Question

A biologist is studying growth in space. He wants to simulate Earth's gravitational field, so he positions the plants on a rotating platform in the spaceship. The distance of each plant from the central axis of rotation is r = 0.37 m. What angular speed is required?

Answer 1

Hi! Do you have an attempt at a solution?

Answer 2

not really since i dont even know where to start. they only gave me the radius. dont i need more information?

Answer 3

Nope! Do you know about centripetal force or centrifugal force?

Answer 4

does this have to do with omega? T=2pi/w

Answer 5

Sort of! It's actually asking for \(\omega\)! That is the angular velocity!
The problem concerns centrifugal force. When you're on the edge of something spinning, you feel like something's pushing you away from the center of your revolution. Like, if you grab a friend's hand and you both start rotating, you feel it get harder to hold on!

Answer 6

That, in space, is going to simulate gravitational force for that plant.

Answer 7

ok so is angular speed the same thing as angular velocity?

Answer 8

so omega=2pi/T but i dont have time and im given the radius instead. what should i don in this case?

Answer 9

Well, speed doesn't consider direction, and velocity does, but that's the only difference. They are measured in either degrees per second, radians per second, radians per minute, or things like that.

Answer 10

Well, you're given the radius for a specific reason. The centrifugal force that will be like gravitational force depends on \(\omega\) and the radius \(r\).

Answer 11

What you want is centripetal acceleration to equal what gravitational acceleration would be on Earth.

Centripetal acceleration is
\(a_c=\dfrac{v^2}{r}\)
normally. But we can find it for \(\omega\).

Answer 12

ah ok.

Answer 13

\(\omega=\dfrac{v}{r}\)
So what is \(v\)?

Answer 14

wouldn't that be the pull of gravity? 9.8m/s

Answer 15

w=26.5 rads/s

Answer 16

\(v\)? No, sorry! That is the tangential velocity - tangent to your path. Like, if you're driving in a circle, it's what the car's speedometer would say.

Answer 17

That's not what I got, but I see what you did.

We don't know \(v\). We want \(a_c=\dfrac{v^2}{r}\), so we need to solve for \(v\). But hey, it's related to \(\omega\) by \(\omega=\dfrac{v}{r}\), so we can solve for \(v\) and have an \(\omega\) to work with!

Answer 18

Do you have any questions? I'm sure you do...

Answer 19

lol everything
well im just stuck because we dont have omega to plug in the second equation and we dont know the velocity either. so idk

Answer 20

and if we wanted to solve for velocity we dont have omega....

Answer 21

Right! I figured that the whole thing was a bit much! It's complex!

I can say more if you need, but let's see if one sentence can clear this up.

If we solve for \(v\) in \(\omega=\dfrac{v}{r}\), we can stick it into the other equation.

I'm writing more now.

Answer 22

So \(\omega=\dfrac{v}{r}\implies v=\omega r\)

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

We want \(a_c\), the centripetal acceleration, to be 9.8m/s^2 like gravity. So we know what \(a_c\) is.

Answer 23

We also konw \(r\) since it's given to us, and we want to know \(\omega\).

Answer 24

AHHHHH

Answer 25

So, after we get \(a_c\), \(r\) and \(\omega\) in one equation, we solve for \(\omega\)!

I knew it'd be easier to show than explain :)

Answer 26

Let me know what you get, and let me know if you have any more questions!

I suggest that, after you get the answer and I say I agree, you look to see what you actually did so that you can repeat it in the future.

Answer 27

i got 5.14rads/sec is that right?

Answer 28

That's what I got! Except my calculator said 5.1465, so I would round up to 5.15.

Answer 29

And I agree with your units!

Answer 30

:) Bravo!

Answer 31

THANK YOU, that only took me forever lol. #lifesaver

Answer 32

Haha, glad I could help! Congrats! You worked through it.