# M3 - Circular motion question

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#1
There's something I'm very confused about, when you have a particle moving on the top of the outer surface of a sphere. Why is it that the centripetal force is equal to the normal reaction plus the component of the weight acting towards the centre? I'd've thought that the weight would be 'helping' the centripetal force, and so the reaction would equal the weight plus the centripetal force. Why is this not the case?
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6 years ago
#2
(Original post by Sataris)
There's something I'm very confused about, when you have a particle moving on the top of the outer surface of a sphere. Why is it that the centripetal force is equal to the normal reaction plus the component of the weight acting towards the centre? I'd've thought that the weight would be 'helping' the centripetal force, and so the reaction would equal the weight plus the centripetal force. Why is this not the case?
your question is typical of a student which does maths and physics.

the centripetal force is a "fictitious force" which only exists in the physics vocabulary.
the are only two forces acting on the particle, assuming smooth.

the weight
the normal reaction
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6 years ago
#3
Remember that the centripetal force as TeeEm said is a 'fictitious force' - it is the RESULTANT force towards the centre of a circle.
Well if at the top and on the outside of a sphere, taking inwards as positive, mg-R= centripetal force.
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#4
(Original post by TeeEm)
your question is typical of a student which does maths and physics.

the centripetal force is a "fictitious force" which only exists in the physics vocabulary.
the are only two forces acting on the particle, assuming smooth.

the weight
the normal reaction
Okay, but I still don't quite understand how to relate them in an equation. I'm having trouble with 7a:

https://8a40d6c38bafca75cc407741c0f3...%20Edexcel.pdf
https://8a40d6c38bafca75cc407741c0f3...%20Edexcel.pdf

The mark scheme adds the weight and reaction together, but I don't see why the weight shouldn't be negative there...
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#5
(Original post by andywells)
Remember that the centripetal force as TeeEm said is a 'fictitious force' - it is the RESULTANT force towards the centre of a circle.
Well if at the top and on the outside of a sphere, taking inwards as positive, mg-R= centripetal force.
That seems very fair. But in the mark scheme in my above post, they add the reaction to the weight instead of subtracting it. What's going on?
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6 years ago
#6
(Original post by Sataris)
That seems very fair. But in the mark scheme in my above post, they add the reaction to the weight instead of subtracting it. What's going on?
Yes, R is negative as they've defined inwards as positive but in any case, it really doesn't matter as R=0. Perhaps the (+R) is to indicate the inclusion of a reaction force, rather than an addition.
1
#7
(Original post by andywells)
Yes, R is negative as they've defined inwards as positive but in any case, it really doesn't matter as R=0. Perhaps the (+R) is to indicate the inclusion of a reaction force, rather than an addition.
Ah, that's fine then. Thanks for the help!
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6 years ago
#8
(Original post by Sataris)
Ah, that's fine then. Thanks for the help!

In mathematics the convention is that the mathematical acceleration (r double dot) is directed radially outwards because that is the direction of r increasing.
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6 years ago
#9
(Original post by Sataris)
There's something I'm very confused about, when you have a particle moving on the top of the outer surface of a sphere. Why is it that the centripetal force is equal to the normal reaction plus the component of the weight acting towards the centre? I'd've thought that the weight would be 'helping' the centripetal force, and so the reaction would equal the weight plus the centripetal force. Why is this not the case?
The centripetal force is essentially the overall force of motion in this circular motion situation. So mv2/r (centripetal force)=component of weight - reaction force. If you do A2 physics, then thinking about Unit 4 mechanics should be helpful.

Hope this helps
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6 years ago
#10
(Original post by TeeEm)
your question is typical of a student which does maths and physics.

the centripetal force is a "fictitious force" which only exists in the physics vocabulary.
the are only two forces acting on the particle, assuming smooth.

the weight
the normal reaction
I think that it's at best "non standard" to call the centripetal force "fictitious" and at worst simply wrong.

The centripetal force ("centre seeking force") is the force required to keep an object moving on a curved path. It's a real force generated by real interactions, such as by the pull of a string, or the reaction at a physical surface.

Usually a force is called fictitious if it has to be invoked to explain motion that is seen by a non-inertial observer, such as one who is accelerating linearly or rotating.

So, for example, suppose you see a mass travelling at constant speed about a fixed point, due its being tied to a string. Then a non-rotating observer sees that the velocity vector of the mass is changing, and hence by N2 he realises that the vector sum of the forces acting on it is non-zero; this corresponds with the fact that he can see that the string produces the required force.

On the other hand, an observer rotating in the same direction and at the same rate as the mass sees the mass as stationary. However, he also sees a string, which must be pulling it toward the centre. By N2, he realises that since it is stationary, there must be another force acting on the mass of equal size to that produced by the string, but in the opposite direction.

He is aware that he and the mass are rotating, so he therefore says that when a mass rotates it experiences two forces, one of which pulls it away from the centre; he calls this the centrifugal force ("centre fleeing force") but it's fictitious since it doesn't arise due to any real interaction or object, and he only has to invoke it due to the fact that he is in a rotating frame.

Fictitious forces are only needed to make Newton's laws work from the point of view of an accelerating observer, and the centripetal force in this example doesn't fall into that category.
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6 years ago
#11
(Original post by atsruser)
(..)
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6 years ago
#12
(Original post by atsruser)
I think that it's at best "non standard" to call the centripetal force "fictitious" and at worst simply wrong.

The centripetal force ("centre seeking force") is the force required to keep an object moving on a curved path. It's a real force generated by real interactions, such as by the pull of a string, or the reaction at a physical surface.

Usually a force is called fictitious if it has to be invoked to explain motion that is seen by a non-inertial observer, such as one who is accelerating linearly or rotating.

So, for example, suppose you see a mass travelling at constant speed about a fixed point, due its being tied to a string. Then a non-rotating observer sees that the velocity vector of the mass is changing, and hence by N2 he realises that the vector sum of the forces acting on it is non-zero; this corresponds with the fact that he can see that the string produces the required force.

On the other hand, an observer rotating in the same direction and at the same rate as the mass sees the mass as stationary. However, he also sees a string, which must be pulling it toward the centre. By N2, he realises that since it is stationary, there must be another force acting on the mass of equal size to that produced by the string, but in the opposite direction.

He is aware that he and the mass are rotating, so he therefore says that when a mass rotates it experiences two forces, one of which pulls it away from the centre; he calls this the centrifugal force ("centre fleeing force") but it's fictitious since it doesn't arise due to any real interaction or object, and he only has to invoke it due to the fact that he is in a rotating frame.

Fictitious forces are only needed to make Newton's laws work from the point of view of an accelerating observer, and the centripetal force in this example doesn't fall into that category.
I certainly know what is "fictitious"as my background is mathematical physics having lectured analytical dynamics for 3 consecutive semesters in the 90s.
(I know the exact standard terminology used with rotating frames of reference and indeed this is where the word fictitious is used as a compensary force).

The context in which I used the word is simply "non distinct" as oppose to weight, tension, reaction etc but merely a resultant ...

I sincerely appreciate the "correction" which I am sure was done for the benefit of the understanding of the student.

As I am now a retired, old fool, I hope you put up with my inaccuracies as they are not deliberate ...

Regards as always.
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6 years ago
#13
(Original post by TeeEm)
The context in which I used the word is simply "non distinct" as oppose to weight, tension, reaction etc but merely a resultant ...
Right. The centripetal force here arises as a resultant, rather than having a single physical cause, but to describe it as fictitious could be very confusing in the context of rotational motion. It's clear that you understand the term well enough to use it loosely but this is such a tricky area to discuss I prefer to be very precise in the language that is used.
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