91Ó°ÊÓ

g = acceleration due to gravity$=9.81\frac{m}{s^2}$

a = acceleration$=\frac{1}{8}g$

d = diameter of the wheel$=2.50$

r = radius of the wheel$=\frac{2.50}{2}$

v = speed of elevator$=1.25=25\frac{cm}{s}$

here we solve our question first we need to find angular velocity.

We know that the angular velocity formula,

$\mathrm{Ó\neg }=\frac{v}{r}$

$$\begin{gathered} \mathrm{Ó\neg }=\frac{0.25}{1.25}\frac{rad}{s} \\ =0.2×\frac{60}{2\mathrm{Ï€}} \end{gathered}$$

We know that the value of$\mathrm{Ï€}=3.14$

$$\begin{gathered} =0.2×\frac{60}{2×3.14} \\ =\frac{12}{6.28} \\ =1.91rad \end{gathered}$$

The angular speed of the wheel is $1.91rad$.

g = acceleration due to gravity$=9.81\frac{m}{s^2}$

a = acceleration$=\frac{1}{8}g$

d = diameter of the wheel$=2.50$

r = radius of the wheel$=\frac{2.50}{2}$

We know that the formula of the angular acceleration,

$\mathrm{Î\pm }=\frac{a}{r}$

We know that value of $g=9.81$and $r=1.25$.

Put the value in the expression,then we get

$$\begin{gathered} \mathrm{Î\pm }=\frac{{\displaystyle \frac{1}{8}}×g}{1.25} \\ \mathrm{Î\pm }=\frac{0.125×9.81}{1.25} \\ =\frac{1.22625}{1.25} \\ =0.981\frac{rad}{s^2} \end{gathered}$$

The angular acceleration of the wheel is $0.981\frac{rad}{s^2}$.

r = radius of the wheel$=\frac{2.50}{2}$

s = displacement = 3.25

We know that $\mathrm{θ}=\frac{s}{r}$.

$$\begin{gathered} =\frac{3.25}{1.25} \\ =2.6rad \end{gathered}$$

we know that$1rad=\frac{360}{2\mathrm{Ï€}}$.

$$\begin{gathered} =2.6×\frac{360}{2\mathrm{π}} \\ =2.6×\frac{36}{2×3.14} \\ =2.6×\frac{36}{6.25} \\ =\frac{936}{2.25} \\ =149.{76}^0 \end{gathered}$$

The angular displacement of the wheel $149.{76}^0$.