Question

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?

Short answer

1. The magnitude of the angular velocity is$3.00\text{\hspace{0.17em}rad}/\text{s}$.
2. The magnitude of the linear velocity is$30\text{\hspace{0.17em}m}/\text{s}$.
3. The magnitude of the tangential acceleration is $6.0\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}$.
4. The magnitude of the radial acceleration is $90\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}$.

Step-by-step solution

Understanding the given information 

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

Concept and Formula used for the given question

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.

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

(a) Calculation for the magnitude of angular velocity

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, $3.00\text{\hspace{0.17em}rad}/\text{s}$.

(b) Calculation for the magnitude of linear velocity

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, $30\text{\hspace{0.17em}m}/\text{s}$.

(c) Calculation for the magnitude of tangential acceleration

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, $6.0\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}$.

(d) Calculation for the magnitude of radical acceleration

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, $90\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}$.