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Angular motion with constant acceleration

  1. Oct 14, 2007 #1
    Consider two particles A and B. The angular position of particle A, with constant angular acceleration, depends on time according to theta(t)=theta_0+omega_0t+alpha t^2. At time t=t_1, particle B, which also undergoes constant angular accelaration, has twice the angular acceleration, half the angular velocity, and the same angular position that particle A had at time t=0.
    How long after the time t_1 does the angular velocity of B have to be to equal A's?
    ___
    ok, here we go:
    i know that i have to write expressions for the angular velocity of A and B as functions of time, and solve for t-t1???
    For A, omega(t) = omega_0 + alpha(t) and
    for B, omega(t) = 0.5 omega_0 + 2alpha(t-t_1)
    so when i equate them and solve for t-t_1, i get something that is not the right answer!

    please help :(
     
  2. jcsd
  3. Oct 14, 2007 #2

    Astronuc

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    Staff: Mentor

    One wants expressions for [itex]\theta_A(t)[/itex] and [itex]\theta_B(t)[/itex], which are set equal at time t, so one can solve for time t, then find t - t1.

    at t = 0, [itex]\theta_A(0)[/itex] = [itex]\theta_0[/itex], and at t=t1, [itex]\theta_B(t_1)[/itex] = [itex]\theta_0[/itex], which is the starting position of A at t=0.

    In the expression for B, one has to address the time lag t1.
     
  4. Oct 15, 2007 #3
    can anyone please just lay it out for me .. because i can't see it
     
    Last edited: Oct 15, 2007
  5. Oct 15, 2007 #4

    Astronuc

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    Staff: Mentor

    One has for A, [tex]\theta(t)[/tex] = [tex]\theta_0\,+\,\omega_0t+\alpha t^2/2[/tex],

    and for B, one must use t-t1 since it starts at t1, with

    angular acceleration [itex]2\alpha[/itex] and angular velocity [itex]0.5\omega[/itex].

    At the same position [tex]\theta_A(t)[/tex]=[tex]\theta_B(t)[/tex]
     
    Last edited: Oct 15, 2007
  6. Oct 15, 2007 #5
    how do i get to the answer. i know that the answer is (omega_0 + 2alpha(t_1)) / (2alpha) but i want to know how they got that. and i need to know what i did wrong in #1 box because it looks okay to me
     
  7. Oct 15, 2007 #6
    i KNOW theta A equation and theta B equation.
     
  8. Oct 15, 2007 #7

    Astronuc

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    Staff: Mentor

    Oops, sorry, I was solving for the same position.

    If theta(t)=theta_0+omega_0t+ 0.5alpha t^2 for A, then

    omega_A(t) = omega_0 + alpha *t

    for B

    omega_B(t) = 0.5*omega_0 + 2 alpha *(t-t1)

    when one differentiates t2, the derivative is 2t.
     
    Last edited: Oct 15, 2007
  9. Oct 15, 2007 #8
    you are wrong. i am sorry. i will change what i said in #1

    Consider two particles A and B. The angular position of particle A, with constant angular acceleration, depends on time according to theta(t)=theta_0+omega_0t+0.5alpha t^2. At time t=t_1, particle B, which also undergoes constant angular accelaration, has twice the angular acceleration, half the angular velocity, and the same angular position that particle A had at time t=0.
    How long after the time t_1 does the angular velocity of B have to be to equal A's?
    ___
    ok, here we go:
    i know that i have to write expressions for the angular velocity of A and B as functions of time, and solve for t-t1???
    For A, omega(t) = omega_0 + alpha(t) and
    for B, omega(t) = 0.5 omega_0 + 2alpha(t-t_1)
    so when i equate them and solve for t-t_1, i get something that is not the right answer!

    please help :(

    IT IS 0.5 ALPHA T^2
     
  10. Oct 15, 2007 #9
    THEREFORE For A, omega(t) = omega_0 + alpha(t) and
    for B, omega(t) = 0.5 omega_0 + 2alpha(t-t_1)

    now what do i do with these equations?
     
  11. Oct 15, 2007 #10

    Astronuc

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    Staff: Mentor

    I suspect the problem is simply algebraic.

    Equate the two expressions for angular velocity.

    [tex]\omega_A(t) = \omega_B(t)[/tex]

    [tex]\omega_0\,+\,\alpha{t}\,=\,\omega_0/2\,+\,2\alpha{(t-t_1)}[/tex]

    which leads to

    [tex]\omega_0/2 \,=\, 2\alpha{(t-t_1)}\,-\,\alpha{t}[/tex]

    or


    [tex]\omega_0/2 \,= \,\alpha{t} - 2 \alpha{t_1}[/tex]

    take an alpha t_1 to the other side

    [tex]\omega_0/2 + \alpha{t_1} \,= \,\alpha{t} - \alpha{t_1}[/tex]

    and see where that leads one
     
    Last edited: Oct 15, 2007
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