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Torque: using modified balistic pendulum with spring

  1. Apr 1, 2006 #1
    Hi everyone i really need help with this, unfortunately my "signals and systems" professor dropped this on us without even covering it in class. From my earlier physics class i can solve for theta in a ballistic pendulum, but since this one has a string attached to it, i really have no clue how to write the torque equation for this.

    Please any help or hints will be appreciated. Thanks


    A pendulum is initially at rest. It is fixed to a horizontal axis by a rigid rod length of l= 100cm (both masses of the axis and rod are negligible). A spring of torsional stiffness K=196n*m/rad and a torsional dashpot of damping coefficient B=88.54n*m*sec/rad have been connected to the axis. A bullet of mass m=100g traveling with a speed v is fired into the pendulum with a bob mass M = 19.9 kg at time t = 0 and remains lodged therein.

    1.Find the differential equation relating the angular displacement θ to the input v. What is the order of the system? Determine the state and output equations in matrix form. Use Matlab to find and plot the unit impulse response and determine the total output response for the inputs of v(t): 200 m/s and 800 m/s
    2.Discretize the system and repeat question 1.
    Last edited: Apr 2, 2006
  2. jcsd
  3. Apr 2, 2006 #2

    Andrew Mason

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    What have you got so far?

    What is the differential equation of motion without the spring and damping? Then add the spring and damping. Assume the bullet takes a very short time to stop in the block so that the initial angular velocity is:

    [tex]v_0/R = \dot\theta_0 = \frac{mv_b}{R(m+M)}[/tex]

  4. Apr 2, 2006 #3
    so far i was able to get the final velocity of the system as

    [tex]\frac{M+m}{2g} v_b^2 = (M+m) h [/tex] where h = l(1-cos θ)

    therefore [tex] v_b^2 = 2 g l (1-cos theta) [/tex]
    Last edited: Apr 2, 2006
  5. Apr 2, 2006 #4

    Andrew Mason

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    Try analysing the forces on the block tangential to the rod:


    [tex]F_g = - (M+m)gsin\theta \approx - (M+m)g\theta[/tex]


    [tex]F_s = - k\theta[/tex]

    Dashpot damper:

    [tex]F_d = - B\dot\theta[/tex]

    So the torque on the system is:

    [tex]F \times R = I\alpha = (F_g + F_s + F_d)R = (m+M)R^2\ddot\theta[/tex]

    So set up the differential equation and use the initial condition for [itex]v_0[/itex] at time t=0.

    Last edited: Apr 2, 2006
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