Y12_FurtherMaths
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#1
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This question is really confusing me. My friend told me that the centripetal force >=frictional force but I dont know why this is the case. It doesn't help that I'm unsure how to draw a force diagram. All I really know is that the centripetal force acts towards the centre, other than that I'm so confused!
http://imgur.com/a/BdoggLh
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DS Steve Arnott
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#2
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ah, the karate kid taught me all i need to know about circular motion, you just have to wax on and wax off and you'll be fine.
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jenitanushi
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(Original post by DS Steve Arnott)
ah, the karate kid taught me all i need to know about circular motion, you just have to wax on and wax off and you'll be fine.
ahh yes.
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Y12_FurtherMaths
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RDKGames
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old_engineer
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As far as I can see, you have correctly determined that F (the force needed to maintain the particle in horizontal circular motion) = mg Newtons. Now you need to reason that the only possible thing that can be providing that force is the disc, and that the force must be in the form of friction. Next, the particle will slip if F is greater than (mu)R, where R is the vertical reaction force exerted by the disc on the particle. Can you pick it up from there?
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Y12_FurtherMaths
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(Original post by old_engineer)
As far as I can see, you have correctly determined that F (the force needed to maintain the particle in horizontal circular motion) = mg Newtons. Now you need to reason that the only possible thing that can be providing that force is the disc, and that the force must be in the form of friction. Next, the particle will slip if F is greater than (mu)R, where R is the vertical reaction force exerted by the disc on the particle. Can you pick it up from there?
Is F the centripetal force? Which acts towards the centre? Which way does friction act?
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old_engineer
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In this instance, the frictional force IS the centripetal force, so it acts towards the centre of rotation.
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Y12_FurtherMaths
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(Original post by old_engineer)
In this instance, the frictional force IS the centripetal force, so it acts towards the centre of rotation.
So if that's our frictional force then what is muR going to calculate?
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old_engineer
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(mu)R is the maximum frictional force that the disc can exert.
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Y12_FurtherMaths
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#10
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(Original post by old_engineer)
(mu)R is the maximum frictional force that the disc can exert.
Could you please draw a force diagram for me? I think that might help me understand
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old_engineer
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Horizontally, there is only one force acting on the particle, namely the centripetal force acting towards the centre of rotation. It just happens that in the situation depicted by the question, the centripetal force is frictional in nature. The catch is that if the disc spins too fast, or if the value of mu is too low, the disc will be unable to provide a sufficient frictional force to maintain the particle in horizontal circular motion, and the particle will then slip outwards instead.
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Y12_FurtherMaths
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(Original post by old_engineer)
Horizontally, there is only one force acting on the particle, namely the centripetal force acting towards the centre of rotation. It just happens that in the situation depicted by the question, the centripetal force is frictional in nature. The catch is that if the disc spins too fast, or if the value of mu is too low, the disc will be unable to provide a sufficient frictional force to maintain the particle in horizontal circular motion, and the particle will then slip outwards instead.
Thank you I appreciate all this
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