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Ladder of mass M and length L, find kinetic energy in terms

  1. Dec 27, 2011 #1
    1. The problem statement, all variables and given/known data

    A ladder of length L and mass M leaning against a wall. Assuming the wall and the floor are frictionless, the ladder will slide down the wall and along the floor intl the left end closes contact with the wall. Before the ladder loses contact with the wall there is one degree of freedom θ. Express the knetic energy of the ladder in terms of θ and θ' (derivative of θ).


    2. Relevant equations

    dm = M/L ds

    3. The attempt at a solution

    I think that the problem should be solved combining the kinetic energy of rotation (sliding) of the ladder + the ladder displacing horizontally.

    Dividing the ladder into infinitesimal slices of ds, each of it has its kinetic energy.
    I first try to calculate the kinetic energy of the rotation about the axis (like it would not move).

    Each part of ladder moves with velocity of v_i= θ'*s_i

    where s_i is the distance from the origin to the slice of i'th slice, θ' - angular velocity.

    Now summing those kinetic energies of each slice:

    1/2 Ʃm_i*(v_i)^2 → 1/2 ∫[from 0 to L] M/L ds * (θ'*s_i)^2 = M/2L *(θ')^2 ∫s^2 ds =
    M(θ')^2/(2L) * L^3/3 = 1/6 * ML^2 (θ')^2


    The second part is to calculate kinetic energy of the ladder only assuming that it moves away from the wall with no rotation. However I could not find the solution yet.
     
  2. jcsd
  3. Dec 29, 2011 #2

    ehild

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    Gold Member

    You can find the whole KE of the ladder by squaring the velocity of all slices and integrating. The components of velocity are the time derivatives of the coordinates. Write out the coordinates in terms of theta and the distance from one end of the ladder, and differentiate: vx=[itex]\dot{x}[/itex] and vy=[itex]\dot{y}[/itex]. You will find that the KE of the ladder is the same expression you have got: 1/2 (mL2/3) (θ')2. There is no other term.

    ehild
     
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