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A project that needs direction, on gravity, friction, acceleration & deceleration

  1. May 19, 2009 #1
    First off I don't believe this would be classed as a homework question, as this is an engineering design project problem and I haven't been to school for over ten years, however if I am wrong I apologise.

    My problem is:-

    The object I want to knock down is a skittle:

    h = distance from centre of gravity to tipping point (makes the hypoteneuse of the triangle)
    x = length of centre of gravity from base of the skittle (makes the vertical)
    y = the radius of the object (completing the horizontal and therefore the triangle)

    I need to knock down a skittle of certain mass, dimensions etc, so by using triangles, pythag. I worked out h

    then by saying that the potential energy to knock it down is the energy in the distance that it takes to move the centre of gravity directly above the tipping point, which makes h vertical so therefore the energy in the difference between h and x would be the absoluet minimum I need and therefore used this E = (h - x)W...... (think that works).

    (Using a hockey ball to knock it down)

    Now here is where I can not see the woods from the trees........... I am using gravity as the accelerator and therefore height, on a curve so as to keep dirctional change losses minimal.......also the distance from the end of the curve to the skittle is a metre, so the ball will be rolling and declerating for a metre.

    How do I work backwards using the energy that I worked out on the skittle to eventually end up with a height that will give me a force = to the one I need at the skittle?

    What equations do I need and please annotate the definitions of the equations i.e. v = velocity otherwise I will probably get confused.

    I appreciate all answers to this problem and look forward to some direction on this.

    Thank you all in advance
     
    Last edited: May 19, 2009
  2. jcsd
  3. May 19, 2009 #2

    berkeman

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  4. May 19, 2009 #3
    Yep its a bowling pin USBC standard I believe
     
  5. May 19, 2009 #4

    berkeman

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    Ah, thanks. Your setup is still a bit for me to understand, though. Where does this meter separation come into it? Are you rolling a ball to knock over the pin, and want to know how fast the ball has to roll to have enough energy to knock over the pin?

    A sketch of your setup would help immensely. Thanks.
     
  6. May 19, 2009 #5
    Hi yeah its all a bit confusing. Basically we are designing a machine that will knock a skittle down, to build it we have to meet certain requirments/constraints etc. One of those requirements is that the device we build will be one metre away from the skittle so that means that the exit of our curve must be one metre away from the skittle.

    Yes basically rolling a hockey ball to knock over the pin, and yep need to know how fast the ball has to roll to knock over the pin.

    Then need to know how high the ball needs to be on release to accelerate it, so that when it reaches the pin it is at the correct speed.

    Hope the diagram helps, the blue indicates the device we are trying to construct, the hockey ball is in red. I have used z to indicate the height that I hope to eventually get.
     
  7. May 19, 2009 #6
    The diagram that I forgot to put in
     

    Attached Files:

  8. May 19, 2009 #7

    berkeman

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    What's a hockey ball? In the US, we use a puck for hockey. This is all so confusing!!! :smile:

    I'm assuming that you want to calculate the minimum initial height of the ball, since anything higher will also knock over the pin, right?

    What are the mass and diameter of the hockey ball? What is the mass of the pin? What is the height of the contact point between the ball and pin?

    There are a bunch of factors that come into the real calculation, including multiple coefficients of friction, etc. What is this project for? How accurate and real-life do the calculations need to be? What's the prize!
     
  9. May 19, 2009 #8
    lol it is a bit confusing yes,

    ok hockey ball is 71mm in diameter, and on a weighing scales 157.9g or base unit of mass 0.1579Kg.

    skittle/Pin = 1.595Kg

    Point of impact I dont know that distance but can put it up on here tomorrow morning, obviously we have a time difference being midnight here but I will put it up and you can see at your leisure.

    Basically the project is a kind of test, I am in the Royal Navy as an engineer and this is to test our understanding of how a project is put together, we have been given as statement of requirements and have to build a device that meets those requirements. Including mathmatically proving it, which is proving difficult as I have not done these types of equations in a long time, but we are allowed to research and get help, hence I thought here might be good.

    The prize basically is a qualification in the Royal Navy and that means I go off to sea where ever command sees fit.

    I appreciate your interest in this dilema of mine
     
  10. May 19, 2009 #9
    as to how accurate and real life, not really just something pretty close that would give a good indication of what we needed.


    and yep minimum height is what we are after, just to prove mathmatically we need a height higher than it

    The hockey ball is because as well as hockey on ice we have some hockey played outside on astroturf, normal ground, mainly played in schools and things. ALl crazy stuff I forget how different two countries can be :)
     
  11. May 19, 2009 #10

    berkeman

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    As background reading, skim the wikipedia.org entry for Moment of Inertia:

    http://en.wikipedia.org/wiki/Moment_of_inertia

    You don't have to worry about the calculus bit in the middle. After reading the start of the article, click on the link at the bottom of that page for "List of moments of inertia", and note the value of I for a solid shpere (your ball).


    Next, read some about Rotational Kinetic Energy:

    http://en.wikipedia.org/wiki/Rotational_kinetic_energy

    When the ball is rolling without slipping, you have both translational kinetic energy KE = 1/2 M * V^2 (center of mass motion), as well as that rotational kinetic energy. You initially have stored potential energy PE from holding the ball some distance up the curve/incline, and that gets converted to translational and rotational KE at the bottom of the incline. The ball then travels to the pin, and has a collision with it.... Depending on the mass of the ball, the height of the contact, the mass of the pin, the coefficient of friction between the pin and the floor, etc., the pin may or may not tip over....
     
  12. May 20, 2009 #11
    SO does that mean that the rotational and translational energy of the ball must be equal to or greater than the potential energy of the bowling pin that I worked out earlier.

    so far I have worked out the Inertia of the ball to be 79.597

    ok so now I think that the minimum energy would look something like this:

    Energy of pin(worked out as explained in first post) = K(rotational of ball) + K(translational of ball)........to me that would give me the minimum energy needed to knock it over. However not sure how the point of contact would play into that.

    So because the KE(rotational) equation uses angular velocity, which I don't know, looking at the following website
    http://en.wikipedia.org/wiki/Angular_velocity

    I can replace angular velocity in the K(rotational equation) with rv/r^2

    Therefore I end up with a modified
    energy of pin = KE(rotational of ball) + KE(translational of ball) to which the only thing I don't know is velocity.

    Transposing the modified equation for velocity, will then enable me to work backwards and use an equation to eventually get to the height to get that velocity.

    If all the above sounds about right then I think that is what I need.
     
  13. May 20, 2009 #12
    thinking the angular velocity is incorrect as time is not part of the equation in that post which is important
     
  14. May 20, 2009 #13
    ok so that didnt go to well haha what I am thinking now is could I use PE = mgh assuming negligble losses as the ball changes its energy at the bottom of the curve. So I would now use the energy of the pin to replace the PE in PE = mgh that gives me height
     
  15. May 20, 2009 #14

    berkeman

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    Since you are assuming the ball stops with the hit, then yes, you may be able to ignore the fact that there is a transfer from PE to both linear KE and rotational KE during the roll. Note that the dynamics of the hit may dominate -- that is, with the slowly rolling ball, rolling forward, the contact patch will push down and forward on the pin. This counteracts the tipping motion that you are shooting for, and will tend to push the pin back rather than tip it. If you can minimize the coefficient of friction between the pin and ball, and maximize the coefficient of friction between the pin and ground, then it's more of a tipping rather than a hit...

    Also, in your calculations be sure to carry along the units of the numbers (like meters per second, or m/s). It's too easy to make a mistake by mixing units (like cm and meters).
     
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