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Mechanics ynamics (Newton's 2nd Law?)

  1. Oct 22, 2009 #1
    Mechanics:Dynamics (Newton's 2nd Law?)

    1. The problem statement, all variables and given/known data
    ic0ot2.png

    A sphere S with a mass of 5 kg is attached by a rigid rod to 1 kg block B which is free to slide with no friction in a horizontal slot. The system is released from rest. At the instant when it is released, find the tension in the rod and the accelerations of both blocks.

    2. Relevant equations

    Sum of the forces in x = ma
    Sum of the forces in y = ma

    3. The attempt at a solution

    I am not exactly sure how to tackle this problem. Would I need to solve this using newton's 2nd law? Or would this question require conservation of energy?
     
  2. jcsd
  3. Oct 23, 2009 #2
    Re: Mechanics:Dynamics (Newton's 2nd Law?)

    Using Newton's 2nd Law would definitely help.
    And yes, your equations are correct.

    Can u show your approach and the answer you are getting?
     
  4. Oct 23, 2009 #3
    Re: Mechanics:Dynamics (Newton's 2nd Law?)

    I have another question then. Since the problem is asking "the instant it is released from rest" can you assume the velocity is 0 at that instant? Therefore the component of normal acceleration is 0?
     
  5. Oct 23, 2009 #4
    Re: Mechanics:Dynamics (Newton's 2nd Law?)

    Yes, you can say that the initial velocity would be zero.. and yes, the normal acc.(due to circular motion) is also zero.
    I guess you also have to use Work-Energy Conservation to solve.
     
  6. Oct 23, 2009 #5
    Re: Mechanics:Dynamics (Newton's 2nd Law?)

    I don't think work and energy can be used here since you are looking for the accelerations.

    Would relative acceleration need to be considered for this problem since the acceleration of the sphere is moving while the block is moving.
     
  7. Oct 24, 2009 #6
    Re: Mechanics:Dynamics (Newton's 2nd Law?)

    Umm.. yeah.
    But we would have used work-energy in case we needed acceleration after some time interval.

    YES.

    Can u post your complete approach?
     
  8. Oct 24, 2009 #7
    Re: Mechanics:Dynamics (Newton's 2nd Law?)

    You need to relate the forces that act between the block and the sphere. I would pick the side of the wall as your reference frame, x.
    I think you should get a pair of coupled differential equations with tension in there somewhere. Then use your initial conditions.
     
  9. Oct 24, 2009 #8
    Re: Mechanics:Dynamics (Newton's 2nd Law?)

    Would tangential acceleration of S = (m_s)*g*cos(30)=(m_s)(a_st)
    so a_s tangential = 8.50 m/s^2?
     
  10. Oct 24, 2009 #9
    Re: Mechanics:Dynamics (Newton's 2nd Law?)

    and also since system is initially at rest wouldn't T=(m_s)*(g)*(sin(30)) = 24.525N
    then solving for T in the horizontal dir. T_x=T(cos(30))

    so T_x=(m_s)*(g)*(sin(30))*(cos(30)) =21.239
    (T_x)/(m_b)= (a_b) = 21.239 m/s^2
     
  11. Oct 24, 2009 #10
    Re: Mechanics:Dynamics (Newton's 2nd Law?)

    You need to take relative acceleration into account I am pretty sure.
     
  12. Oct 24, 2009 #11
    Re: Mechanics:Dynamics (Newton's 2nd Law?)

    Let X, Y be the position of the sphere, and x the x-position of the block

    [itex] A_x [/itex] is the acceleration in the x direction of the sphere.
    [itex] A_y [/itex] is the acceleration in the y direction of the sphere.
    [itex] a_x [/itex] is the acceleration in the x direction of the block.

    Find [itex] A_x [/itex]. [itex] A_y [/itex], [itex] a_x [/itex] as a function of T, the tension in
    the rope.

    Find another relation between [itex] A_x [/itex]. [itex] A_y [/itex]and [itex] a_x [/itex]
    by differentiating (X(t)-x(t))^2 + Y(t)^2 = L^2.

    Solve for T
     
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