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PHYSICS(HELP)

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A pendulum of mass m= 0.8 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= 7 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|=

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

|α|=


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

T=

  • PHYSICS(HELP) - ,

    Is this an 8:01 problem ? If not I will do it.

  • Is this an 8:01 problem? - ,

    If it is not I will help.

  • PHYSICS(HELP) - ,

    no it isn't

  • PHYSICS(HELP) - ,

    Thanks

  • PHYSICS(HELP) - ,

    Lets do moments of inertia about the pivot point

    First the moment of inertia of this thin shell about a horizontal line through its center is (2/3) mR^2
    for derivation see http://hyperphysics.phy-astr.gsu.edu/hbase/isph.html
    This has to be transferred up to the pivot point using the parallel axis theorem
    I = m L^2 + (2/3) m R^2
    here
    m = .8
    L = 1
    R = .4
    so
    I = .8 + .085 = .885

  • PHYSICS(HELP) - ,

    but how do you get alpha?

  • PHYSICS(HELP) - ,

    Now let's do Moments about the pivot point
    let angle theta = T
    restoring Force = - k x - m g sin T
    but sin t is about T in radians for small T
    so
    F = -k x - m g T
    x is about L T
    so
    F = -k L T - m g T
    moment = M = F L = -(k L^2 + mgL)T
    moment = I alpha
    -(k L^2 + mgL)T = .885 d^2T/dt^2
    well calculate the coefficient of T on the left
    -(7 + 8) T = .885 d^2T/dt^2
    .885 d^2T/dt^2 = - 15 T
    let T = A sin ( w t)
    then d^2T/dt^2 = - w^2 T
    so
    .885 w^2 = 15
    w = 2 pi f = 2 pi/period = 4.11

  • PHYSICS(HELP) - ,

    Hey, I am kind of slow and can not do everything at once.
    Check arithmetic!

  • PHYSICS(HELP) - ,

    when you do these, use T in RADIANS
    5 degrees *pi/180


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

    Torque = -15T where T = 5 * pi/180

    |τP|=

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

    alpha = d^2T/dt^2 = - 15 T /.885

    |α|=

  • PHYSICS(HELP) - ,

    Hey Damon, i just want to say thank you for making me believe in myself one more time. i got this but i didn't think my workings were right,so i was scared to key in my answers. Now,i am sooo confident in myself. thank you.

  • PHYSICS(HELP) - ,

    how is the period found?

  • PHYSICS(HELP) - ,

    this is an exercise of the exam, somoone posted in my name. no problem though . cheaters are fooling themselves.

  • PHYSICS(HELP) - ,

    I really pity these fools, who do not have a clue about anything...

  • PHYSICS(HELP) - ,

    are you the real shaka? or are you a saboture

  • PHYSICS(HELP) - ,

    the fact that your browsing these forums means you are a cheater as well

  • PHYSICS(HELP) - ,

    At this point anyone who needs help finding the period is in serious trouble.

  • PHYSICS(HELP) - ,

    hahaha

  • PHYSICS(HELP) - ,

    hey DAMON can u please tell about C part of the question

  • PHYSICS(HELP) - ,

    I did.

  • PHYSICS(HELP) - ,

    This problem is really all about an expression summarizing harmonic motion.

    So your period for this problem is oddly similar to

    T = 2pi * sqrt (m/k)

    but now m = moment of inertia
    and k = expanded coefficient, that was given above by Mr. Damon

    Also make sure that you end up with seconds as units

  • PHYSICS(HELP) - ,

    this will surely hurt more than help but here you go

    Period = 2*pi sqrt([kl^2+mgl]/[(2mr^2)/3+ ml^2)

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