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!

Springy Problem

  1. Oct 2, 2004 #1
    Hi fellow physicists, I have some homework problems i'm trying to work through and i need to know if i'm on the right track.

    This is the problem:

    A Spring is upright (vertical) and has a constant k and a length Lo, a tray of M mass is attached to the spring and on the tray is placed a particle of m mass, suppose you initially compress the spring a distance d from the equilibrium point of the spring-tray-particle system. Calculate the maximum height above the ground the particle will reach.

    First i calculated the equilibrium point of the spring-tray-particle system:

    (Lo-L1)k = (M+m)g => L1 = Lo - (M+m)g/k

    Then I used the conservation of energy, Potencial energy when the spring is compressed = to kinetic energy at equilibrium point.

    k(d^2)/2=(M+m)(v^2)/2 => v=sqrt(k/(M+m))d

    Then i used the velocity to calculate the height above L1

    h= k(d^2)/2g(M+m)

    So the height above the ground is

    L1 + h = Lo + (M+m)g/k +k(d^2)/2g(M+m)

    Is this right? i'd appreciate it if some one could double check it for me.

    Thanks in advance. By the way how do you use the special font for numbers and mathematical symbols?
     
  2. jcsd
  3. Oct 2, 2004 #2

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Good try! (and great that you provided your calculations)
    However, it is incorrect to assume that the particle will leave the tray when the system is at the equilibrium point of the system!

    I'll break this into a few steps:
    1. Condition of leaving plate:
    Normal force N acting on particle is zero (agreed?)
    In particular, regarding Newton's second law for the particle, what does this mean that the acceleration of the particle must be then?
    2. System of tray+particle
    At the point when the particle is about to leave the tray, this system has the acceleration from 1.); what does this imply about the length of the spring?
    3. Proceed to find the initial velocity as you've done; your approach here looks fine.
     
  4. Oct 2, 2004 #3
    Lets see.. N=0 when a =0 is that right? does that mean the particle will come off the tray at Lo?
     
  5. Oct 2, 2004 #4
    Okay another problem...

    A ring (mass = M) is suspended by an ideal cord from the ceiling, two equal masses (mass = m) are released from the top of the ring and slide down, one on either side of the ring, without friction. The question is:
    What condition must the masses meet in order for the ring to move up?

    I'm having trouble imagining it, I cant see which force could make the ring go up, possibly the normal of the sliding masses but i dont know how to balance the ecuations.

    I hope it doesn't look like im trying to get someone else to do my homework, i just need some help getting started. Thanks in advance for any help.
     
  6. Oct 2, 2004 #5

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Newton's 2.law for the particle reads:
    [tex]N-mg=ma[/tex]
    That is, N=0 implies a=-g
     
  7. Oct 2, 2004 #6
    Okay i hope i've got this right, does that mean that the particle will come off the tray at Lo + (M+m)g/k i.e when the force applied by the spring is equal to -(M+m)g?
     
  8. Oct 2, 2004 #7

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Mm..no!
    I didn't say your suggestion of L0 was wrong (it is in fact right!)
    Reason:
    Let's look at the system tray+particle.
    Up to the point of separation, they move and accelerate as a single object.
    Newton's 2.law reads:
    [tex]-k(L(t)-L_{0})-(M+m)g=(M+m)a[/tex]
    Hence, when a=-g, [tex]L(t)=L_{0}[/tex]
     
  9. Oct 2, 2004 #8
    Right i forgot to put -(M+m)g in the force equation, thanks for the help.
    I wonder if you could help me with the second problem i posted i cant't really get my head around it. P.S. How do you use the special font for numbers and mathematical symbols?
     
  10. Oct 2, 2004 #9

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Just click on the line to see how the LATEX code is generated.
    (Was that the second problem?)
     
  11. Oct 2, 2004 #10
    The second problem is a bit futher up in this thread.
     
  12. Oct 2, 2004 #11

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    All right, I'll have a look at it..
     
  13. Oct 2, 2004 #12

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    OK, now I've thought about it:
    1. You're right about the forces which may push the circle upwards.
    If the normal force acting upon a particle from the circle provides a part of the necessary centripetal acceleration of the particle (that is, the particle's gravity is not enough), then the particles will exert an upwards directed force on the circle if they are on the upper half of the circle.

    2. If the circle is to be lifted, a necessary condition is that at some time, the tension in the cord becomes 0.
    Prior to this moment, the circle will remain at rest.
    This will give us one equation.
    3. The two particles' path are, of course, symmetric about the vertical axis.
    4. Up to the point where the circle might start to move upwards, a given particle's mechanical energy must remain constant.
    This provides us with a relation between the angle displacement of the particle (measured with respect to the vertical) and its (angular) velocity about the circle.
    5. Now, consider the radial component of a particle's Newton's 2.law:
    Clearly, we must require in order for a raising of the circle, that the expressions gained from the equations found in 2. and 4., is consistent with that Newton's 2.law for that particle must have a solution!
    This will give you a desired relation between M and m.
    I'd like you to try this out by yourself first; I got [tex]M\leq\frac{2}{3}m[/tex]
    If you found any of this unclear; just post a reply, I'll get back with the maths later on.
     
  14. Oct 2, 2004 #13
    That's the right answer, i'll try my best to work it out tonight.
    I'd appretiate you posting your workings though!!
     
  15. Oct 2, 2004 #14
    Let's see.. the centripedal acceleration has to be v^2/r of which cos(alpha)g is given by gravity, so N =m(v^2/r -cos(alpha)g) perpendicular to the circle and cos(alpha)N is the force that could lift the circle. From conservation of energy for the particles i get v^2= 2rg(1-cos(alpha)). N=m(2g-3gcos(alpha)). Is that right so far...
     
  16. Oct 3, 2004 #15

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    What you've written seems almost right, I'll post my own:
    1. Normal forces acting on the circle
    Clearly, we may write the normal forces from the particles as:
    [tex]N(\pm\sin\theta\vec{i}+\cos\theta\vec{j})[/tex],
    where N is the magnitude, and the sign in the horizontal component signifies which particle produce the force.
    Hence, net normal force on the circle is:
    [tex]\vec{N}_{net}=2N\cos\theta\vec{j}[/tex]
    2. The cord tension must be zero:
    For the circle at rest, the vertical component of Newton's 2.law reads:
    [tex]T+2N\cos\theta-Mg=0[/tex]
    or, when T=0,
    [tex]N=\frac{Mg}{2\cos\theta}[/tex]
    3. Conservation of a particle's mechanical energy:
    This is simply, while the circle is at rest:
    [tex]\frac{1}{2}m\vec{v}^{2}+mgR\cos\theta=mgR[/tex]
    where the reference level for potential energy has been chosen at [tex]\theta=\frac{\pi}{2}[/tex]
    Or:
    [tex]\frac{m\vec{v}^{2}}{R}=2mg(1-\cos\theta)[/tex]

    4.Radial component of Newton's 2.law for the particle:
    This is, in general:
    [tex]-N-mg\cos\theta=-m\frac{\vec{v}^{2}}{R}[/tex]
    Inserting relations from 2+3, and eliminating "g":
    [tex]\frac{M}{2\cos\theta}=2m-3m\cos\theta[/tex]

    Rearranging, we have:
    [tex]3m\cos^{2}\theta-2m\cos\theta+\frac{M}{2}=0[/tex]
    Or:
    [tex]\cos\theta=\frac{1}{3}(1\pm\sqrt{1-\frac{3M}{2m}})[/tex]
    Clearly, we must require real solutions, so:
    [tex]M\leq\frac{2}{3}m[/tex]
     
  17. Oct 3, 2004 #16
    I actually got the result on my own in the end which is very satisfying. Thanks for taking so much time to help me out.
     
  18. Oct 3, 2004 #17

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Great that you managed it on your own!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Springy Problem
  1. Springy question (Replies: 2)

  2. Problems with problems (Replies: 1)

  3. Springy stuff (Replies: 3)

  4. A problem (Replies: 2)

Loading...