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Mass and time question using calculus

  1. Nov 30, 2003 #1
    I am working on an AP physics question that I am supposed to solve by using an integrated force(not an average) and I am a bit stumped. Please help!

    An 8 kg turkey is placed 1 m(from their centers of mass) from an 8 kg ham on a friction less table. How much time passes before the turkey and ham touch?
    (A picture that is above the question shows both the turkey and ham to be .2m wide, so it is only a distance of .8 m before their edges touch)

    I have started by finding the integral of F=GMm/r^2 dr (from 0 to .4) by plugging in the gravitational constant and both 8 kg masses for M and m. I found the indefinite integral to be (4.27*10^-9)/r and the definite integral to be -1.07*10^-8 N Is this the work that I've found, and if so(or if not) where do I go next?
  2. jcsd
  3. Nov 30, 2003 #2
    Okay I am rethinking this. What can I find the integral of to give me a force?
  4. Nov 30, 2003 #3


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    Integrating a force with respect to a distance gives energy.
  5. Nov 30, 2003 #4
    Okay thanks for that information. So I don't need to integrate the gravitiational force, but what can I integrate to get the gravitiational force? What is measured in Newtons per time? Please tell me if I'm going in the wrong direction here.
  6. Nov 30, 2003 #5


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    Well, I'm not sure that differentiating force with respect to time will be helpful here.

    I've thought a bit about this. You can find the acceleration of each particle as a function of their distance apart, although you need them as functions of time so you can integrate twice to get their positions as functions of time.

    I'm unsure how to do this.
  7. Nov 30, 2003 #6
    I have figured this question out without using calculus but it is only with an average force. Maybe if I explain what I did, you'll see something that I haven't and be able to help me find the true force.

    I found the gravitational force to be 4.27*10^-11 by doing F=GMm/r^2

    Then I set that equal to F=ma where m=8 and I solved for a
    a=5.34*10^-12 m/s^2

    Then using d=vt+.5at^2 (where v is the initial velocity which is 0)
    d=.4 (the distance one of the particles must move before touching the other)
    a= answer from above
    so t= 387,000 sec (approxiamately)

    Did you think of a was to integrate to find this force?

    Thank you for trying this!
  8. Nov 30, 2003 #7


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    Your answer is incorrect because you didn't integrate!

    You calculated the intial force, used that to find the initial acceleration and then assumed that acceleration until they hit: the force of gravity between the two objects becomes greater as they get closer so the acceleration will increase.

    You know F= ma so m dv/dt= -GmM/r2. But you need "dv/dr" not dv/dt since you have accelration as a function of r, not time. Use the chain rule: dv/dt= dv/dr dr/dt= v dv/dr:
    m vdv/dr= -GmM/r2 so you can integrate v dv= -GM/r2 (this is a general method called "quadrature".)

    After you know v as a function of r, v= dr/dt so dt= (1/v)dr.

    You will need to take into account that both objects are moving.
  9. Nov 30, 2003 #8
    I can follow and understand everything up to the quadrature part, how does that work?
  10. Nov 30, 2003 #9
    I think what Halls was saying was that the two masses 'M' and 'm' are moving towards each other. The quadrature method will solve for the position of only one mass. And you wanted to work out the time for the masses to come together so you've got set each of their positions (with respect to time) equal to each other and solve for time. Just be wary of one thing, the mass on the RHS is travelling with a negative velocity from the equations that have been setup, but the algebra should sort that out:smile: !!
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