QuestionFind the rotation rate in revolutions per second for a $32.5 \mathrm{ft}$ radius centrifuge to achieve $20.0 \mathrm{~g}$ acceleration.

Studdy Solution

STEP 1

Assumptions1. The length of the centrifuge is $=65.0 \, \mathrm{ft}$ . . The astronaut is sitting at one end, facing the axis of rotation $32.5 \, \mathrm{ft}$ away.
3. The required centripetal acceleration is $20.0 \, g$ , where $g$ is the acceleration due to gravity.

STEP 2

First, we need to understand that the centripetal acceleration $a_c$ of an object moving in a circle of radius $r$ at a speed $v$ is given by the formula $a_c = \frac{v^2}{r}$

STEP 3

We also need to know that the speed $v$ of an object moving in a circle of radius $r$ with a rotation rate $\omega$ (in revolutions per second) is given by the formula $v =2\pi r \omega$

STEP 4

Substitute the expression for $v$ from step3 into the formula for $a_c$ from step2 to get an expression for $a_c$ in terms of $r$ and $\omega$ $a_c = \frac{(2\pi r \omega)^2}{r}$

STEP 5

implify the expression obtained in step4 $a_c =4\pi^2 r \omega^2$

STEP 6

We want to find $\omega$ , so we rearrange the formula from step5 to solve for $\omega$ $\omega = \sqrt{\frac{a_c}{4\pi^2 r}}$

STEP 7

Now, plug in the given values for $a_c$ and $r$ to calculate $\omega$ . Note that $a_c =20.0g$ and $r =32.5 \, \mathrm{ft}$ . We also need to convert $g$ and $r$ to consistent units. We'll use $g =32.174 \, \mathrm{ft/s^2}$ and keep $r$ in feet $\omega = \sqrt{\frac{20.0 \times32.174 \, \mathrm{ft/s^2}}{4\pi^2 \times32.5 \, \mathrm{ft}}}$

STEP 8

Calculate the value of $\omega$ $\omega = \sqrt{\frac{643.48 \, \mathrm{ft/s^2}}{407.15 \, \mathrm{ft}}} =1.26 \, \mathrm{s^{-1}}$ The rotation rate required to give the astronaut a centripetal acceleration of $20.0 \, g$ is $1.26 \, \mathrm{s^{-1}}$ or $1.26$ revolutions per second.

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QuestionFind the rotation rate in revolutions per second for a $32.5 \mathrm{ft}$ radius centrifuge to achieve $20.0 \mathrm{~g}$ acceleration.

Studdy Solution

STEP 1

Assumptions1. The length of the centrifuge is $=65.0 \, \mathrm{ft}$ . . The astronaut is sitting at one end, facing the axis of rotation $32.5 \, \mathrm{ft}$ away.
3. The required centripetal acceleration is $20.0 \, g$ , where $g$ is the acceleration due to gravity.

STEP 2

First, we need to understand that the centripetal acceleration $a_c$ of an object moving in a circle of radius $r$ at a speed $v$ is given by the formula $a_c = \frac{v^2}{r}$

STEP 3

We also need to know that the speed $v$ of an object moving in a circle of radius $r$ with a rotation rate $\omega$ (in revolutions per second) is given by the formula $v =2\pi r \omega$

STEP 4

Substitute the expression for $v$ from step3 into the formula for $a_c$ from step2 to get an expression for $a_c$ in terms of $r$ and $\omega$ $a_c = \frac{(2\pi r \omega)^2}{r}$

STEP 5

implify the expression obtained in step4 $a_c =4\pi^2 r \omega^2$

STEP 6

We want to find $\omega$ , so we rearrange the formula from step5 to solve for $\omega$ $\omega = \sqrt{\frac{a_c}{4\pi^2 r}}$

STEP 7

STEP 8

Calculate the value of $\omega$ $\omega = \sqrt{\frac{643.48 \, \mathrm{ft/s^2}}{407.15 \, \mathrm{ft}}} =1.26 \, \mathrm{s^{-1}}$ The rotation rate required to give the astronaut a centripetal acceleration of $20.0 \, g$ is $1.26 \, \mathrm{s^{-1}}$ or $1.26$ revolutions per second.

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QuestionFind the rotation rate in revolutions per second for a 32.5ft32.5 \mathrm{ft}32.5ft radius centrifuge to achieve 20.0 g20.0 \mathrm{~g}20.0 g acceleration.

Studdy Solution

STEP 1

STEP 2

First, we need to understand that the centripetal acceleration aca_cac​ of an object moving in a circle of radius rrr at a speed vvv is given by the formulaac=v2ra_c = \frac{v^2}{r}ac​=rv2​

STEP 3

We also need to know that the speed vvv of an object moving in a circle of radius rrr with a rotation rate ω\omegaω (in revolutions per second) is given by the formulav=2πrωv =2\pi r \omegav=2πrω

STEP 4

Substitute the expression for vvv from step3 into the formula for aca_cac​ from step2 to get an expression for aca_cac​ in terms of rrr and ω\omegaωac=(2πrω)2ra_c = \frac{(2\pi r \omega)^2}{r}ac​=r(2πrω)2​

STEP 5

implify the expression obtained in step4ac=4π2rω2a_c =4\pi^2 r \omega^2ac​=4π2rω2

STEP 6

We want to find ω\omegaω, so we rearrange the formula from step5 to solve for ω\omegaωω=ac4π2r\omega = \sqrt{\frac{a_c}{4\pi^2 r}}ω=4π2rac​​​

STEP 7

STEP 8

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QuestionFind the rotation rate in revolutions per second for a $32.5 \mathrm{ft}$ radius centrifuge to achieve $20.0 \mathrm{~g}$ acceleration.

First, we need to understand that the centripetal acceleration $a_c$ of an object moving in a circle of radius $r$ at a speed $v$ is given by the formula $a_c = \frac{v^2}{r}$

We also need to know that the speed $v$ of an object moving in a circle of radius $r$ with a rotation rate $\omega$ (in revolutions per second) is given by the formula $v =2\pi r \omega$

Substitute the expression for $v$ from step3 into the formula for $a_c$ from step2 to get an expression for $a_c$ in terms of $r$ and $\omega$ $a_c = \frac{(2\pi r \omega)^2}{r}$

implify the expression obtained in step4 $a_c =4\pi^2 r \omega^2$

We want to find $\omega$ , so we rearrange the formula from step5 to solve for $\omega$ $\omega = \sqrt{\frac{a_c}{4\pi^2 r}}$