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Pre-U paper questions: Rotations and circular motion watch

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    Hi, I've been doing some Pre-U papers during the holiday, and there were a few rotations/circular motion questions I didn't fully understand:

    In Q1b(iii) in 9792/03/M/J/11(http://www.acethem.com/pastpapers/pr...-3-20079.html/), the mark-scheme states that the resultant force is "smaller than force D" (and therefore always this case), but in a case like when both forces on the object are acting in the same direction, how can this be true?

    In question 10d(iii) (same paper), the mark scheme says that the kinetic energy loss is assumed to be equal to the work done against friction. I don't see how this is any more an assumption than an explanation.

    Finally, Q8f in 9792/03/SP/10(http://www.cie.org.uk/images/163482-...en-paper-3.pdf), I used the equation E=(1/2)Iw^2 since the question asks for the rotational kinetic energy. The mark scheme decides it's more appropriate to use E=(1/2)mw^2r^2, which I'm quite sure gives the linear kinetic energy E=(1/2)mv^2. Please help me on this.

    As a general question on exam technique, if it's not written in the mark scheme, what can be assumed to be a valid answer when it comes to purely qualitative questions?
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    For the first one, try drawing the weights of the object in. This will help visualise the forces.
    the weights' direction is always in the same direction, downwards. This means that the resultant force will change on the object as it goes round. The force D is given as
    D=\dfrac{mv^2}{r}

    But the weight of the object is mg and always is downwards. in the second picture, the object is at the top but is still held to the drum through centrifugal force.This means D \ge W

    For the second one, the only method of energy transfer here is work done against friction. Therefore it is safe to assume that this is true. This is only an assumption, and is stated below that you should give the assumption to the calculation above.

    The third one i cant see an 8f) :O sorry if im missing it...

    A general point I use is to write as many valid 'marking points' for the question as i can. If the question is worth 2 marks, then 2 points answering that question should be sufficient.
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    (Original post by The-Spartan)
    For the first one, try drawing the weights of the object in. This will help visualise the forces.
    the weights' direction is always in the same direction, downwards. This means that the resultant force will change on the object as it goes round. The force D is given as
    D=\dfrac{mv^2}{r}

    But the weight of the object is mg and always is downwards. in the second picture, the object is at the top but is still held to the drum through centrifugal force.This means D \ge W

    For the second one, the only method of energy transfer here is work done against friction. Therefore it is safe to assume that this is true. This is only an assumption, and is stated below that you should give the assumption to the calculation above.

    The third one i cant see an 8f) :O sorry if im missing it...

    A general point I use is to write as many valid 'marking points' for the question as i can. If the question is worth 2 marks, then 2 points answering that question should be sufficient.
    Hi Spartan, thanks for your reply.

    I understand the first image well enough, the resultant will definitely be smaller than D as long as mg>0 .However, if this resultant is supposed to stay constant as long as the object keeps a constant velocity (therefore making a=(v^2)/r constant) then the resultant is therefore always smaller than D. In the second image, D and mg are both in the same direction, so the resultant force couldn't possibly be greater than D. In the third image, the only force in the direction of the centripetal force is D, so the resultant would therefore be equal to D, wouldn't it?

    Thanks for your answer for the second question, I think I understand it now.

    I can't find any images of 8f online, though I do have an image of it on my computer, is it alright to post it here?
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    (Original post by Deepseawork)
    Hi Spartan, thanks for your reply.

    I understand the first image well enough, the resultant will definitely be smaller than D as long as mg>0 .However, if this resultant is supposed to stay constant as long as the object keeps a constant velocity (therefore making a=(v^2)/r constant) then the resultant is therefore always smaller than D. In the second image, D and mg are both in the same direction, so the resultant force couldn't possibly be greater than D. In the third image, the only force in the direction of the centripetal force is D, so the resultant would therefore be equal to D, wouldn't it?

    Thanks for your answer for the second question, I think I understand it now.

    I can't find any images of 8f online, though I do have an image of it on my computer, is it alright to post it here?
    Ah, i see the issue here You are in fact correct in your statement, the mark scheme is very vague (and in fact i think its actually wrong)
    This is because the mark scheme takes no account for mass of the object (and/or takes the average force on the object)
    Lets look at this through vectors.

    First picture:
    D=20N, W=mg\approx 2N
    The resultant in this case is the addition of these two vectors, note W is in the opposite direction so it is negative.
    20+(-2)=18N.

    Second picture:
    D_{net}=20+2=22N

    Third picture:
    Note weight has no component aligned with the force (they are perpendicular)
    D_{net}=20N

    Now the mark scheme takes resultant to be either not involving the weight, or to take the average force on the object per cycle

    \frac{18+20+22}{3} = 20 obviously, which D=20N

    You are correct, the mark scheme is infact wrong. The resultant involves the weight of the object also.

    EDIT: yes please do post it here
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    (Original post by The-Spartan)
    Ah, i see the issue here You are in fact correct in your statement, the mark scheme is very vague (and in fact i think its actually wrong)
    This is because the mark scheme takes no account for mass of the object (and/or takes the average force on the object)
    Lets look at this through vectors.

    First picture:
    D=20N, W=mg\approx 2N
    The resultant in this case is the addition of these two vectors, note W is in the opposite direction so it is negative.
    20+(-2)=18N.

    Second picture:
    D_{net}=20+2=22N

    Third picture:
    Note weight has no component aligned with the force (they are perpendicular)
    D_{net}=20N

    Now the mark scheme takes resultant to be either not involving the weight, or to take the average force on the object per cycle

    \frac{18+20+22}{3} = 20 obviously, which D=20N

    You are correct, the mark scheme is in fact wrong. The resultant involves the weight of the object also.

    EDIT: yes please do post it here
    Well, if the mark scheme didn't account for the weight, then it wouldn't have stated that the resultant was slightly less than D for the first image. Also, the point of the question was to prove the answer to the previous question, and the equation a=(v^2)/r. The resultant should be constant throughout the entire cycle, my only issue is why the mark scheme seems to suggest that it is always smaller than D.

    Here is an attachment for 8f (I can't get the exact image, please scroll down to the final couple of pages)
    Attached Images
  1. File Type: pdf 7_RotationAndCircularMotion.pdf (624.5 KB, 156 views)
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    (Original post by Deepseawork)
    Well, if the mark scheme didn't account for the weight, then it wouldn't have stated that the resultant was slightly less than D for the first image. Also, the point of the question was to prove the answer to the previous question, and the equation a=(v^2)/r. The resultant should be constant throughout the entire cycle, my only issue is why the mark scheme seems to suggest that it is always smaller than D.

    Here is an attachment for 8f (I can't get the exact image, please scroll down to the final couple of pages)
    I think what the markscheme is driving at is that the force D varies in magnitude as the socks (or whatever) go around, but that R is constant... though it's been rather confusingly written.
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    (Original post by Joinedup)
    I think what the markscheme is driving at is that the force D varies in magnitude as the socks (or whatever) go around, but that R is constant... though it's been rather confusingly written.
    Could you please explain how D varies through the rotation? I expected the same thing, but couldn't think of a convincing explanation.
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    (Original post by Deepseawork)
    Could you please explain how D varies through the rotation? I expected the same thing, but couldn't think of a convincing explanation.
    May I just start by saying wow, the mark schemes to these papers are abysmal.
    The overall force D cannot physically change unless the angular velocity of the drum changes or the radius of the object from the drum changes.
    The mark scheme as such must be wrong, as even though the reaction force is less than D in the first picture it is obviously greater in the second. It is in fact just D in the third as the forces are perpendicular.

    As to 8f, I cannot see why they have used linear either. They even ask for you to find the moment of inertia beforehand, pointing to rotational kinetic energy of the drum as being as you have stated.
    Very peculiar.
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    (Original post by Deepseawork)
    Could you please explain how D varies through the rotation? I expected the same thing, but couldn't think of a convincing explanation.
    Well I've been using D for the reaction force between the wet socks and the drum, W for the weight of the wet socks and R for the resultant force on the socks... since we're told the drum and socks are moving together at a constant angular velocity I'm reasoning that R must be a constant centripetal force.

    if the angular velocity is rather low the socks are barely experiencing any reaction force D from the drum at the top of the circle... at lower angular velocities they'd be weightless for the instant they pass the top and at an even lower velocity they'd fall off the inside of the drum and follow a parabolic trajectory till they hit the inside of the drum again.

    seems it's quite similar to the old 'bucket of water over the head' demo... http://www.education.com/science-fai...il-water-head/

    ---
    I don't understand what the MS is saying about the stored energy of the flywheel though, seems like you've got a correct answer and should get the marks. Maybe they have very thorough examiners meetings and the MS is produced as an afterthought?
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    (Original post by Joinedup)
    Well I've been using D for the reaction force between the wet socks and the drum, W for the weight of the wet socks and R for the resultant force on the socks... since we're told the drum and socks are moving together at a constant angular velocity I'm reasoning that R must be a constant centripetal force.

    if the angular velocity is rather low the socks are barely experiencing any reaction force D from the drum at the top of the circle... at lower angular velocities they'd be weightless for the instant they pass the top and at an even lower velocity they'd fall off the inside of the drum and follow a parabolic trajectory till they hit the inside of the drum again.

    seems it's quite similar to the old 'bucket of water over the head' demo... http://www.education.com/science-fai...il-water-head/

    ---
    I don't understand what the MS is saying about the stored energy of the flywheel though, seems like you've got a correct answer and should get the marks. Maybe they have very thorough examiners meetings and the MS is produced as an afterthought?
    If I look at it from image to image: in pic 1 D is greater than mg, in pic 2, mg must be greater than D (though, this makes me ask why the socks do not fall), in pic 3, the resultant is simply D? Is this correct?
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    (Original post by Deepseawork)
    If I look at it from image to image: in pic 1 D is greater than mg, in pic 2, mg must be greater than D (though, this makes me ask why the socks do not fall), in pic 3, the resultant is simply D? Is this correct?
    MS says
    W and D directions correct (1)
    resultant smaller than D (1)

    W and D directions correct (1)
    same size resultant (1)

    D correct direction (1)
    resultant horizontal (1) [6]

    Size of resultant may be indicated by relative sizes of arrows, or described in words or with mathematical relationship.
    Which from the way they're grouped I think is 2 marks per diagram

    the mark for R<D should, I think, apply to the first diagram only.

    the mark for 'same size resultant' is for showing the resultant has the same size as in the first diagram (i.e. D has changed)

    at the horizontal position in the 3rd diagram I think adding W will result in a slightly non centrepetal resultant force unless you have a vertical component from friction with the drum (doesn't seem to be any explicit marks for this though)

    fwiw my back of envelope estimation is that at typical full spin speed the centripetal force would be ~400 times greater than weight, so weight is only making a small difference to the resultant. You'd need to exaggerate your vector diagram to make this clear.
 
 
 
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