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Hammer striking an anvil with a velocity of 50ft/sec

  1. Oct 20, 2014 #1
    1. The problem statement, all variables and given/known data
    A hammer strikes an anvil with a velocity of 50ft/sec. The hammer weights 12 lb and the anvil weighs 100 lb. The anvil is supported on four springs with k = 100 lb/in. Find the motion if
    (a) the hammer stays in contact with the anvil
    (b) the hammer doesn't remain in contatct

    2. Relevant equations
    ##1 lb = 4.45 N##
    ##1 m = 3.28ft##
    ##k_{eq} = 4k = 400\cdot 4.45\cdot 12\cdot 3.28 = 70060.8## N/m
    ##W = mg## so ##m_h = 12*4.45/9.8 = 5.45## kg and ##m_a = 100*4.45/9.8 = 45.41## kg
    ##\dot{x}(0) = 50/3.28 = 15.24## m/s

    3. The attempt at a solution
    (a)
    $$
    M\ddot{x} + k_{eq}x = 0
    $$
    where ##M = m_h + m_a = 50.86## kg.
    $$
    \ddot{x} + \omega_n^2x = 0
    $$
    where ##\omega_n^2 = \frac{70060.8}{50.86} = 1377.52##
    Let ##x(t) = A\cos(\omega_nt) + B\sin(\omega_nt)##. Then ##\ddot{x} = -A\omega_n^2\cos(\omega_nt) - B\omega_n^2\sin(\omega_nt)##.
    $$
    \cos(\omega_nt)[-A\omega_n^2 + A\omega_n^2] + \sin(\omega_nt)[-B\omega_n^2 + B\omega_n^2] = 0
    $$
    The coefficient zero out so this cant be correct. The RHS has to be zero since there is no driving force.
     
    Last edited: Oct 20, 2014
  2. jcsd
  3. Oct 20, 2014 #2

    haruspex

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    Leave out all the numbers and conversions for now, and just do everything symbolically. Hammer mass m moving vertically down at speed u strikes anvil of mass M supported on four springs of coefficient k. What is the speed of the anvil immediately after impact?
     
  4. Oct 20, 2014 #3
    I have been thinking about this. Should the initial velocity for the inelastic collision be found as
    $$
    m_h(50)^2 = v^2(m_h + m_a)
    $$
    Then v would be ##\dot{x}(0) = v##?

    For part b, since the hammer comes off immediately, that would be view as a delta spike and ##\dot{x}(0) = 50## the velocity of the hammer, correct?
     
  5. Oct 21, 2014 #4

    haruspex

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    Since it is an inelastic collision, work is not conserved. What is conserved?
     
  6. Oct 21, 2014 #5

    rude man

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    Try instead using momentum conservation for both cases, then relate decreasing kinetic energy to increasing potential energy after contact.
     
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