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M3 Jun 2009 Question 6a

This question has been confusing me for at least a week :s I've got that K.E.=mgl(cosθ14)K.E. = mgl(\cos \theta - \frac{1}{4}), but I don't see how that leads to the solution...

m3_06_09_q6a.JPG

Any help is appreciated.
Original post by jonatan18
This question has been confusing me for at least a week :s I've got that K.E.=mgl(cosθ14)K.E. = mgl(\cos \theta - \frac{1}{4}), but I don't see how that leads to the solution...

m3_06_09_q6a.JPG

Any help is appreciated.


Then resolve inwards along the radius:
T-mgcos theta=mv^2/l and substitute for v^2 from your KE
Reply 2
Is mv^2/l from circular motion? If so, why?
Original post by jonatan18
Is mv^2/l from circular motion? If so, why?


For circular motion, acceleration inwards along the radius is (v^2)/r or r(omega)^2
Reply 4
Original post by tiny hobbit
For circular motion, acceleration inwards along the radius is (v^2)/r or r(omega)^2


I'm still failing to see how this is an example of circular motion...
Original post by jonatan18
I'm still failing to see how this is an example of circular motion...


Try this out, physically. Tie something to the end of a piece of string, do as the question says and look at the path along which the object travels. It will be an arc of a circle.
Reply 6
Original post by tiny hobbit
Try this out, physically. Tie something to the end of a piece of string, do as the question says and look at the path along which the object travels. It will be an arc of a circle.


I thought circular motion was only true if the acceleration of the body is perpendicular to the motion. Where is this acceleration in this example?
Original post by jonatan18
I thought circular motion was only true if the acceleration of the body is perpendicular to the motion. Where is this acceleration in this example?


When a particle is moving in a vertical circle, there will also be acceleration along the tangent, but you never need to use it in M3.
Reply 8
Original post by tiny hobbit
When a particle is moving in a vertical circle, there will also be acceleration along the tangent, but you never need to use it in M3.


Is this a centrifugal acceleration?

(BTW thanks a lot tiny hobbit; this has been a huge help with my understanding of mechanics)
Original post by jonatan18
Is this a centrifugal acceleration?

(BTW thanks a lot tiny hobbit; this has been a huge help with my understanding of mechanics)


No, it's a tangential one.

Centrifugal is a word associated with being flung outwards. There isn't a separate force/acceleration causing this. It's the feeling you get because you would carry on in a straight line if there were not an inwards force causing you to go round in a circle.
Reply 10
Original post by tiny hobbit
No, it's a tangential one.

Centrifugal is a word associated with being flung outwards. There isn't a separate force/acceleration causing this. It's the feeling you get because you would carry on in a straight line if there were not an inwards force causing you to go round in a circle.


So does this mean that in a pendulum the acceleration can be described as either tangential or in the same plane as the displacement from the centre (x¨=ω2x\ddot{x}=-\omega^2 x)? If that's the case, how would I know which one would be appropriate for different problems?

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