Question

An astronaut is rotated in a horizontal centrifuge at a radius of $\mathbf{5}\mathbf{.}\mathbf{0}\mathit{m}$. (a) What is the astronaut’s speed if the centripetal acceleration has a magnitude of$\mathbf{7}\mathbf{.}\mathbf{0}\mathit{g}$? (b)How many revolutions per minute are required to produce this acceleration? (c)What is the period of the motion?

Short answer

Answer:

1. The speed of the astronaut is $19m/s$

2. The number of revolutions per minute to produce this acceleration is $35$.

3. The period of the motion is$1.7s$.

Step-by-step solution

The given data

1. Radius of a horizontal centrifuge,$r=5.0m$

2. Magnitude of the centripetal acceleration,$a=7.0gor68.6m/s^2$

Understanding the concept of the acceleration

Using the equation of centripetal acceleration, we can find the velocity of an astronaut. The centripetal acceleration is the acceleration along the radius of the circle and depends on the radius of the circular path and the velocity.

Formulae:

The centripetal acceleration of a body in motion,$a=\frac{v^2}{r}$ …(i)

The time period of a revolution,$T=\frac{2\mathrm{\pi r}}{V}$ …(ii)

$Numberofrevolutionperminute=\frac{60sec}{T}$ …(iii)

(a) Calculation of the astronaut’s speed

By rearranging equation (i), we get the speed of the astronaut by substituting the given values as follows:

$$\begin{gathered} V=\sqrt{a\times r} \\ =\sqrt{\left(68.6m/s^2\right)\times \left(5m\right)} \\ =18.52m/s \\ \approx 19m/s \end{gathered}$$

Hence, the value of the speed is$19m/s$.

(b) Calculation of the revolutions per minute

Using the above value and the given data in equation (ii), we can get the time taken by the astronaut as follows:

$$\begin{gathered} T=\frac{2\mathrm{\pi }\left(5\mathrm{m}\right)}{19m/s} \\ =1.7s \end{gathered}$$

For number of revolution per minute, we can use the given data in equation (iii) as follows:

$$\begin{gathered} N=\frac{60s}{1.7s} \\ =35.3 \\ \approx 35 \end{gathered}$$

Hence, the value of the revolutions per minute is 35.

(c) Calculation of the period of the motion

From the calculations of part (b), we get the period of the motion to be$\mathbf{1}\mathbf{.}\mathbf{7}\mathit{s}$.