# Centripital force

1. Oct 24, 2008

### omggg

hi, I am new to physics and math. This my first year taking a physics course.

We just had a lab done for centripital force, and I really got confused about this. The lab was not for marks, but I am still very confused

I couldn't figure out the relationship between the radius and the Fc, if the speed is always constant

what happens when radius is 0?
doesn't that produce an infinite calculation?

anyways my graph was very weird, because i could not find a relationship, maybe my data from the experiment was messed up..

and one more thing, is there anything special for its slope?

2. Oct 25, 2008

### phyzmatix

$$F=ma$$

but for centripetal acceleration,

$$a=\frac{v^2}{r}$$

so

$$F_c=m\frac{v^2}{r}$$

Thus Fc is inversely propportional to r. Was that what you are unsure about? What were your known values for the practical? What did your linearised equation look like?

Lastly, it's been some time since I watched this, but if I remember, this is Prof Walter Lewin's Lecture on Uniform Circular Motion

3. Oct 25, 2008

### schroder

Looking at the relationship a = v^2 / r you get the immediate impression that the centripetal acceleration is inversely proportional to r. But that is wrong! Remember that v is itself a function of r, and v is in fact directly proportional to r. Therefore, when v is squared, there is an r^2 term in the numerator. It is more clearly seen when you look at the angular definition of centripetal acceleration, which is angular velocity squared multiplied by r. Now you can see that centripetal acceleration is actually directly proportional to r. At the center of the circle, where r is zero, centripetal acceleration is zero.

4. Oct 25, 2008

### omggg

ok so i did some work, and i got this

mv2 = Fc
r

m(2 pie r)2 = Fc
T2r

so final eqn is

(m) (4) (pie2) (r) = Fc
T2

I am sure my math is correct, but this means that radius actually increases the Fc, when it should actually decrease

T= time for one rotation
Pie = 3.14

wow, i understand it now.

(m) (4) (pie2) (r) = Fc
T2

I just derived this newtons formula, and I understand that there actually are no r values on the bottom, therefore the denominator can never be 0..
so i therefore as radius increases, the Fc should also increase.

BUT

T, is based on radius.. so wont that make a difference to the EQN?

Can someone tell me the exact relationship between radius and Fc, considering that did my practical correctly, and got correct values.
my known values are T and radius, on those the entire experiment can be based on, but if they are wrong, then I wont get correct relationship.

thanks

Last edited: Oct 25, 2008
5. Oct 25, 2008

### LURCH

Very true. In fact, you can't even have rotational motion without a radius.
Yes. You see, as r increases and T remains constant, Fc will increase. This is because, as the radius increases, the circumference must also increase (because the circumference is 2pi x r). So, if the circumference is increasing, and the time to complete one rotation remains constant, v must be increasing.

However, if radius increases and velocity remains constant, then T must increase, giving you a larger number in your denominator. This causes Fc to decrease.

I actually have a similar thread regarding orbital dynamics in another Forum. I will link you to it the next time I'm in this thread.

6. Oct 25, 2008

### omggg

thanks so much, your explaining helped alot.

For our practical, we were supposed to keep velocity constant, so that means that out T would decrease, and Fc would also decrease each time.

this is what i got in my data, for the first 2 radius, and then the data creates error, i am guessing this is a human error, because we were rotating using our hands, and it is really easy to not keep the same speed.

thanks, you guys helped explain this very well, and i actually understand this. i wish i had come to this forum earlier :(
i would have gotten perfect on my test, which was mainly on circular motions..
thanks again

7. Oct 26, 2008

### phyzmatix

Oh crap! Sorry, of course...