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Centripital Force? Help

  • #1
I currently did a lab where we go to swing a rubber stopper around our head and see how tension, radius and mass of the object(rubber stopper) effects frequency.

http://www.batesville.k12.in.us/Physics/PhyNet/Mechanics/Circular Motion/labs/cf_and_speed.htm

Im being asked what is the relationship between the frequency and the tension,radius and mass of object(rubber stopper). So i know that increase in radius/mass of the object would cause a decrease in frequency, while increase in tension will increase frequency. But WHY though? I need some kind of reasoning why.

Also im asked to draw a free body diagram of the situation.(shown in the link above), how would it look like? Isnt there only the tension force and force of gravity?

Lastly how does this investigation illustrate all three of newton's laws of motion?

Help plz
 

Answers and Replies

  • #2
AEM
Gold Member
360
0
Here's some hints: The frequency is the number of revolutions per second (or fractions of a revolution per second). That depends on the velocity of the mass because the faster it goes the more complete revolutions it can make in a given time period. But you have this nice little equation that says that the centripetal force is given by

F = M V^2/R

What provides the centripetal force??

That should get you started.
 
  • #3
107
0
The relationship of centripetal force (which is equal to tension) and frequency is as follows:
F[tex]_{centri}[/tex] = m v^2 / R

[tex]\omega[/tex] = 2[tex]\pi[/tex]f , where [tex]\omega[/tex] is angular speed and f is frequency.

v = [tex]\omega[/tex]R

Therefore:
F[tex]_{centri}[/tex] = m ([tex]\omega[/tex]R)^2 / R
= m [tex]\omega[/tex] R
= m (2[tex]\pi[/tex]f) R

The tension force that is created by the weight (F=mg , which is constant) of the weight hanger is equal to the centripetal force on the rubber stopper. This can be seen by the free-body diagram. In order to maintain this same tension force, if f increases, then R must decrease by the same magnitude, and vice versa.
Since f = F[tex]_{centri}[/tex] / m (2[tex]\pi[/tex]) R, f decreases when mass m of the stopper or radius R increase, and f increases when F[tex]_{centri}[/tex] increases (if more mass was added to the weight hanger and the radius was fixed).

The inertia (Newton's 1st) of the rubber stopper is what is causing the reaction force (centrifugal force, Newton's 3rd) that counteracts the tension force, which is caused by the weight (Newton's 2nd) of the hanger, keeping the stopper at a fixed radius for a given frequency.
 

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