College Physics by Openstax Chapter 6 Problem 29

The centripetal acceleration of a large centrifuge as experienced in rocket launches and atmospheric reentries of astronauts

Problem:

A large centrifuge, like the one shown in Figure 6.34(a), is used to expose aspiring astronauts to accelerations similar to those experienced in rocket launches and atmospheric reentries.

(a) At what angular velocity is the centripetal acceleration 10g10g if the rider is 15.0 m from the center of rotation?

(b) The rider’s cage hangs on a pivot at the end of the arm, allowing it to swing outward during rotation as shown in Figure 6.34(b). At what angle θ\theta below the horizontal will the cage hang when the centripetal acceleration is  10g10g? (Hint: The arm supplies centripetal force and supports the weight of the cage. Draw a free body diagram of the forces to see what the angle 10g10g should be.)

Figure 6.34 (a) NASA centrifuge used to subject trainees to accelerations similar to those experienced in rocket launches and reentries. (credit: NASA) (b) Rider in cage showing how the cage pivots outward during rotation. This allows the total force exerted on the rider by the cage to always be along its axis.

Solution:

Part A

The centripetal acceleration, aca_c, is calculated using the formula ac=rω2a_c = r \omega ^2. Solving for the angular velocity, ω\omega, in terms of the other variables, we should come up with

ω=acr\omega = \sqrt{\frac{a_c}{r}}

We are given the following values:

  • centripetal acceleration, ac=10g=10(9.81 m/s2)=98.1 m/s2a_c = 10g = 10 \left( 9.81\ \text{m/s}^2 \right) = 98.1\ \text{m/s}^2
  • radius of curvature, r=15.0 mr = 15.0\ \text{m}

Substituting the given values into the equation,

ω=acrω=98.1 m/s215.0 mω=2.5573 rad/secω=2.56 rad/sec  (Answer)\begin{align*} \omega & = \sqrt{\frac{a_c}{r}} \\ \\ \omega & = \sqrt{\frac{98.1\ \text{m/s}^2}{15.0\ \text{m}}} \\ \\ \omega & = 2.5573\ \text{rad/sec} \\ \\ \omega & = 2.56\ \text{rad/sec} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

The free-body diagram of the force is shown

The free-body diagram of the rider’s cage that hangs on a pivot at the end of the arm of a large centrifuge. College Physics Problem 6-29
The free-body diagram of the rider’s cage hangs on a pivot at the end of the arm of a large centrifuge.

Summing forces in the vertical direction, we have

∑Fy=0Farmsin⁡θ−w=0Farm=wsin⁡θ Equation 1\begin{align*} \sum_{}^{} F_y & = 0 \\ \\ F_{arm} \sin \theta-w & = 0 \\ \\ F_{arm} & = \frac{w}{\sin \theta} \ \quad \quad \color{Blue} \text{Equation 1} \end{align*}

Now, summing forces in the horizontal direction, taking into account that FcF_c is the centripetal force which is the net force. That is,

Fc=mac\begin{align*} F_c & = m a_c \end{align*}

We know that FcF_c is equal to the horizontal component of the force FarmF_{arm}. That is Fc=Farmcos⁡θF_c = F_{arm} \cos \theta. Therefore,

Farmcos⁡θ=mac\begin{align*} F_{arm} \cos \theta & = m a_c \\ \end{align*}

Now, we can substitute equation 1 into the equation, and the value of the centripetal acceleration given at 10g10g. Also, we note that the weight ww is equal to mgmg. So, we have

Farmcos⁡θ=macwsin⁡θcos⁡θ=m(10g)mgcos⁡θsin⁡θ=10mg\begin{align*} F_{arm} \cos \theta & = m a_c \\ \\ \frac{w}{\sin \theta} \cos \theta & = m (10g) \\ \\ \frac{mg \cos \theta}{\sin \theta} & = 10 mg \\ \\ \end{align*}

From here, we are going to use the trigonometric identity tan⁡θ=sin⁡θcos⁡θ\displaystyle \tan \theta = \frac{\sin \theta}{\cos \theta}. We can also cancel mm, and g since they can be found on both sides of the equation.

1tan⁡θ=10tan⁡θ=110θ=tan⁡−1(110)θ=5.71∘  (Answer)\begin{align*} \frac{1}{\tan \theta} & = 10 \\ \\ \tan \theta & = \frac{1}{10} \\ \\ \theta & = \tan ^{-1} \left( \frac{1}{10} \right) \\ \\ \theta & = 5.71 ^\circ \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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