1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Acceleration of a point on square

  1. Jul 11, 2015 #1
    1. The problem statement, all variables and given/known data

    ?temp_hash=dd05ed32e7a7e111396fc0f1b81af655.png

    2. Relevant equations



    3. The attempt at a solution

    The origin is placed to the right of point P and left is taken to be the positive direction.Let us call mass of each rod as M and the angle which rods make with the horizontal as θ .

    Coordinates of A = (##x_P+\frac{L}{2}cosθ,\frac{L}{2}sinθ## )

    Coordinates of B = (##x_P+\frac{3L}{2}cosθ,\frac{L}{2}sinθ## )

    Coordinates of C = (##x_P+\frac{3L}{2}cosθ,-\frac{L}{2}sinθ##)

    Coordinates of D = (##x_P+\frac{L}{2}cosθ,-\frac{L}{2}sinθ## )


    Now writing some constraint equations ,

    ##x_q = x_p + 2Lcosθ##

    ##v_q = v_p -2L(sinθ\dot{θ})##

    ##a_q = a_p -2L(cosθ\dot{θ}^2+sinθ\ddot{θ})##

    ##a_{cm} = \frac{1}{2}(a_p+a_q)##

    The force applied on the square F= 4Macm . This force produces displacement of P ,xP in a very short time interval 't' . ##x_P = \frac{1}{2}a_Pt^2##

    Work done on the square =FxP .

    Now we will calculate the velocities of the CM of the rods .

    ##v_{A,x} = v_{P,x} - \frac{L}{2}sinθ\dot{θ}##
    ##v_{A,y} = \frac{L}{2}cosθ\dot{θ}##
    ##v_{B,x} = v_{P,x} - \frac{3L}{2}sinθ\dot{θ}##
    ##v_{B,y} = \frac{L}{2}cosθ\dot{θ}##

    Similarly we can write speeds of the two bottom rods .

    Magnitude of vertical component of velocity of CM of all four rods would be equal . Magnitude of horizontal component of velocity of CM of two right side rods would be equal whereas that of two left side rods would be equal.Angular speeds of all four rods would be same .

    Now applying Work Energy theorem ,

    $$4M\frac{1}{2}(a_p+a_q)(x_P) = 2(\frac{1}{2}M){(v_{P,x}- \frac{L}{2}sinθ\dot{θ})}^2+2(\frac{1}{2}M){(v_{P,x}- \frac{3L}{2}sinθ\dot{θ})}^2 + 4(\frac{1}{2}M){(\frac{L}{2}cosθ\dot{θ})}^2 + 4\frac{1}{2}(\frac{ML^2}{12})\dot{θ}^2$$

    In the above expression we can put ##x_P = \frac{1}{2}a_Pt^2## , ##\dot{θ} = \ddot{θ}t ## and ##v_{P,x} = a_Pt##.

    $$4M\frac{1}{4}(a_p+a_q)(a_Pt^2) = 2(\frac{1}{2}M){(a_Pt- \frac{L}{2}sinθ(\ddot{θ}t))}^2+2(\frac{1}{2}M){(a_Pt- \frac{3L}{2}sinθ(\ddot{θ}t))}^2 + 4(\frac{1}{2}M){(\frac{L}{2}cosθ(\ddot{θ}t))}^2 + 4\frac{1}{2}(\frac{ML^2}{12})(\ddot{θ}t)^2$$

    Is it correct till here ? If yes, how do I eliminate ##\ddot{θ}## ?

    I would be grateful if someone could help me with the problem .
     

    Attached Files:

  2. jcsd
  3. Jul 11, 2015 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    aq can be also expressed by ##\ddot \theta##. Solve for ##\ddot \theta##, and determine aq with it.
     
  4. Jul 11, 2015 #3
    Is there another expression between aq and ##\ddot \theta## besides ##a_q = a_p -2L(cosθ\dot{θ}^2+sinθ\ddot{θ})## which I mentioned in the OP?
     
    Last edited: Jul 11, 2015
  5. Jul 11, 2015 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    It is enough.
    The velocity terms vanish. You can substitute sinθ and cosθ with their valus at t=0.
     
  6. Jul 11, 2015 #5
    Thank you very much :oldsmile: .
     
    Last edited: Jul 11, 2015
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Acceleration of a point on square
Loading...