1. Not finding help here? Sign up for a free 30min 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!

I really need help !- angular speed and speed at equilibrium position

  1. May 5, 2004 #1
    I really need help !!!!!!- angular speed and speed at equilibrium position

    Problem 1. A 5.00-kg block free to move on a horizontal, frictionless surface is attached to one end of a light horizontal spring. The other end of the spring is held fixed. The spring is compressed 0.100m from equilibrium and released. The speed of the block is 1.20m/s when it passes the equilibrium position of the spring. The same experiment is now repeated with the frictionless surface replaced by a surface for which the coefficient of kinetic friction is 0.300. Determine the speed of the block at the eqilibrium position of the spring.

    The second problem I have is: The reel has radius R and moment of inertia I. One end is connected to a spring of force constant k, and the other end is fastened to a cord wrapped around the reel. The reel axle and the incline are frictionless. The reel is wound counterclockwise so that the spring stretches a distance d from its unstretched position and is then released from rest. Find the angular speed of the reel when the spring is again unstretched.
    :confused: :confused: :confused: If someone could be help me I would really appreciate it
  2. jcsd
  3. May 5, 2004 #2
    What have you done? Where are you stuck?
  4. May 5, 2004 #3
    getting started..I lost my physics book while I was packing and now I'm having trouble starting these two problems. I can remember how to start them.
  5. May 5, 2004 #4
    Is that can or can't?

    Some useful facts (not all necessarily needed for these problems):

    [tex]x = A\cos(\omega t + \phi)[/tex]
    [tex]v = -\omega A \sin(\omega t + \phi)[/tex]
    [tex]U = \frac{1}{2}kx^2[/tex]
    [tex]K = \frac{1}{2}mv^2[/tex]
    [tex]K = \frac{1}{2}I\omega^2[/tex]
    [tex]\omega = 2\pi f[/tex]
    [tex]s = r\theta[/tex]
    [tex]v_t = r\omega[/tex]

    A hint for both problems: what can you determine about the total energy of the system?
  6. May 5, 2004 #5
    :confused: :confused: :confused: :confused:
  7. May 5, 2004 #6
    In the first setup, there is no friction, and you can ignore gravity since acts perpendicular to the direction of motion. So the only force you need to consider is the spring tension. This is a conservative force, meaning that TOTAL energy is conserved -- i.e. constant. Total energy in this case is kinetic energy plus potential energy.

    When the spring has sprung back to its equilibrium position (it is not compressed or stretched), how much potential energy is there? How much kinetic energy? (You are given the mass and the velocity at this point.) Now you know the total energy.

    When the spring is at the maximum compression or extension and the object is NOT MOVING, how much kinetic energy does it have? How much potential energy does it have? (You can figure this out since you know the total energy, which is constant as long as there is no friction.) Now, knowing the amount of potential energy, and the amount of displacement (0), you could figure out the spring constant k (but you don't have to to solve this problem). What's important is to know how much energy the system has, and what kind.

    But on the second run, there is friction. You can figure out how much work is done by friction as the block slides back towards the equilibrium point, and then ...

    (I don't want to do the whole thing for you. Work on it.)
  8. May 5, 2004 #7
    sorry you're so confused...I'll already figure out the first one.The second is the one I'm having trouble with
  9. May 7, 2004 #8
    Do you have any numbers for the second problem?

    Here's a couple equations that might help your thinking

    alpha = a / R, (a is the tangental acceleration, not the radial acceleration)

    angular speed = alpha * time

    Think about the spring. The spring constant's always a biggy.

    U = 1/2*k*x^2 could come in very handy. Like gnome said, energy's handy.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: I really need help !- angular speed and speed at equilibrium position
  1. I really need help (Replies: 2)

  2. I really need help (Replies: 3)

  3. I really need help (Replies: 2)