1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Force & Acceleration

  1. Apr 29, 2009 #1
    1. The problem statement, all variables and given/known data

    A 10lb bar is pinned at its center O and connected to a torsional spring. The spring has a stiffness k = 5 lb ft/rad so that the torque develped is M = 5(theta) lb ft where theta is in radians. If the bar is released from rest when it is vertical at theta = 90 degrees, determine its angular velocity at the instand theta = 45 degrees


    2. Relevant equations

    I(o) = .5mr^2
    Moment = I(o)a where a is angular acceleration
    w^2 = w(o)^2 + 2a(theta - theta(initial))

    3. The attempt at a solution

    10lb = .3108 slug

    I = .5(.3108)(1^2) = 1.554 sluf ft^2

    Moment = Ia 5theta = 1.554a
    a = 32.1750theta

    90 degrees = 1.5708 radians
    45 degrees = .7854 radians

    w^2 = 0 + 2(32.1750)(.7854)(.7854 - 1.5708)


    w = 6.3004 rad/s


    Should the inital theta be 0.
     
  2. jcsd
  3. Apr 29, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper

    First of all your I for the bar is incorrect.

    I = 1/12*M*L2

    You have used 1 for the length of the bar. Is that part of the given?
     
  4. Apr 30, 2009 #3
    Sorry i forgot the picture. My text says r, which i assumed was radius.
     

    Attached Files:

  5. May 1, 2009 #4
    am i still inncorect
     
  6. May 1, 2009 #5

    LowlyPion

    User Avatar
    Homework Helper

    Your drawing shows L = 2

    This means your I is

    I = 1/12*(.3108)*22 = .3108/3 = .1036
     
  7. May 1, 2009 #6

    LowlyPion

    User Avatar
    Homework Helper

    Consider using the potential energy in the spring, and determining the rotational kinetic energy imparted by the time it reaches 1/2 θ

    PE = 1/2*k*θ2 = 1/2*k*(θ/2)2 + 1/2*I*ω2

    Simplifying I get:

    ω2 = k*θ2/ (2*I)
     
  8. May 1, 2009 #7
    w^2 = (k*theta^2)/(2I) where k = 5 lb ft, theta = .7854 radians, I = .1036

    w = 3.86 rad/s


    why did u use 1/2 theta
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Force & Acceleration
  1. Force and Acceleration (Replies: 0)

  2. Force and Acceleration (Replies: 5)

Loading...