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Velocity and Acceleration

  1. Dec 6, 2011 #1
    1. The problem statement, all variables and given/known data
    Determining the constants.

    After having trouble with differenciation, I think I finally got there. However now I have to determine some constants.

    I need to find numerical values for the constants B and C, which are velocity and acceleration.

    A ball is dropped down a shaft on an asteroid from position = R at t=0s.

    the x-component of the ball (it's position) has this form.

    x(t) = A = Bt + Ct^2

    Hence using differenciation, i've found

    v(t) = dx/dt = B + 2ct

    and

    a(t) = d^2x/dt^2 = 2C

    I've already found that A=R, but now need to find NUMERICAL values for B and C

    All i've been given to start are the balls mass = 0.10kg, and it weight at the surface - 1.2 N.

    The heading for the question reads;
    The cosmonaut performs an experiment by dropping a small ball down the shaft, and uses a laser to accurately determine the ball’s position as a function of time. The ball is released from x = R at time t = 0. At small times, she finds that the x-component of the position of the ball has the following form:

    part one reads) Determine the constant A (this part I have completed)
    part two reads) Given that the ball is released from rest, derive an expression for the velocity, and hence determine the value of constant B.
    (I don't know about you, but that to me sounds like we just need a velocity expression, which will give me the answer v, no differenciation required here)
    part three reads) the ball has a mass and weight (given above) at the surface of the asteroid. Determine the acceleration due to gravity at the surface of the asteroid. Hence use differenciate to detemine the value of constant C.
    (To me this sounds like we need an equation to find acceleration, and then differenciate to find how it plays in this context)



    2. Relevant equations

    W = mg, rearranged to g = W/m

    I thought this equation might help, but since I haven't got the values of rho (the density of the asteroid) or r (the radius) I can't see how it can.

    gif.latex?F%20=%20\frac{4G\pi%20\rho%20}3{}%20m%20r\hat{%20r}.gif

    So now we have the gravity force of the asteriod.

    3. The attempt at a solution

    g = W/m = 1.2 N/0.10kg = 12 N kg^-1 (Not sure on the units)

    Not sure where to begin with finding these numerical values. Help please!
     
  2. jcsd
  3. Dec 6, 2011 #2
    Also, forgot to say the equation for v

    v = u + at

    And although we dont yet know a, it's neglible since it's x 0, so makes it 0 anyway. hence v = 0 m/s
     
  4. Dec 6, 2011 #3
    Since
    x(t) = A = Bt + Ct^2

    then x(0) = A = B*0 + C*0^2 = 0, Hence R = A = 0.

    Also since v(t) = dx/dt = B + 2ct and 'that the ball is released from rest'
    then v(0) = dx/dt = B + 2C*0 = B = 0.

    since a(t) = d^2x/dt^2 = 2C = g = 12 as your correctly worked out, then C = ...
     
  5. Dec 6, 2011 #4
    Ah, so 2C = g = 12?

    So it would be 2/C = 2/12 = 6 m/s.
     
  6. Dec 6, 2011 #5
    C = 12/2 = 6m/s^2
     
  7. Dec 6, 2011 #6
    oh yes, of course, it's an acceleration!

    So, thats it then? I've been going over this for about two days now... it seems a little simple now!

    Many thanks.
     
  8. Dec 6, 2011 #7
    By the way, did I have my units correct for w/m = g? = 12 N kg^-1
     
  9. Dec 6, 2011 #8
    Weight W is a force and hence in units of N.
    Mass is M in units of kg.
    g is an acceleration in units of m/s^2.

    W = Mg
    N = kgms[itex]^{-2}[/itex]
     
  10. Dec 6, 2011 #9
    I thought g was the gravitational force, and we were looking for the acceleration from that?
     
  11. Dec 6, 2011 #10
    Since W = mg then g is the acceleration due to gravity.

    If m = 1 then g is the gravitational force on 1kg.

    So g can be called the 'acceleration due to gravity' but it can also be called the 'gravitational force on 1kg'.
     
  12. Dec 6, 2011 #11
    Ah ok, I get it. Thanks very much
     
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