Calculating Centripetal Force: From 1m to 100m in 100s with a 1kg Mass

AI Thread Summary
The discussion revolves around calculating the centripetal force required for a 1 kg mass to transition from a 1m to a 100m radius while maintaining a speed of 100m/s, specifically after switching tension from 10,000N to 100N. Participants highlight the complexities of the problem, including the need for precise equations and the challenges of real-world application, such as reeling in or out a tether. There is confusion regarding the terminology used, particularly between "spinning" and "revolving," which affects the understanding of the physics involved. The conversation suggests that a better approach might involve framing the scenario within orbital mechanics to clarify the dynamics at play. Overall, the discussion emphasizes the importance of accurately applying physics principles to solve the problem effectively.
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how long would it take a spinning 1 kg mass to go from 1m to 100m radius at 100m/s by switching tension from 10000N to 100N?

the object is spinning and i would like to know how many revolutions would it take to go from a radius of 1 m to a radius of 100 m starting to count in the moment the tension is switched from 10000N to 100N

i would apreciate any formula
 
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Your question makes no sense. Please rephrase. Is this another tetherball question, related to your previous thread? If so, the answer depends on the length of the tether and the ball's rate of rotation about the center point - and you already know the equations involved.

Technically, the problem with your thought experiment is how you could do it in reality - being able to reel-in or reel-out a tether in the way you suggest is not an easy thing to accomplish.
 
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Why don't try to solve Newton's equation :

The force on the point is along the rope, at 100 N...just use polar coordinates :

x=r(cos(\phi),sin(\phi)) -> v=r'(cos(\phi),sin(phi))+r\phi '(-sin(\phi),cos(\phi))

so that the accel. is a=(r''-r\phi^{'2})(cos(\phi),sin(\phi))+(2r'\phi '+r\phi '')(-sin(\phi),cos(\phi))

So that in your case : 2r'\phi'+r\phi ''=0

and r''-r\phi^{'2}=100

Just solve the sys. of equation...this should be not so difficult I think
 
im afraid it seems dificult to me

i was expecting an answer like a quarter of revolution or something like that

i don't even know what stands for what in your equations

i just have one university year on physics and another in mechanics

i was expecting some equation that related the minor tension with the bigger tension and the constant speed and radius but i don't know which is which in your equations
 
if i have the ball spinning with a radius of 1 m at a speed of 100m/s to make it go from a radius of 1 m to a radius of 100m how many revolutions will it take after switching tension from 10000N to 100N?

i acomplish this by gearing the spin with the extension or retraction of the cable so the force of spin makes the cable extend or retract

on this way if the ball is going with a speed of 100m/s and a radius of 100 m and i gear it so the radius becomes 1 m the speed will keep constant of 100 m/s so kinetik energy is kept

what would happen if i make the radius 0 what would have happen to the kinetic energy of the ball?

shouldnt the ball be spinning in the axe to keep right the law of conservation of kinetic energy?

but then this means that linear speed can be transform into spin which makes no sense
 
aviator said:
shouldnt the ball be spinning in the axe to keep right the law of conservation of kinetic energy?

but then this means that linear speed can be transform into spin which makes no sense
I never actually noticed this in your previous posts, but it seems to me from this one that you're failing to differentiate between rotating and revolving when you use the word 'spinning'. If the thing that you're talking about is tethered to an axe (and I've always taken this to mean axle), then its rotation and revolution periods would match. It would essentially be tide locked to the axle. It's the same basic idea that explains the Moon keeping the same side toward us. In fact, you might be better off if you put your entire hypothesis into an orbital mechanics problem to start with, then work out whatever Earthbound modifications are required.
 
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