# Homework Help: Centripetal force experiment

1. Nov 12, 2005

### SN1987a

I did this experiment on centripetal forces, where we rolled a ball onto a turntable and it left dots at equal intervals of time.

For every dot I was asked to measure the component of its position (from the center) parallel to its track (the perpendicular disance, along the path, between the point and a radius of the turntable). Plotting these, i get a radius vs time graph that looks like a sine curve (shifted by Pi/2).

I know that the centripetal acceleration is given by
$$a(t)=\omega^2r(t)$$, with $\omega$ known

So if I fit a curve through my points, and then integrate the function, I am supposed to get the change in speed (which i know).
$$v_0-v_i=\omega^2 \int ^{r_0}_{r_i}r(t) dt$$

Now the first thing that's misterious to me is why am I asked to plot the parallel radius versus time, since I'm looking for centripital acceleration, which is radial? I think I should rather measure the perpendicular component of r.

My question however is what should that function be, theoretically? Because I don't think it is a sine function. This again, relates to the question above. If it's the prallel component I should graph, then it makes sense to obtain a sine function. But if we're talking about the perpendicular component, then the function should have no uper bound, i.e., if the ball were unconstrained, it would fly off the turntable.

PS: The function I fitted is a completely unrelated one, and there is no conceptual justification for using it except the fact that i think it fits nice :tongue2:

#### Attached Files:

• ###### centripetal.jpg
File size:
24.3 KB
Views:
89
Last edited: Nov 13, 2005
2. Nov 13, 2005

### lightgrav

You haven't said what would have *caused* a centripetal acceleration.
If the acceleration is zero, then the radial distance from the table center
(the hypotenuse of a rt triangle, sqrt(y^2 + x^2) , with x=vt and y=const).

Usually we'd use components parallel to the velocity and perpendicular to v.
The "centripetal" component is the one perp to v - it does NOT need to
point toward the center of some coordinate system.

Components that are radial from the turntable center and perp to the r_vec
would only be really useful if a force was directed in the radial direction.

3. Nov 13, 2005

### SN1987a

I know it does not need to point in towards the center, but only needs to be perpendicular to the v vector.

That is precisely my question. Isn't the centrifugal force directed radially outward?