Iconic_panda
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Circular motion.
What does it mean when the book said 'the force directed towards the centre does no work as it is perpendicular to the velocity' does it mean that the particle is not moving towards the centre and it stays a constant distance from the centre of the circle?
Also, what happens to the acceleration tangential to the circle, how come we never really talk about it and use it in calculations? Does this force do work on the particle?
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Pangol
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(Original post by Iconic_panda)
What does it mean when the book said 'the force directed towards the centre does no work as it is perpendicular to the velocity' does it mean that the particle is not moving towards the centre and it stays a constant distance from the centre of the circle?
I wouldn't say that it means that, but that is the explanation. Work done is the product of the force and the distance moved in the direction of the force, but because the force is tangential to the direction of motion, there is never any distance moved in the direction of the force, and hence no work is done.

(Original post by Iconic_panda)
Also, what happens to the acceleration tangential to the circle, how come we never really talk about it and use it in calculations? Does this force do work on the particle?
If you have circular motion with a constant speed, then this component is zero in any case. But if you have non-constant speed, which is often seen in vertical circle questions, then yes, the force responsible for the tangential component of the acceleration does do work on the particle, which is why it changes speed (equivalently, changes kinetic energy in accordance with the amount of work done). You don't tend to get problems that use this component in M3 (I believe, I'm not an expert in the content of all units), but it does get used a lot in problems about rotational motion that include consideration of moments of inertia (I think this is in Edexcel M5, so don't worry about it!).
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Iconic_panda
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Thank you very much buddy!
(Original post by Pangol)
I wouldn't say that it means that, but that is the explanation. Work done is the product of the force and the distance moved in the direction of the force, but because the force is tangential to the direction of motion, there is never any distance moved in the direction of the force, and hence no work is done.


If you have circular motion with a constant speed, then this component is zero in any case. But if you have non-constant speed, which is often seen in vertical circle questions, then yes, the force responsible for the tangential component of the acceleration does do work on the particle, which is why it changes speed (equivalently, changes kinetic energy in accordance with the amount of work done). You don't tend to get problems that use this component in M3 (I believe, I'm not an expert in the content of all units), but it does get used a lot in problems about rotational motion that include consideration of moments of inertia (I think this is in Edexcel M5, so don't worry about it!).
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okayplease
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can someone please help me with this question, m3 edexcel june 2008
Name:  Screen Shot 2018-04-29 at 21.49.52.png
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Size:  138.7 KB so for part a i understand up untill the end where the normal reaction is 0 why is that? and part b im confused as to how the extension is h/4
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tiny hobbit
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(Original post by okayplease)
can someone please help me with this question, m3 edexcel june 2008
Name:  Screen Shot 2018-04-29 at 21.49.52.png
Views: 71
Size:  138.7 KB so for part a i understand up untill the end where the normal reaction is 0 why is that? and part b im confused as to how the extension is h/4
Theparticle is in contact with the table so R > or = 0. =0 is whn it is just about to lift off, but is still touching the table.

Cos theta = h/l. If tan theta is 3/4, cos theta = 4/5 which gives the length of the string as 5/4 h
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okayplease
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(Original post by okayplease)
can someone please help me with this question, m3 edexcel june 2008
Name:  Screen Shot 2018-04-29 at 21.49.52.png
Views: 71
Size:  138.7 KB so for part a i understand up untill the end where the normal reaction is 0 why is that? and part b im confused as to how the extension is h/4
thank you, so even if it says its on the table when its turning in a horizontal circle its not really in contact with the table?
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tiny hobbit
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(Original post by okayplease)
thank you, so even if it says its on the table when its turning in a horizontal circle its not really in contact with the table?
It's the same idea as the M1 situation when a plank is "on the point of tilting". When it's just about to tip, the reaction is 0.
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okayplease
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(Original post by tiny hobbit)
It's the same idea as the M1 situation when a plank is "on the point of tilting". When it's just about to tip, the reaction is 0.
okay thank you !
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