# Homework Help: Centripetal motion

1. Oct 15, 2007

### kateman

1. The problem statement, all variables and given/known data
this is a three part question (this i just a general thing for equations without numbers) for an object (the same as explained on this link: https://www.physicsforums.com/showthread.php?t=167283) but the string is not quite horizontal i.e. the r will be r cos theta because it will be a right angle triangle:

1. calculate the magnitude of the centripetal acceleration acting on a mass moving in a circular path at constant speed.
2. revise "components of forces"
3. verify the relation between the centripetal force acting on a body and the angular speed of the body.

and would i also be right is saying that a varying radius for a constant velocity isnt a problem since v = rw = 2pier/rw = 2pie/w

2. Relevant equations

3. The attempt at a solution
here's how i read them, please let me know if iam wrong or right!

1.centripetal force is usually a_c=v^2/r but since it isnt exactly at the full radius length since it forms (what we will call) a perfect right angle triangle, wouldn't it be the same radius and would be instead a_c = v^2/r cos theta

2. "components of forces", i guess this means all the forces acting upon the moving weight i.e. gravity, centripetal force, acceleration and how they all equate to the the weight just spinning there.

3. This iam not sure how to start. help would be appreciated :)

2. Oct 15, 2007

### Hootenanny

Staff Emeritus
Your green relationship is correct but I'm not sure where your going with the rest of it...
Sounds good to me
Components [of vectors] means the values of these vectors in some specific direction; usually this is the horizontal and vertical directions. So I would examine all the forces, and split them up into their horizontal and vertical components and see how they relate to each other.
Well, do you know the relationship between the centripetal force and the angular velocity?

3. Oct 15, 2007

### kateman

1. ahh my bad. see that was a comb equation from v=rw and v= 2pier/t
and hence t = 2 pie r/rw

2. that makes sense, i just wanted to re-confirm, cheers

3. F_c= mv^2/r

angular velocity = w = theta/time

just dont see how that relates

4. Oct 15, 2007

### Hootenanny

Staff Emeritus
Do you know a relation between angular velocity and linear velocity?

5. Oct 16, 2007

### kateman

v=s/t
w=theta/t

t=s/v
t=theta/w

s/v=theta/w

v= s x w/ theta

F_c=mv^2/r

= m(s x w/theta)^2/r

is that what your getting at? if so, then is that really a relation between centripetal force and angular velocity?

just to clarify for question two; are the only forces acting upon this gravity and centripetal or are there others?
i could see how (if we assume friction negligible with a glass rod and nylon thread) that there are other forces in action.

6. Oct 17, 2007

### kateman

i asked my teacher about the relation i had between centripetal motion and angular velocity and he said that you can't compare angular velocity and linear velocity.

edit: wait, i just found this equation; v=rw

okay, i understand that but i really need to know how that relates to centripetal force tonight. unless of course you mean F_c= mv^2/r = m (rw)^2/r

also a reminder about my second question; is centripetal, weight and gravitational the only forces acting upon this whole thing (if friction and air resistance not included)?

Last edited: Oct 17, 2007
7. Oct 17, 2007

### Hootenanny

Staff Emeritus
That's exacatly what I mean
The centripetal force isn't actually a force in itself, in this case the centripetal force is provided by the horizontal component of the tension in the string. In addition, the weight of the object is the same as the gravitational force.

I apologise if this is too late to help you [different time zones].

8. Oct 17, 2007

### kateman

right so the: F_c=m (rw)^2/ = mr^2w^2/r= mrw^2

am i right to say that arnt i?

as for the other question; that all makes sense. so the centripetal force is the tension in the string (your better at explaining than my teacher).

and no, you have helped me just in time
thank-you very much hootenanny :)

9. Oct 17, 2007

### Hootenanny

Staff Emeritus
Yup, spot on
Almost, the horizontal component of the tension provides the centripetal force, i.e. the 'bit' of the tension thats pointing towards the centre of the circle.
A pleasure