Slight confusion about centripetal acceleration

AI Thread Summary
Centripetal acceleration increases with distance from the axis of rotation when angular velocity is fixed, as described by the formula a_c = ω²r. The confusion arises from mixing angular and linear velocities; while linear velocity increases with radius, centripetal acceleration is directly proportional to radius under constant angular velocity. In the carousel example, all points share the same angular velocity, leading to greater linear velocities and centripetal acceleration further from the center. Conversely, in circular orbital motion, centripetal acceleration decreases with increasing radius, being proportional to 1/r². Clarifying these distinctions helps resolve the initial confusion.
ace123
Messages
250
Reaction score
0
I just read in my book that the farther you are from the axis of rotation the greater the centripetal acceleration. But when dealing with circular motion the centripetal acceleration decreases as you increase the radius. Am I missing something? Or did I just confuse 2 different things?

Thanks for the clarification.
 
Science news on Phys.org
ace123 said:
I just read in my book that the farther you are from the axis of rotation the greater the centripetal acceleration.
Could you quote the passage in question from the book and perhaps give us some idea of the context in which the comment was made?
 
Well the textbook just gave the centripetal acceleration =(omega^2)r. Then it says thus the centripetal acceleration is greater the farther you are from the axis of rotation. It then talks about a carousel. Is that enough?
 
If you're thinking of centripetal acceleration in terms of \frac{v^2}{r} then you have to remember that v is also a function of r.
 
ace123 said:
Well the textbook just gave the centripetal acceleration =(omega^2)r. Then it says thus the centripetal acceleration is greater the farther you are from the axis of rotation. It then talks about a carousel. Is that enough?
Yes that is enough, thank you. The text is correct, but could perhaps be a little clearer by specifying that the angular velocity is fixed. For example, the quote could be re-written thus:

Given that the centripetal acceleration may be written as a_c = \omega^2 r, then for a fixed \omega, increasing the radius results in an increased centripetal acceleration.

In the case of the carousel, each point on the floor of the carousel has the same angular velocity (since the angular velocity is uniquely defined for a rigid body). Hence, the further you are from the centre of the carousel, the faster your linear velocity and the greater your centripetal acceleration. This can perhaps be seen more explicitly if you note that \omega = v r \Rightarrow v= \omega/r.

Does that make sense?

Edit: Kurdt beat me to it.
 
Ah I got it. I thought of angular velocity and linear velocity as interchangeable for some reason. Thanks for the help guys. And thanks for the help yesterday Hootenanny.
 
The other situation you might be thinking about is circular orbital motion. In the this case, centripetal acceleration is proportional to 1/r^2, and thus decreases as r increases.

As your book, Kurdt, and Hootenanny have stated, in the carousel case, centripetal acceleration is proportional to r, and thus increases as r increases.

[EDIT]I see I'm wrong.[/EDIT]
 
ace123 said:
Thanks for the help guys. And thanks for the help yesterday Hootenanny.
A pleasure :smile:
 
Hootenanny said:
Yes that is enough, thank you. The text is correct, but could perhaps be a little clearer by specifying that the angular velocity is fixed. For example, the quote could be re-written thus:

Given that the centripetal acceleration may be written as a_c = \omega^2 r, then for a fixed \omega, increasing the radius results in an increased centripetal acceleration.

In the case of the carousel, each point on the floor of the carousel has the same angular velocity (since the angular velocity is uniquely defined for a rigid body). Hence, the further you are from the centre of the carousel, the faster your linear velocity and the greater your centripetal acceleration. This can perhaps be seen more explicitly if you note that \omega = v r \Rightarrow v= \omega/r.

Does that make sense?

Edit: Kurdt beat me to it.

Isn't v=r times omega. Not omega over r?
 
  • #10
ace123 said:
Isn't v=r times omega. Not omega over r?
You are indeed correct, an embarrassing typo. I need to stop posting at 1am... :zzz:
 
Last edited:
Back
Top