An astronaut is tested in a centrifuge with radius $\mathbf{10}\mathbf{\text{\hspace{0.17em}m}}\mathbf{}$and rotating according to $\mathbf{\theta }\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.30}{\mathbf{t}}^{\mathbf{2}}$. At $\mathbf{t}\mathbf{=}\mathbf{}\mathbf{5.0}\mathbf{\text{\hspace{0.17em}s}}\mathbf{}$, what are the magnitudes of the

(a) angular velocity,

(b) linear velocity,

(c) tangential acceleration,

and (d) radial acceleration?

1. The angular position of the astronaut$\mathrm{\theta }=0.30{\mathrm{t}}^2$
2. The radius of centrifuge r is, $10\text{\hspace{0.17em}m}$.

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.

$\begin{array}{c}\mathbf{\omega }\mathbf{=}\mathbf{}\frac{\mathbf{d\theta }}{\mathbf{dt}}\\ \mathbf{v}\mathbf{=}\mathbf{}\mathbf{r}\mathbf{}\mathbf{\omega }\\ {\mathbf{a}}_{\mathbf{t}}\mathbf{=}\mathbf{}\mathbf{\alpha }\mathbf{}\mathbf{r}\\ {\mathbf{a}}_{\mathbf{r}}\mathbf{=}\mathbf{}\mathbf{r}\mathbf{}{\mathbf{\omega }}^{\mathbf{2}}\end{array}$

The definition of angular velocity helps us determine it as follows

$\begin{array}{c}\mathrm{\omega }=\frac{\mathrm{d\theta }}{\mathrm{dt}}\\ =\frac{\mathrm{d}}{\mathrm{dt}}\left(0.30{\mathrm{t}}^2\right)\\ =0.30\times 2\mathrm{t}\\ =\frac{\mathrm{d}}{\mathrm{dt}}\left(0.30{\mathrm{t}}^2\right)\\ =0.30\times 2\mathrm{t}\end{array}$

At$\mathrm{t}=5.0\text{\hspace{0.17em}s}$, we get

$\begin{array}{c}\mathrm{\omega }=0.30\times 2\mathrm{t}\\ =0.30\times 2\times 5.0\text{\hspace{0.17em}s}\\ \mathrm{\omega }=3.00\text{\hspace{0.17em}rad}/\text{s}\end{array}$

Hence the magnitude of the linear velocity is, $3.00\text{\hspace{0.17em}rad}/\text{s}$.

Now we calculate linear velocity using the value of angular velocity

$\mathrm{v}=\mathrm{r}\mathrm{\omega }$

Substitute all the value in the above equation.

$\begin{array}{c}\mathrm{v}=10\text{\hspace{0.17em}m}\times 3.00\text{\hspace{0.17em}rad}/\text{s}\\ \mathrm{v}=30\text{\hspace{0.17em}m}/\text{s}\end{array}$

Hence the magnitude of the linear velocity is, $30\text{\hspace{0.17em}m}/\text{s}$.

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

$\begin{array}{c}\mathrm{\alpha }=\frac{{\mathrm{d}}^2\mathrm{\theta }}{\mathrm{d}{\mathrm{t}}^2}\\ =\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{d\theta }}{\mathrm{dt}}\right)\\ \mathrm{\alpha }=\frac{\mathrm{d\omega }}{\mathrm{dt}}\end{array}$

$\begin{array}{c}\mathrm{\alpha }=\frac{\mathrm{d}}{\mathrm{dt}}(0.30\times 2\mathrm{t})\\ \mathrm{\alpha }=0.60\text{\hspace{0.17em}rad}/{\text{s}}^{\text{2}}\end{array}$

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

Now, the tangential component of the acceleration is given as

${\mathrm{a}}_{\mathrm{t}}=\mathrm{\alpha }\mathrm{r}$

Substitute all the value in the above equation.

$\begin{array}{c}{\mathrm{a}}_{\mathrm{t}}=0.60\text{\hspace{0.17em}rad}/{\text{s}}^{\text{2}}\times 10\text{\hspace{0.17em}m}\\ {\mathrm{a}}_{\mathrm{t}}=6.0\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\end{array}$

Hence the magnitude of tangential acceleration is, $6.0\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}$.

The radial component of the acceleration is calculated as

${\mathrm{a}}_{\mathrm{r}}=\mathrm{r}{\mathrm{\omega }}^2$

At $\mathrm{t}=5.0\text{\hspace{0.17em}s},\mathrm{\omega }=3.0\text{\hspace{0.17em}rad}/\text{s}$

${\mathrm{a}}_{\mathrm{r}}=\mathrm{r}{\mathrm{\omega }}^2$

Substitute all the value in the above equation.

$\begin{array}{c}{\mathrm{a}}_{\mathrm{r}}=\left(10\text{\hspace{0.17em}m}\right)\times {(3.0\text{\hspace{0.17em}rad}/\text{s})}^2\\ {\mathrm{a}}_{\mathrm{r}}=90\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\end{array}$

Hence the magnitude of radial acceleration is, $90\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}$.