Turn on thread page Beta
    • Thread Starter
    Offline

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

    Name:  m3_06_09_q6a.JPG
Views: 53
Size:  51.3 KB

    Any help is appreciated.
    Offline

    15
    ReputationRep:
    (Original post by jonatan18)
    This question has been confusing me for at least a week :s I've got that K.E. = mgl(\cos \theta - \frac{1}{4}), but I don't see how that leads to the solution...

    Name:  m3_06_09_q6a.JPG
Views: 53
Size:  51.3 KB

    Any help is appreciated.
    Then resolve inwards along the radius:
    T-mgcos theta=mv^2/l and substitute for v^2 from your KE
    • Thread Starter
    Offline

    1
    ReputationRep:
    Is mv^2/l from circular motion? If so, why?
    Offline

    15
    ReputationRep:
    (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
    • Thread Starter
    Offline

    1
    ReputationRep:
    (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...
    Offline

    15
    ReputationRep:
    (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.
    • Thread Starter
    Offline

    1
    ReputationRep:
    (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?
    Offline

    15
    ReputationRep:
    (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.
    • Thread Starter
    Offline

    1
    ReputationRep:
    (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)
    Offline

    15
    ReputationRep:
    (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.
    • Thread Starter
    Offline

    1
    ReputationRep:
    (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 (\ddot{x}=-\omega^2 x)? If that's the case, how would I know which one would be appropriate for different problems?
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: January 7, 2013
Poll
Do you think parents should charge rent?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.