Angular Velocity/Speed

[MCAT guy]

So I was doing a problem that asked how the acceleration was affected as a person rotating
around a circular space state (basically a person standing on the inside of a rotating circle,
I found a similar pic below).

Anyway, I went to solve with a = v^2/r and reasoned that as the radius shrunk I would get
more and more acceleration. Well, as I went to the answer key, BR said that the angular speed
W would be the variable to use, they reasoned that since the angular speed is constant that
they would have the relation a = W^2 r, which means as the radius shrunk the acceleration
would increase.

How in the world are there 2 equations that represent angular velocity or velocity, but
when I use one when the radius decreases I increase my acceleration, yet the other
one when the radius decreases I decrease my acceleration???

on top of that, it also seems like this acceleration would lift you off of the
circle, not press you towards? Why do they say that the spinning is
causing people to cling to the side instead of fly to the middle?

From the passage:

The artificial gravity is provided by a normal force acting on the
residents and pointing toward the central axis of the space station.
the station rotates at a constant angular velocity about its central axis.

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[Rayhaan]

a = v^2 /r
omega = v / r
therefore:
a = omega^2 x r

omega is the angular velocity. for a CD to spin, or any other object for that matter the angular velocity must be the same at all points (unless I am changing the distance or time of my spin) Just think of omega as being revolution per minute (the whole thing must revolve at the same time or it would split apart).

You can not use a = v^2 /r for this problem as tangential velocity changes depending on how far you are from the circle. Why? because tangential velocity is the velocity directed tangent to the circle, and different tangent points at different radii do not need to have the same speed to cover the same distance (eg. a point farthest from the circle needs to move faster then one at 1/2 radii because it must cover a greater arc length)

Based of this equation you now see the proper relationship between a and r. A good analogy is a spinning teacup ride. If you sit at the edge of the teacup you feel huge acceleration, but at the center not as much. Decreasing radius decreases centripetal acceleration.

 

[Rayhaan]

And for the second part. I assume it is because of centrifugal force, by newton's third law the centrifugal force opposes the centripetal force inwards. Think about this: if you tied a ball to a string and swung it around, does the ball move inwards like you are suggesting? Nope, it stays as far away as possible, AND you feel that tension in the rope pulling away (the centrifugal force)

 

[MCAT guy]

a = v^2 /r
omega = v / r
therefore:
a = omega^2 x r

omega is the angular velocity. for a CD to spin, or any other object for that matter the angular velocity must be the same at all points (unless I am changing the distance or time of my spin) Just think of omega as being revolution per minute (the whole thing must revolve at the same time or it would split apart).

You can not use a = v^2 /r for this problem as tangential velocity changes depending on how far you are from the circle. Why? because tangential velocity is the velocity directed tangent to the circle, and different tangent points at different radii do not need to have the same speed to cover the same distance (eg. a point farthest from the circle needs to move faster then one at 1/2 radii because it must cover a greater arc length)

Based of this equation you now see the proper relationship between a and r. A good analogy is a spinning teacup ride. If you sit at the edge of the teacup you feel huge acceleration, but at the center not as much. Decreasing radius decreases centripetal acceleration.

• Centripetal accelereation
  • acentripetal = v2/r
  • acentripetal =
    
    2r

I guess this makes sense, but I'm still confused as to why the centripetal acceleration will DECREASE as I shorten the radius, but INCREASE as I shorten the radius according to the linear speed. On the v^2/r equation, I will have more centripetal acceleration as the radius decreases...

Your example makes sense but I can't reconcile it with the v^2/r equation.

 

[MCAT guy]

I guess the odd idea is this...

on earth, the acceleration points down and I feel the normal force pushing up on me, hence I feel gravity.

on this circle, by creating an acceleration to the center (opposite the way it is if I standing on earth), there is a normal force pointing to the center, hence I feel gravity rotating on a circle.

So although the sum of the forces are in the exact opposite direction, I feel the same force of gravity... I guess it's weird because the vectors are in 180 degree opposite direction.

 

[Rayhaan]

The reasoning is because you can not use an equation where there are multiple variables changing.

In a = v^2/r, all THREE variables are changing! As you change the radius both acceleration and tangential velocity will change!

But in a = omega^2 x r we know that omega (the angular velocity) is constant! The key to this is to remember this one point: angular velocity is constant unless you are told otherwise. Whereas the tangential velocity is different depending how far you are from the center.

 

[MCAT guy]

The reasoning is because you can not use an equation where there are multiple variables changing.

In a = v^2/r, all THREE variables are changing! As you change the radius both acceleration and tangential velocity will change!

But in a = omega^2 x r we know that omega (the angular velocity) is constant! The key to this is to remember this one point: angular velocity is constant unless you are told otherwise. Whereas the tangential velocity is different depending how far you are from the center.

Thanks man.