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Q6.3-29 PE

Expert-verifiedFound in: Page 222

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**A large centrifuge, like the one shown in Figure (a), is used to expose aspiring astronauts to accelerations similar to those experienced in rocket launches and atmospheric re-entries. (a) At what angular velocity is the centripetal acceleration \(10g\) if the rider is \(15.0{\rm{ m}}\) from the centre 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 (b). At what angle \(\theta \) below the horizontal will the cage hang when the centripetal acceleration is \(10g\)? (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 \(\theta \) should be.)**

***Figure (a) NASA centrifuge used to subject trainees to accelerations similar to those experienced in rocket launches and re-entries. (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 be along its axis at all times.**

**The angular velocity is \(2.55{\rm{ rad}}/{\rm{s}}\).****The angle below the horizontal is \(5.71^\circ \).**

**Angular velocity may be defined as the speed at which an item rotates or revolves around an axis. The SI unit for angular velocity is radians per second because it is expressed as an angle per unit of time.**

The centripetal acceleration is calculated as follows:

\({a_c} = 10g\) The acceleration owing to gravity is denoted by \(g\).

Substitute \({\rm{9}}{\rm{.8 m/}}{{\rm{s}}^{\rm{2}}}\) for \(g\),

\(\begin{aligned}{}{a_c} = 10 \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right)\\ = 98{\rm{ m}}/{{\rm{s}}^2}\end{aligned}\)

The centripetal acceleration is,

\({a_c} = {\omega ^2}r\)

Here, \({a_c}\) is the centripetal acceleration, \(\omega \) is the angular velocity, and \(r\) is the distance of the rider from the center.

To generate an expression for the angular velocity, rearrange the previous equation.\(\omega = \sqrt {\frac{{{a_c}}}{r}} \)

Substitute \(98{\rm{ m}}/{{\rm{s}}^2}\) for \({a_c}\), and \(15.0{\rm{ m}}\) for \(r\),

\(\begin{aligned}{}\omega = \sqrt {\frac{{98{\rm{ m}}/{{\rm{s}}^2}}}{{15.0{\rm{ m}}}}} \\ = 2.55{\rm{ rad}}/{{\rm{s}}^2}\end{aligned}\)

Hence, the angular velocity is \(2.55{\rm{ rad}}/{\rm{s}}\).

The vertical component of the force on the arm will counter the weight i.e.,

\(\begin{aligned}{}{F_{arm}}\sin \theta = W\\ = mg\end{aligned}\) \({\rm{(1}}{\rm{.1)}}\)

Here, \({F_{arm}}\) is the force on the arm, \(\theta \) is the angle below the horizontal, \(W\) is the weight, \(m\) is the mass, and \(g\) is the acceleration due to gravity.

The horizontal component of the force on the arm will provide necessary centripetal force i.e.,

\(\begin{aligned}{}{F_{arm}}\cos \theta = {F_c}\\ = m{a_c}\end{aligned}\) \({\rm{(1}}{\rm{.2)}}\)

Here, \({F_c}\) is the centripetal force, and \({a_c}\) is the centripetal acceleration.

Dividing equation \({\rm{(1}}{\rm{.1)}}\) by equation \({\rm{(1}}{\rm{.2)}}\),

\(\begin{aligned}{}\frac{{{F_{arm}}\sin \theta }}{{{F_{arm}}\cos \theta }} = \frac{{mg}}{{m{a_c}}}\\\tan \theta = \frac{g}{{{a_c}}}\end{aligned}\)

Rearranging the above equation to get an expression angle below the horizontal.

\(\theta = {\tan ^{ - 1}}\left( {\frac{g}{{{a_c}}}} \right)\)

Substitute \(9.8{\rm{ m}}/{{\rm{s}}^2}\) for \(g\), and \(98{\rm{ m}}/{{\rm{s}}^2}\) for \({a_c}\),

\(\begin{aligned}{}\theta = {\tan ^{ - 1}}\left( {\frac{{9.8{\rm{ m}}/{{\rm{s}}^2}}}{{98{\rm{ m}}/{{\rm{s}}^2}}}} \right)\\ = 5.71^\circ \end{aligned}\)

Hence, the angle below the horizontal is \(5.71^\circ \).

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