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    (Original post by Euclidean)
    I think I f****d up. :facepalm:

    Anyone have any numerical answers?

    Q1: -6 i + 25/2 j

    Q2: cos (t) i - 2sin (2t) j

    Q3: (-3, 0, 1) but this was different to the resultant x the vector they give so I probably messed up

    Q4: Really messed up

    Q5: Assuming this is the variable mass

    v = u•ln (5) - 19.6

    Q6: a) omega = 3J/2mga I think
    b) T = 20pi/3 • root {g/5a}

    Q7: b) Y = 7/8 mg
    But I think I was supposed to take 3mg/8 away from mg/2 not add it... FFS

    Posted from TSR Mobile
    Pretty sure for q7 the impulse was at the bottom of the shape not in the middle (2a from axis not a) so mg/4 for last part?
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    (Original post by Bolts)
    Pretty sure for q7 the impulse was at the bottom of the shape not in the middle (2a from axis not a) so mg/4 for last part?
    b) Resultant moment about PQ axis:

    I theta double dot = amgsin (pi/6)

    4/3 ma^2 theta double dot = amg/2

    theta double dot = 3g/8a

    c) Resultant perpendicular to PQ:

    ma theta double dot = mgsin (pi/6) - F

    F = mgsin (pi/6) - ma • 3g/8a

    F = mg/2 - 3mg/8 = mg/8

    Edit: Sorry TSR bugged out on me.

    This was my method bar the last line. I did:

    F = mg/2 + 3mg/8 = 7mg/8


    Posted from TSR Mobile
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    What are the odds that the board will accept both cases for cylindrical shell? The answer I got for closed ends is
    (M(3r^3+6hr^2+6rh^2+4h^3))/(12(r+h))
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    (Original post by borgunblue)
    What are the odds that the board will accept both cases for cylindrical shell? The answer I got for closed ends is
    (M(3r^3+6hr^2+6rh^2+4h^3))/(12(r+h))
    I don't think they made it very clear. I actually did the same as you and crossed it out after.

    I think they should give credit for a method with closed ends but they probably won't award 10/10.
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    (Original post by Euclidean)
    b) Resultant moment about PQ axis:

    I theta double dot = amgsin (pi/6)

    4/3 ma^2 theta double dot = amg/2

    theta double dot = 3g/8a

    c) Resultant perpendicular to PQ:

    ma theta double dot = mgsin (pi/6) - F

    F = mgsin (pi/6) - ma • 3g/8a

    F = mg/2 - 3mg/8 = mg/8

    Edit: Sorry TSR bugged out on me.

    This was my method bar the last line. I did:

    F = mg/2 + 3mg/8 = 7mg/8


    Posted from TSR Mobile
    Yeah that's it (the first one).

    His confusion is over fact that the impulse was at 2a, but in this part we only care about COM which is distance a from axis.
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    (Original post by Euclidean)
    I think I f****d up. :facepalm:

    Anyone have any numerical answers?

    Q1: -6 i + 25/2 j

    Q2: cos (t) i - 2sin (2t) j

    Q3: (-3, 0, 1) but this was different to the resultant x the vector they give so I probably messed up

    Q4: Really messed up

    Q5: Assuming this is the variable mass

    v = u•ln (5) - 19.6

    Q6: a) omega = 3J/2mga I think
    b) T = 20pi/3 • root {g/5a}

    Q7: b) Y = 7/8 mg
    But I think I was supposed to take 3mg/8 away from mg/2 not add it... FFS

    Posted from TSR Mobile
    Q 6 is wrong: can't remember my and a but yours has units of 1/s not seconds
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    (Original post by Random1357)
    Yeah that's it (the first one).

    His confusion is over fact that the impulse was at 2a, but in this part we only care about COM which is distance a from axis.
    Any predictions on how many marks my sign error will cost me? Really annoyed I did that :facepalm:
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    (Original post by Random1357)
    Q 6 is wrong: can't remember my and a but yours has units of 1/s not seconds
    I may have got a/5g instead of g/5a. I remember a 5 on the denominator of the root.

    Edit: Yes it must have been a/5g.
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    (Original post by Euclidean)
    Any predictions on how many marks my sign error will cost me? Really annoyed I did that :facepalm:
    Well it's A1 or A1M1 lost for wrong equation and A1 final ans so 2-3
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    (Original post by Random1357)
    Well it's A1 or A1M1 lost for wrong equation and A1 final ans so 2-3
    Thanks.

    Could still be on 66-68ish then with my mess of a question 4, which isn't too bad.
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    (Original post by Euclidean)
    I may have got a/5g instead of g/5a. I remember a 5 on the denominator of the root.

    Edit: Yes it must have been a/5g.
    I'm pretty sure there was a square root 2 under sort sign as COM was sqrt(2) from axis
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    Name:  ImageUploadedByStudent Room1466508723.642911.jpg
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    (Original post by Random1357)
    I'm pretty sure there was a square root 2 under sort sign as COM was sqrt(2) from axis
    CoM was root10/2 from axis not root2. CoM was at 1/2,1/2 taking the kink in the wire as the origin.
    I had T=pi root(8root10a/3g) or T=2pi root(2root10a/3g)
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    (Original post by 8752)
    CoM was root10/2 from axis not root2. CoM was at 1/2,1/2 taking the kink in the wire as the origin.
    I had T=pi root(8root10a/3g) or T=2pi root(2root10a/3g)
    How? Two wires of equal mass side length 2a each. So at (a,a)???

    Or did I misread and forget a particle?
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    (Original post by Random1357)
    How? Two wires of equal mass side length 2a each. So at (a,a)???
    i meant combined centre of mass for the entire kinked wire. should have clarified what I meant.
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    (Original post by 8752)
    i meant combined centre of mass for the entire kinked wire. should have clarified what I meant.
    Still don't understand where that number comes from? Was there only one wire and nothing else stuck on?
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    (Original post by Random1357)
    Still don't understand where that number comes from? Was there only one wire and nothing else stuck on?
    sum of moments= moments of sum

    m(0,a)+m(a,0)=2m(x,y)

    (x.y)=1/2a, 1/2a
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    (Original post by 8752)
    sum of moments= moments of sum

    m(0,a)+m(a,0)=2m(x,y)

    (x.y)=1/2a, 1/2a
    Ah I just looked at it and didn't think .
    How many marks was that?
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    (Original post by 8752)
    I think 6 marks I'm not quite sure. You couldhave done it using 2 separate centres of mass on each wire but I chose to combine the CoM to make it easier.Attachment 554191
    Yeah I know; tried to do that but obviously didn't work! Ah well probably only lost 3.
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    (Original post by 8752)
    I think 6 marks I'm not quite sure. You couldhave done it using 2 separate centres of mass on each wire but I chose to combine the CoM to make it easier.Attachment 554191
    I found the com like you and theb made a mistake in my working on pythag.


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