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Parallel axis theorem problem

  1. Apr 30, 2013 #1
    1. The problem statement, all variables and given/known data
    311uvyb.jpg


    2. Relevant equations

    I=Bmr^2
    Parallel axis theorem = Icm + MD^2
    WET, KE, PE equations


    3. The attempt at a solution

    So far I've only done parts a and b and I wanted to post this up as soon as possible, I want to make sure if I'm on the right path so far.

    part A) I know for this question I need to use the parallel axis theorem. What I did is I first found the moment of inertia of the rod first using I=(1/12)ML^2 , and also found the moment of inertia of the head (I=1/4MR^2 + 1/12ML^2). Then to find the moment of inertia of the mallet when held by the player at a distance of LP from the center of mass of the hammer, I derived the expression Ip = (Irod + Ihead + (mg + mh)(Lp + 1/2Lg)

    part B) For this part I'm thinking the work done by gravity is just the change in potential? where the total length (Lp + Lh + 1/2Rh - 1/2 if we were to just calculate up until the center of the mallet head) is used in a cosine equation to find h.

    so

    x = some change in height

    (total length / total length + x) = cos65°

    and I just solve for x and use that in the potential energy equation?
     
  2. jcsd
  3. Apr 30, 2013 #2

    Simon Bridge

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    Your thought processes seem OK. Just be sure to get the correct distances.
    Note: gravity acts at the center of mass.
     
  4. Apr 30, 2013 #3
    are you saying part a) is correct?

    or for part b) I need to find the center of mass of the mallet and that value will be the length I use to find the change of height?
     
  5. Apr 30, 2013 #4

    Simon Bridge

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    As a policy I try not to commit myself to an answer being correct or not, after all, I can make mistakes too.
    The trick is for you to get used to checking your answers.
    Infuriating I know - however, at some point you'll be facing problems where nobody know the right answer.
    Best get used to it :)

    Your reasoning is good - if your numbers are correct then you can be confident in the result.
    Check the numbers, check the arithmetic. Do you have any reason to think you may have got it wrong?

    From what you know about gravitational PE close to the surface of the Earth, the change in PE depends on the change in position of which part of the object?

    Of course, you could define a small mass dm at position x along the mallet ... each dm moves a different distance under gravity... you'd then add up the contribution of all the bits (hint: integration).


    It's probably easier to notice that the work done by force F(θ) swinging from θ to θ+dθ at distance R from the pivot is given by... what? Therefore the total work swinging from angle A to angle B is given by... what? (What's F(θ)?)
     
  6. Apr 30, 2013 #5
    advice well taken, I am pursuing a career in computer science :)

    I am only unsure of the moment of inertia for the rod, since it is also shaped as a cylinder, I was either thinking whether to use the same formula as for the head, which includes using the given radius also?

    F(θ) i'm thinking it is torque.

    so work done = R* (Fgravity)* (65°) ?
     
  7. May 1, 2013 #6

    Simon Bridge

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    Ah right - the trick with the moment of inertia is to get the right orientation for the rotation when you apply the parallel axis theorem. The shaft and head have different orientations - but they are rotated in the same plane wrt to each other so it doesn't matter: use the same com formula. Be careful to get the distance to the com correct in each case.

    That's not too bad - except that the force of gravity is not providing the same torque throughout the swing.

    F(θ) has to be the force in the direction of the motion for W=Fd to work.

    You can work it as torques - in which case, the work done by a constant torque moving through angular displacement θ is ##W=\tau\theta##

    If the torque is not a constant, then you need to think in terms of calculus:
    Work moving a small angle from θ to θ+dθ is dW= ...
     
  8. May 1, 2013 #7
    Ah I see,

    [itex]\tau[/itex] = |R||F|sinRF --- which I know both,

    R = (i'm thinking from axis of rotation up until center of the wooden head?)
    F = mg

    then I do


    W = ∫[itex]\tau[/itex]dθ
    W = [itex]\tau[/itex](θf - θo)
     
  9. May 1, 2013 #8
    Part c) I believe is just W = R * F using the new given values?

    Part d) hmm this is a bit difficult, I know that the system is a perfect elastic collision with a free axis of rotation so that

    linear momentum is conserved, angular momentum is conserved and kinetic energy is conserved.

    For linear momentum :
    I'm thinking right before impact, the mallet is traveling a Vo while the ball is at rest, and after impact, the ball has momentum and also the mallet has momentum.

    so my equation for this is

    Po = Pf

    (Mg + Mh)Vomallet = MballVFball + (Mg + Mh)VFmallet

    For angular momentum :

    Lo = Lf

    Imalletωo = Imalletωf + Rball * MballVFball


    For conservation of kinetic energy:

    KEo = KEf


    Imalletωo2 = Imalletωf2 + MballVFball2


    and it is also given to us that ωf = 3/20ωo

    I am having trouble isolating an unknown variable for me to solve. I'm thinking there are some values I have listed that should not be there,

    I am unsure if the mallet has linear momentum before impact? or if the ball has angular momentum after impact?
     
  10. May 1, 2013 #9

    Simon Bridge

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    Note - $$\int_{\theta_i}^{\theta_f}\ \tau(\theta)d\theta \neq \tau(\theta_f-\theta_i)$$ ... just like $$\int_a^b f(x)dx \neq f(b-a)$$ ... though that could be ambiguity about the parenthesis ... so $$\int_{\theta_i}^{\theta_f}\ \tau(\theta)d\theta = (\theta_f-\theta_i)\tau$$ only if ##\tau## is not a function of ##\theta##.

    For (c) where does the force act?
    What is the torque?
    What is the relationship between torque and work?

    For (d) - this looks like conservation of momentum... though, if the collision is elastic, you can use conservation of energy easily enough.
     
    Last edited: May 1, 2013
  11. May 1, 2013 #10
    so is it like

    ∫[itex]\tau[/itex](θ)dθ = [itex]\tau[/itex](θ)(θf - θo)


    part c)

    Force is acting perpendicular to the axis of rotation at the end of the mallet,

    [itex]\tau[/itex] = |R||F|sin[itex]\phi[/itex]RF
    W = ∫[itex]\tau[/itex]dθ


    (I am thinking since force is constant, the torque would be constant as well, so I can just do)

    W = [itex]\tau[/itex](θf - θo)


    part d) , oh that makes it a lot easier. I can find the initial by finding Vo of the mallet right before impact, and relating that to a ωo using the equation

    Rω = V

    I can find Vo by using trigonometry and conservation of energy pre-collision

    Cos65° = R / (R-X) --- x is will be the height that the mallet displaces.
    --- R will be the length of mallet head and shaft and persons arm

    then just use conservation of energy -

    mg(x) = 1/2MVf2 to find V and then solve for Vf ball
     
  12. May 1, 2013 #11

    haruspex

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    I don't understand that last bit. Don't you have to add different parallel axis distances for rod and head? And it looks dimensionally wrong, the last terms just being mass * distance.
    Yes, I would use the change in energy, but I don't think that gives the right distance. if the mass centre is a distance d from the axis, how far below the axis is it when at top of swing?
    Where sinϕRF = 1, right?

    For part d, you are not told the collision is elastic. Indeed, you are given enough information to determine whether it is. Use conservation of momentum only, but be careful. You are told the ball commences to roll without slipping straight away, which is so far removed from what would really happen that you can get misled. There will be two instantaneous impulses concurrently.
     
  13. May 1, 2013 #12

    Simon Bridge

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    I think you need to concentrate on the information provided in the question.
    Also on making connections between your past learning and the present exercise - you seem to have lost focus. i.e. pat lessons: you have learned how to integrate before I hope - i.e. in math class?
    No. You need to review the rules for integration. That would only happen if ##\tau(\theta)## was a constant. Is it a constant?

    Imagine, for the sake of an example, that ##\tau = \cos\theta##, then $$\int_{A}^{B} \tau(\theta)d\theta = \int_A^B \cos(\theta)d\theta = \sin(B)-\sin(A) \neq \tau(\theta)(B-A)$$... because ##\frac{d}{d\theta}sin\theta = \cos\theta##

    What does the question say?
    Is that what it says in the question or are you guessing?

    I'll echo Haruspex for part d.
     
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