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    Question 6.

    I get that you can probably work out a) through Pythagoras but I'm confused at where to go with this knowledge. And b) simply confuses me.
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    (Original post by 2014_GCSE)
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    Question 6.

    I get that you can probably work out a) through Pythagoras but I'm confused at where to go with this knowledge. And b) simply confuses me.
    Force diagram of forces acting on mass. Note: threaded bead means that the tension is constant throughout the string .. i.e. a force of T towards the centre of the circular motion and a force of T acting at an angle theta. Other forces acting on mass are mg and R (normal reaction).resolve vertically to get a relationship involving T, mg and R.
    resolve towards the centre using newtons 2nd law with acceleration as w^2 r.Remove instances of T from this second equation by using the equation obtained from vertical motion.Make R the subject. Now state R>= 0 to remain in contact with table. hence solve.
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    (Original post by dpm)
    Force diagram of forces acting on mass. Note: threaded bead means that the tension is constant throughout the string .. i.e. a force of T towards the centre of the circular motion and a force of T acting at an angle theta. Other forces acting on mass are mg and R (normal reaction).resolve vertically to get a relationship involving T, mg and R.
    resolve towards the centre using newtons 2nd law with acceleration as w^2 r.Remove instances of T from this second equation by using the equation obtained from vertical motion.Make R the subject. Now state R>= 0 to remain in contact with table. hence solve.
    Amazing reply! Here's what I have got down so far then, but I think I've gone wrong somewhere.

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    (Original post by 2014_GCSE)
    Amazing reply! Here's what I have got down so far then, but I think I've gone wrong somewhere.

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    Put the value of sin (theta) in ...
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    (Original post by Muttley79)
    Put the value of sin (theta) in ...
    Oh right, yeah. So sin theta = 4/5.

    Also I was wondering, doesn't R have to be bigger than 0. It can't be 0.

    So I have

    R = g / (12/5 x w^2)
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    (Original post by 2014_GCSE)
    Oh right, yeah. So sin theta = 4/5.

    Also I was wondering, doesn't R have to be bigger than 0. It can't be 0.

    So I have

    R = g / (12/5 x w^2)
    Where did the last line come from? Check your rearrangement for R after you eliminate T.
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    (Original post by Muttley79)
    Where did the last line come from? Check your rearrangement for R after you eliminate T.
    Tsin(theta) + R = mg

    T = m(3w^2)

    sin(theta) x (m x 3w^2) + R = mg

    4/5m*3w^2 = mg

    w^2 = 49/12

    w = 2.02

    Wrong answer.

    Just getting dpm 's attention, just in case he hasn't seen this.
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    (Original post by 2014_GCSE)
    Tsin(theta) + R = mg

    T = m(3w^2)

    sin(theta) x (m x 3w^2) + R = mg

    4/5m*3w^2 = mg

    w^2 = 49/12

    w = 2.02

    Wrong answer.
    Sorry I'm trying to mark a load of books and not focusing on this very well as it's late.

    When you used F = ma horizontally what about the component of T in the other part of the string?
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    (Original post by Muttley79)
    Sorry I'm trying to mark a load of books and not focusing on this very well as it's late.

    When you used F = ma horizontally what about the component of T in the other part of the string?
    YES! That's what I was missing! Brilliant, thank you.
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    (Original post by 2014_GCSE)
    YES! That's what I was missing! Brilliant, thank you.
    No problem - sorry it took so long to figure where you'd gone wrong
 
 
 
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