91Ó°ÊÓ

1. The angular position of the astronaut$\mathrm{θ}=0.30{\mathrm{t}}^2$
2. The radius of centrifuge r is, $10\text{″¾}$.

The astronaut in the centrifuge undergoes rotational motion. Hence, we need to use the equations relating angular and linear variables to determine the required quantities which are given below.

$\begin{array}{c}\mathbf{Ó\neg }\mathbf{=}\mathbf{}\frac{\mathbf{»åθ}}{\mathbf{dt}}\\ \mathbf{v}\mathbf{=}\mathbf{}\mathbf{r}\mathbf{}\mathbf{Ó\neg }\\ {\mathbf{a}}_{\mathbf{t}}\mathbf{=}\mathbf{}\mathbf{Î\pm }\mathbf{}\mathbf{r}\\ {\mathbf{a}}_{\mathbf{r}}\mathbf{=}\mathbf{}\mathbf{r}\mathbf{}{\mathbf{Ó\neg }}^{\mathbf{2}}\end{array}$

The definition of angular velocity helps us determine it as follows

$\begin{array}{c}\mathrm{Ó\neg }=\frac{\mathrm{»åθ}}{\mathrm{dt}}\\ =\frac{\mathrm{d}}{\mathrm{dt}}\left(0.30{\mathrm{t}}^2\right)\\ =0.30×2\mathrm{t}\\ =\frac{\mathrm{d}}{\mathrm{dt}}\left(0.30{\mathrm{t}}^2\right)\\ =0.30×2\mathrm{t}\end{array}$

At$\mathrm{t}=5.0\text{ s}$, we get

$\begin{array}{c}\mathrm{Ó\neg }=0.30×2\mathrm{t}\\ =0.30×2×5.0\text{ s}\\ \mathrm{Ó\neg }=3.00\text{ r²¹»å}/\text{s}\end{array}$

Hence the magnitude of the linear velocity is, $3.00\text{ r²¹»å}/\text{s}$.

Now we calculate linear velocity using the value of angular velocity

$\mathrm{v}=\mathrm{r}\mathrm{Ó\neg }$

Substitute all the value in the above equation.

$\begin{array}{c}\mathrm{v}=10\text{″¾}×3.00\text{ r²¹»å}/\text{s}\\ \mathrm{v}=30\text{″¾}/\text{s}\end{array}$

Hence the magnitude of the linear velocity is, $30\text{″¾}/\text{s}$.

First, we will determine the angular acceleration of the point on the object. It is given by the equation.

$\begin{array}{c}\mathrm{Î\pm }=\frac{{\mathrm{d}}^2\mathrm{θ}}{\mathrm{d}{\mathrm{t}}^2}\\ =\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{»åθ}}{\mathrm{dt}}\right)\\ \mathrm{Î\pm }=\frac{\mathrm{dÓ\neg }}{\mathrm{dt}}\end{array}$

$\begin{array}{c}\mathrm{Î\pm }=\frac{\mathrm{d}}{\mathrm{dt}}(0.30×2\mathrm{t})\\ \mathrm{Î\pm }=0.60\text{ r²¹»å}/{\text{s}}^{\text{2}}\end{array}$

It is clear that this angular acceleration is constant, i.e., independent of time

Now, the tangential component of the acceleration is given as

${\mathrm{a}}_{\mathrm{t}}=\mathrm{Î\pm }\mathrm{r}$

Substitute all the value in the above equation.

$\begin{array}{c}{\mathrm{a}}_{\mathrm{t}}=0.60\text{ r²¹»å}/{\text{s}}^{\text{2}}×10\text{″¾}\\ {\mathrm{a}}_{\mathrm{t}}=6.0\text{″¾}/{\text{s}}^{\text{2}}\end{array}$

Hence the magnitude of tangential acceleration is, $6.0\text{″¾}/{\text{s}}^{\text{2}}$.

The radial component of the acceleration is calculated as

${\mathrm{a}}_{\mathrm{r}}=\mathrm{r}{\mathrm{Ó\neg }}^2$

At $\mathrm{t}=5.0\text{ s},\mathrm{Ó\neg }=3.0\text{ r²¹»å}/\text{s}$

${\mathrm{a}}_{\mathrm{r}}=\mathrm{r}{\mathrm{Ó\neg }}^2$

Substitute all the value in the above equation.

$\begin{array}{c}{\mathrm{a}}_{\mathrm{r}}=\left(10\text{″¾}\right)×{(3.0\text{ r²¹»å}/\text{s})}^2\\ {\mathrm{a}}_{\mathrm{r}}=90\text{″¾}/{\text{s}}^{\text{2}}\end{array}$

Hence the magnitude of radial acceleration is, $90\text{″¾}/{\text{s}}^{\text{2}}$.