Visualizing Centripetal Accelearation

  • Thread starter aconner5
  • Start date
  • #1
3
0
Hey Everyone :),

This is my first post and I'm hoping it will give me some insight into how to view centripedal force. I was following a problem in An Introduction to Mechanics by Kleppner and Kolenkow and made my way through their problem on uniform circular motion. I am having trouble understanding why the acceleration's vector for uniform circular motion is negative. I know it pushes inwards towards the center of rotation but what does this mean physically? I have viewed other forums on this topic but am still confused as to how I should look at acceleration.

When you attempt to spin something such as a bike wheel and you slide you hand down the side to begin accelerating it how is the force translated as a negative vector extending towards the center?

Is this simply the mathematical way of expressing acceleration around a particle spinning or am I missing something? Perhaps I am overthinking this.

Thanks for any replies!
 
Last edited:

Answers and Replies

  • #2
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
17,167
6,973
The force needs to be directed towards the center of a uniform circular motion because this is the direction you need to accelerate in and force is proportional to mass. If the force was directed outwards, the object would accelerate away from the alleged central point.

Also, the correct nomenclature is centripetal force.
 
  • #3
3
0
When it is said the force must be directed towards the center I do not see how this would generate circular motion around the center. When I hear this I imagine pressing down on a car tire at the very top. Obviously this would not cause the tire to spin but I am applying force towards the center. When you swing a ball on a string where is the force being directed towards the center. All I visualize is the force being exerted tangentially and not directed towards the center.
 
  • #4
Nugatory
Mentor
13,401
6,411
When you swing a ball on a string where is the force being directed towards the center. All I visualize is the force being exerted tangentially and not directed towards the center.

A string can only pull along its length. Is its length oriented tangentially or radially when you're swinging the object?
 
  • #5
Drakkith
Staff Emeritus
Science Advisor
21,171
5,029
When it is said the force must be directed towards the center I do not see how this would generate circular motion around the center. When I hear this I imagine pressing down on a car tire at the very top.

That's the problem. You're thinking about a tire that isn't moving. Imagine a block moving to the right. In order to make it move in a circular motion, you need to apply a force perpendicular to the direction of motion. But, as the direction the block is moving changes, so does the direction of the force. That's the key. The direction of the force changes as the object moves so that it always points perpendicular to the direction of motion (in perfectly circular motion).

When you swing a ball on a string where is the force being directed towards the center. All I visualize is the force being exerted tangentially and not directed towards the center.

The centripetal force is the tension in the string that holds the ball in place. Once you have the ball up and spinning, there is no more tangential acceleration and the only force is the centripetal one along the string.
 
  • #6
A.T.
Science Advisor
11,261
2,654
Is this simply the mathematical way of expressing acceleration around a particle spinning or am I missing something?
It follows from the mathematical definition of acceleration, as the second derivative of position.
 
  • #7
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
17,167
6,973
It follows from the mathematical definition of acceleration, as the second derivative of position.

To expand on this. Uniform circular motion on a circle of radius one may be described as ##\vec x(t) = \cos(\omega t) \vec e_1 + \sin(\omega t) \vec e_2##. Taking the second derivative results in ##\ddot{\vec x} = -\omega^2 \vec x##. Since, ##\vec x## is a vector pointing in the direction from the origin to the position and ##\omega^2 > 0##, the resulting acceleration ##\vec a = \ddot{\vec x}## points from the the position to the origin due to the minus sign.
 
  • #8
3
0
Thanks everyone for the help. I think I understand better now but will probably just move on and play around with the notion in my mind. :)
 

Related Threads on Visualizing Centripetal Accelearation

  • Last Post
Replies
14
Views
3K
  • Last Post
Replies
5
Views
7K
  • Last Post
Replies
12
Views
4K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
10
Views
3K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
12
Views
3K
Top