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March 30, 2017

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Suppose that a simple pendulum consists of a small 66.0 g bob at the end of a cord of negligible mass. Suppose that the angle between the cord and the vertical is given by
θ = (0.0800 rad) cos[(4.30 rad/s)t + ϕ]

(a) What is the pendulum's length?

(b) What is its maximum kinetic energy?

  • Physics - ,

    I figured out a) by using w=2pi/T
    T=1.46 then used T to solved T=(2pi*sq root L/g) L = .53 but I can't figure out an equation to give me the velocity. Any ideas?

  • Physics - ,

    (a) w = 4.40 rad/s
    is the angular frequency, sqrt(g/L)

    Solve for L, the pendulum length

    L = g/w^2

    (b) Max velocity Vmax = = L*theta*w

    Maximum KE = (1/2) M Vmax^2

  • Physics - ,

    You are right about L = 0.53. Don't forget the units (meters)

    In simple harmonic motion,
    max velocity = w*(Amplitude) and
    max acceleration = w^2*(Ampitude)

    In your case I had to add an L factor to get linear velocities from the angular amplitude.

  • Physics - ,

    Where did the Vmax = L*theta*w
    come from? How do I find theta?

  • Physics - ,

    Theta should be theta-max, the angular amplitude, which is 0.0800 radians

    Sorry about that

  • Physics - ,

    Vmax= (.53m)(.0800rad)(4.30 rad/s)
    =.182

    KE=1/2mv^2
    =1/2(.066kg)(.182)^2
    =.0011 or 1.09e-03
    That doesn't look right to me. what am I doing wrong? ;_;

  • Physics - ,

    Another question, sorry, trying my bests to understand your thinking. I don't see where you knew to multiple L * theta * W.
    I understand that the 2nd derivative of the equation gives you the Velocity function.

  • Physics - ,

    No, the first derivative of your theta vs t (multiplied by L) gives you the velocity function.

    L*theta_max is the displacement amplitude (in small angle approximation) ; so
    w*L*theta_max
    is the maximum velocity

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