91Ó°ÊÓ

The angular displacement of the body at t = 0 is $\mathrm{θ}=\frac{\mathrm{π}}{4}rad$.

The angular speed of the body at t = 0 is $\mathrm{Ó\neg }=2\mathrm{rad}/\mathrm{s}$.

The angular acceleration of the body at t = 1.5 s is$\mathrm{Î\pm }=1.25\mathrm{rad}/{\mathrm{s}}^2$ .

The angular velocity of the body in the time is the ratio of the angular displacement$\mathit{d}\mathit{θ}$ to dt.

$\mathit{Ó\neg }\mathbf{=}\frac{\mathbf{d}\mathbf{θ}}{\mathbf{d}\mathbf{t}}$

The angular acceleration is given by,

$\mathit{θ}\mathbf{=}\frac{\mathbf{dÓ\neg }}{\mathbf{dt}}$

Consider the given equation of displacement.

$\mathrm{θ}\left(\mathrm{t}\right)=\mathrm{a}+\mathrm{bt}-{\mathrm{ct}}^3$

Substitute $\mathrm{θ}=\frac{\mathrm{π}}{4}\mathrm{rad}$, and t = 0 in equation $\mathrm{θ}\left(t\right)=a+bt-c{t}^3$.

$$\begin{gathered} \frac{\mathrm{Ï€}}{4}rad=a+b\left(0\right)-c{\left(0\right)}^3 \\ a=\frac{\mathrm{Ï€}}{4}rad \end{gathered}$$

Find the expression for angular velocity.

$$\begin{gathered} \mathrm{Ó\neg }=\frac{\mathrm{d}}{\mathrm{dt}}\left[\mathrm{a}+\mathrm{bt}-{\mathrm{ct}}^3\right] \\ =\mathrm{b}-3{\mathrm{ct}}^2.........\left(1\right) \end{gathered}$$

Substitute$\mathrm{Ó\neg }=2rad/s$, and t = 0 in equation (1).

$$\begin{gathered} 2\mathrm{rad}/\mathrm{s}=\mathrm{b}-3\mathrm{c}\left(0\right) \\ \mathrm{b}=\mathrm{rad}/\mathrm{s} \end{gathered}$$

Find the expression for angular acceleration.

$$\begin{gathered} \mathrm{Î\pm }=\frac{\mathrm{d}}{\mathrm{dt}}\left[\mathrm{b}-3{\mathrm{ct}}^2\right] \\ =-6\mathrm{ct}........\left(2\right) \end{gathered}$$

Substitute$\mathrm{Î\pm }=1.25\mathrm{rad}/{\mathrm{s}}^2$, and t = 1.5 s in equation (2).

$$\begin{gathered} 1.25\mathrm{rad}/{\mathrm{s}}^2=-6\mathrm{c}\left(1.5\mathrm{s}\right) \\ \mathrm{c}=-\frac{1.25\mathrm{rad}/{\mathrm{s}}^2}{9\mathrm{s}} \\ =-0.139\mathrm{rad}/{\mathrm{s}}^3 \end{gathered}$$

Therefore, the values of a, b, and c are $\frac{\mathrm{Ï€}}{4}\mathrm{rad}$, 2 rad/s , and $-0.139\mathrm{rad}/{\mathrm{s}}^3$, respectively.

It is known that$\mathrm{θ}=\frac{\mathrm{π}}{4}rad$ at t = 0 .

Find the angular acceleration at t = 0 as follows.

Substitute t = 0 in equation (2).

$$\begin{gathered} \mathrm{Î\pm }=-6\mathrm{c}\left(0\right) \\ =0\mathrm{rad}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the required angular acceleration is 0 rad/${\mathrm{s}}^2$.

Substitute $\mathrm{Î\pm }=3.5\mathrm{rad}/{\mathrm{s}}^2$and$\mathrm{c}=0.139\mathrm{rad}/{\mathrm{s}}^3$ in the equation (2).

$$\begin{gathered} 3.5\mathrm{rad}/{\mathrm{s}}^2=-6(-0.139\mathrm{rad}/{\mathrm{s}}^3)\left(\mathrm{t}\right) \\ \mathrm{t}=\frac{3.5\mathrm{rad}/{\mathrm{s}}^2}{6\left(0.139\mathrm{rad}/{\mathrm{s}}^3\right)} \\ =4.2\mathrm{s} \end{gathered}$$

Substitute $\mathrm{t}=4.2\mathrm{s},\mathrm{b}=2\mathrm{rad}/\mathrm{s}$and$\mathrm{c}=0.139\mathrm{rad}/{\mathrm{s}}^3$ in the equation (1).

$$\begin{gathered} \mathrm{Ó\neg }=2\mathrm{rad}/\mathrm{s}-3\left(0.139\mathrm{rad}/{\mathrm{s}}^3\right){\left(4.2\mathrm{s}\right)}^2 \\ =9.36\mathrm{rad}/\mathrm{s} \end{gathered}$$

Substitute $\mathrm{t}=4.2\mathrm{s},\mathrm{a}=\frac{\mathrm{Ï€}}{4}\mathrm{rad},\mathrm{b}=2\mathrm{rad}/\mathrm{s}$ and$\mathrm{c}=0.139\mathrm{rad}/{\mathrm{s}}^3$ in the equation .

$$\begin{gathered} \mathrm{θ}\left(\mathrm{t}\right)=\mathrm{a}+\mathrm{bt}-{\mathrm{ct}}^3 \\ \mathrm{θ}\left(\mathrm{t}\right)=\frac{\mathrm{π}}{4}\mathrm{rad}+\left(2\mathrm{rad}/\mathrm{s}\right)\left(4.2\mathrm{s}\right)-\left(-0.139\mathrm{rad}/{\mathrm{s}}^3\right){\left(4.2\mathrm{s}\right)}^3 \\ =19.5\mathrm{rad} \end{gathered}$$

Therefore, the value of$\mathrm{θ}$ is 19.5 rad, and angular speed is 9.36 rad/s .