Why Does the Measured Force Decrease as Distance Increases in Circular Motion?

AI Thread Summary
In circular motion, the force measured at different points along a ruler connected to a spring decreases as the distance from the center increases due to the relationship between torque and radius. The torque exerted by the spring is non-zero, allowing the ruler to rotate, which means the force at any point can be calculated using the equation f = t/r. As the radius (r) increases, the force (f) measured decreases proportionally; for instance, if the radius doubles, the force is halved. Understanding this relationship clarifies why the force decreases with distance from the center of rotation. This principle is fundamental in analyzing forces in rotational dynamics.
Ry122
Messages
563
Reaction score
2
I have a ruler which is connected to a spring that causes the ruler to move in a circular motion around the spring.
When I measure the force with a Newtonmeter at the top of the ruler it is less than it is at midpoint.

So my question is, why is it that the force exerted decreases as the point where force is being measured from increases.
I thought it might have something to do with torque, however
the force is created in the center of circular motion and not at point r, so i don't know if this equation applies or not:
t=f x r.
 
Physics news on Phys.org
Well, it would help me if you could explain the setup in a little more detail. It sounds as if you have a spring (coil spring?) attached to the ruler very near the center of the rotation. If it's literally at the exact point, then it can't exert a torque, as you've suggested, but that means the ruler wouldn't rotate, either. The fact that it does rotate means there is a non-zero torque, so your spring must be attached at some non-zero r from the center of rotation. r might be very small, but it can't be zero if the spring is exerting a torque, which is necessary for rotational motion.

That said, once you agree that there is a torque on the ruler, then the first part of your question becomes easier to answer. The same torque produces the forces you measure at the two points on the ruler, so the force will be given in each case as f = t/r (assuming you're measuring the force perpendicular to the ruler). If the second r is twice the first, then the force measured there will be half what you measured at the first r.

Make sense?
- Bruce
 
Yep that makes sense thanks
 
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top