Centripetal Force Experiment: Measuring Acceleration on a Turntable

In summary: So why am I being asked to measure the parallel component of r instead of the perpendicular component? In summary, the conversation discusses an experiment on centripetal forces and the confusion about measuring the component of position parallel to the track. It is mentioned that the centripetal acceleration is given by a(t)=\omega^2r(t), but there is uncertainty about what function should be used to measure the perpendicular component of r. The question is raised about why the parallel component is being measured instead of the perpendicular component, since the centrifugal force is directed radially outward.
  • #1
SN1987a
35
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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
[tex] a(t)=\omega^2r(t)[/tex], with [itex]\omega[/itex] 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).
[tex] v_0-v_i=\omega^2 \int ^{r_0}_{r_i}r(t) dt[/tex]

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:
 

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  • #2
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
lightgrav said:
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.
I know it does not need to point in towards the center, but only needs to be perpendicular to the v vector.


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

1. What is centripetal force?

Centripetal force is the force that keeps an object moving in a circular path. It acts towards the center of the circle, pulling the object inward.

2. How can centripetal force be measured?

Centripetal force can be measured using a device called a centripetal force apparatus, which typically consists of a rotating arm and a weight hanger. By varying the speed of rotation and the mass of the weight, the centripetal force can be calculated using the formula F=mω²r, where m is the mass of the weight, ω is the angular velocity, and r is the radius of rotation.

3. What factors affect centripetal force?

The magnitude of centripetal force is affected by the mass of the object, the speed of rotation, and the radius of rotation. The direction of centripetal force is always towards the center of the circle.

4. What is the relationship between centripetal force and centripetal acceleration?

Centripetal force and centripetal acceleration are directly proportional. As the centripetal force increases, the centripetal acceleration also increases. This means that if the force acting on an object is doubled, the acceleration will also double.

5. How does centripetal force relate to Newton's laws of motion?

Centripetal force is a result of Newton's First Law of Motion, also known as the Law of Inertia, which states that an object will remain in motion in a straight line unless acted upon by an external force. In order for an object to maintain a circular motion, a centripetal force is needed to continuously change its direction, in accordance with Newton's Second Law of Motion (F=ma).

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