physics Classical Mechanics

A pendulum of mass m= 0.9 kg and length l=1 m is hanging from the ceiling. The massless string of the pendulum is attached at point P. The bob of the pendulum is a uniform shell (very thin hollow sphere) of radius r=0.4 m, and the length l of the pendulum is measured from the center of the bob. A spring with spring constant k= 12 N/m is attached to the bob (center). The spring is relaxed when the bob is at its lowest point (θ=0). In this problem, we can use the small-angle approximation sinθ≃θ and cosθ≃1. Note that the direction of the spring force on the pendulum is horizontal to a very good approximation for small angles θ. (See figure)

Take g= 10 m/s2

(a) Calculate the magnitude of the net torque on the pendulum with respect to the point P when θ=5∘. (magnitude; in Nm)

|τP|=

unanswered
(b) What is the magnitude of the angular acceleration α=θ¨ of the pendulum when θ=5∘? (magnitude; in radians/s2)

|α|=

unanswered
(c) What is the period of oscillation T of the pendulum? (in seconds)

T=

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  1. 8:01 final question. The easiest one.

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    posted by Damon
  2. damon can u please tell me about the T that is the last part of this question

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    posted by fighter
  3. omega= sqrt((kl^2+mgl)/I)

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    posted by enjoy
  4. please tell me the exact expression for this question.

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    posted by Singh
  5. |ôP|= 1.83

    |á|= 1.84

    T= 1.37

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    posted by Greco
  6. Can you plz give out the formula??

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  7. I = m L^2 + (2/3) m R^2
    -(k L^2 + mgL)T = .885 d^2T/dt^2
    well calculate the coefficient of T on the left e.g.
    -(7 + 8) T = .885 d^2T/dt^2
    .885 d^2T/dt^2 = - 15 T
    Torque = -15T where T = 5 * pi/180
    alpha = d^2T/dt^2 = - 15 T /.885

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    posted by Greco
  8. .885 w^2 = 15
    w = 2 pi f = 2 pi/period = 4.11 for example and you solve fot period

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    posted by Greco

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