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M2 Circular Motion

Question 16

ImageUploadedByStudent Room1381409186.566807.jpg


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Original post by MAyman12
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What have you tried, and where are you stuck?
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MAyman12
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Original post by ghostwalker
What have you tried, and where are you stuck?


I resolved vertically to find the Tension in terms of m, g and cos(alpha) then resolved horizontally and substituted T and got this.ImageUploadedByStudent Room1381410210.663679.jpg

I don't know how to get tan(0.5*alpha)...


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Original post by MAyman12
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Your problem lies in resolving vertically for the ring. Each part of the string has a vertical component.
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Original post by ghostwalker
Your problem lies in resolving vertically for the ring. Each part of the string has a vertical component.


I still don't know how to get tan(0.5 *alpha), sorry.


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Original post by MAyman12
I still don't know how to get tan(0.5 *alpha), sorry.


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So, vertically, T+Tcos⁡α=mgT+T\cos\alpha = mg

As before, dividing one equation by the other.

sin⁡α1+cos⁡α=....\dfrac{\sin\alpha}{1+\cos\alpha}=....

Now use half angle formulae on the left, top and bottom.
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Original post by ghostwalker
So, vertically, T+Tcos⁡α=mgT+T\cos\alpha = mg

As before, dividing one equation by the other.

sin⁡α1+cos⁡α=....\dfrac{\sin\alpha}{1+\cos\alpha}=....

Now use half angle formulae on the left, top and bottom.


Thank you. Can you please also show me how to do the first part of this question too? ImageUploadedByStudent Room1381411446.834424.jpg

Question 13.



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Original post by MAyman12
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Just going out for a short while, but given your postings, I'd like to see you have a go first and see what you can do - you may surprise yourself.

Diagram would help you. The mass will leave the surface at some point above the horizontal plane through the midpoint of the sphere.

Edit: Will respond later if you're still stuck.
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Original post by ghostwalker
Just going out for a short while, but given your postings, I'd like to see you have a go first and see what you can do - you may surprise yourself.

Diagram would help you. The mass will leave the surface at some point above the horizontal plane through the midpoint of the sphere.

Edit: Will respond later if you're still stuck.


Ok I tried and got v^2=ga(2+2sin(alpha)) which is incorrect.


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Original post by MAyman12
Ok I tried and got v^2=ga(2+2sin(alpha)) which is incorrect.

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Post your initial equations+. I suspect your error lies with the motion in a circle equation, but can't be sure without seeing some working.
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Original post by ghostwalker
Post your initial equations+. I suspect your error lies with the motion in a circle equation, but can't be sure without seeing some working.


ImageUploadedByStudent Room1381419902.540491.jpg


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Original post by MAyman12


When the particle leaves the surface, the kinetic energy is not zero.

At that point, it is moving with such a speed that there is no reaction with the sphere, viz. the component of weight towards the centre of the sphere is equal to the force to move in a circle.
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Original post by ghostwalker
When the particle leaves the surface, the kinetic energy is not zero.

At that point, it is moving with such a speed that there is no reaction with the sphere, viz. the component of weight towards the centre of the sphere is equal to the force to move in a circle.


Can you please elaborate more with some equations?



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Original post by MAyman12
Can you please elaborate more with some equations?



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When the particle leaves the sphere:

mgsin⁡θ=mv2amg\sin\theta=\dfrac{mv^2}{a}

To be read in conjunction with my previous post.

That together with the energy equation allows you to resolve things.
(edited 10 years ago)
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Original post by ghostwalker
When the particle leaves the sphere:

mgsin⁡θ=mv2amg\sin\theta=\dfrac{mv^2}{a}

That together with the energy equation allows you to resolve things.


Ok, now I got it. You don't know how much you've helped, thank you so much .


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