrm{rad}/{\mathrm{s}}^3\right)\times {\left(5\mathrm{s}\right)}^3 \\ =3.5\mathrm{rad} \end{gathered}$$

Find the average angular velocity as follows.

$$\begin{gathered} {\mathrm{\omega }}_{\mathrm{av}-\mathrm{z}}=\frac{{\mathrm{\theta }}_2-{\mathrm{\theta }}_1}{{\mathrm{t}}_2-{\mathrm{t}}_1} \\ =\frac{3.5\mathrm{rad}-0\mathrm{rad}}{5\mathrm{s}-0\mathrm{s}} \\ =0.7\mathrm{rad}/\mathrm{s} \end{gathered}$$

Substitute t = 0 s ,$\mathrm{\gamma }=0.400\mathrm{rad}/\mathrm{s}$, and$\mathrm{\beta }=0.0120\mathrm{rad}/{\mathrm{s}}^3$in equation (1) to find instantaneous angular velocity.

role="math" localid="1667970733387" $$\begin{gathered} {\mathrm{\omega }}_{\mathrm{z}}{|}_{\mathrm{t}=0\mathrm{s}}=0.400\mathrm{rad}/\mathrm{s}+3\left(0.0120\mathrm{rad}/{\mathrm{s}}^3\right){\left(0\right)}^2 \\ =0.40\mathrm{rad}/\mathrm{s} \end{gathered}$$

Show that,role="math" localid="1667970866050" ${\omega }_{av-z}$is not equal to the average of the instantaneous angular velocities.

$$\begin{gathered} {\mathrm{\omega }}_{\mathrm{inst}-\mathrm{av}}=\frac{{\mathrm{\omega }}_1+{\mathrm{\omega }}_2}{2} \\ =\frac{1.3+0.4}{2} \\ =0.85\mathrm{rad}/\mathrm{s} \end{gathered}$$

Therefore, it is shown that${\omega }_{av-z}$ is not equal to the average of the instantaneous angular velocities.

The average of the instantaneous angular velocities is larger than the average angular velocity because the instantaneous angular velocities does not increase linearly but it increase faster as $t^2$.

Therefore, the instantaneous angular velocity is , the average angular velocity is 1.3 rad/s , and is 0.7 rad/s and ${\omega }_{av-z}$not equal to the average of the instantaneous angular velocities.