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A few challenging physics problems - .

  1. Sep 22, 2003 #1
    Plz help - # 1 and #2 are due tomorrow

    Im in an engineering physics honors class and I am having trouble figuring out how to go about solving a few of the problems that I have encountered.

    1) A system has 2 blocks m1 and m2 ( on a flat frictionless plane), connected by a massless spring, with spring constant k. m1 is against a wall. If at t0, m2 is pushed towards m1, compressing the spring from its original length of L to L/2, find the motion of the spring in relation to time.

    I am not at all sure how to go about this. My book converts the motion into simple harmonic motion involving coswt and such for a similar problem, but I'm not sure how to use this as I never learned in high school.

    2) A fisherman is sitting at the stern of a boat, and both the boat and the man are at rest. The fishermen than moves towards the bow, and eventually, the boat and the man are at rest again. Find the displacement of the boat assuming that the motion of the boat in the water is turbulent, i.e., it is characterized by a friction force that is proportional to -v2, where v is the velocity of the boat with respect to water. You may assume some simple model for the process if necessary (for example, the man jumps, and after a bit of a ballistic flight, lands on the bow).

    Not sure how the friction affects the situation.

    3) A rope of Mass M and length L lies on a frictionless table. A small portion of the rope, L0 is hanging through a hole in the table. Initially the rope is at rest.

    a) find a general equation for x(t) the length of the rope through the hole.

    The answer they havee is crazy : Ae^($t) + Be^($t) = x, also
    $^2 = g/l where $ is a symbol standing for gamma.
    Im not sure where they pulled this from

    b) evaluate the constants A and B so initial conditions are satisfied

    No clue

    plz help me get this - this class is making me hate physics
    Last edited by a moderator: Sep 23, 2003
  2. jcsd
  3. Sep 23, 2003 #2

    The differential equation of movement :
    Ma=[rho]xG- [mu](L-x)[rho]G
    Ma-x[rho]G(1+[mu])= [mu][rho]LG
    a-acselleration of movement
    G-gravitation constant


    s=[squ]([rho]G(1+[mu])/M) ;
    Last edited by a moderator: Sep 23, 2003
  4. Sep 23, 2003 #3


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    Science Advisor

    Let me see if I have the picture: you're taking an HONORS engineering physics class, you "never learned in highschool", and you have no clue whatsoever about these problems? It is possible you are in over your head.

    1) The force acting on the second mass is -k(x2- x1-L) where x1 is the x coordinate of the first mass and x2 is the x coordinate of the second so you have m d^2x2/dt^2= -k(x2- x1). The force on the first mass is more complicated. Since it is against the wall, it can't move to the "left"- the wall will cancel any force that way. But the wall doesn't "pull" so m d^2x1/dt^2= -k(x1- x2-L) as long as the right hand side is positive but 0 if it is negative.

    2) The friction affects the motion by opposing the force. If there were no friction, you could answer this by using "conservation of momentum" but friction means this is not a closed system.

    3) The book gives an answer involving "gamma" but doesn't say what gamma is? Are you sure you read it correctly?

    The point of theproblem is that the force on the rope is the weight of the part hanging down. Since the rope has mass M and length L, it "linear density" is M/L. If "x" is the length of the section hanging down, then its mass is (M/L)x and its weight is (M/L)xg.

    The equation governing the motion (force= mass*acceleration) is
    M d^2x/dt^2= (M/L)xg or d^2x/dt^2= gx/L.
    (Notice that this is not negative. As the rope slides DOWN, the length x increases.)

    That's a "linear homogeneous differential equation with constant coefficients" and, as you should already know if you are taking a course like this, the general solution is
    x(t)= C1 exp([sqrt](g/L)t)+ C2exp(-[sqrt](g/L)t)

    (I'll bet that you book does, somewhere mention that "gamma" is [sqrt](g/L).)

    Now, I, personally, don't consider that "crazy".

    Initially, the length x(t)= L0 and its speed, dx/dt= 0.

    Solve the equations
    C1 exp([sqrt](g/L)t)+ C2exp(-[sqrt](g/L)t)= L0 and
    C1 [sqrt](g/L)exp([sqrt](g/L)t)-[sqrt](g/L)C2exp(-[sqrt](g/L)t)= 0.
    for C1 and C2.

    Since this is an engineering course, it is possible that, instead of expecting you to solve the differential equation, they are just giving you the formulas. Not the kind of course I would want for an "honors" course, but then, I, personally, never wanted to be an engineer!
    You talk about never having learned (what?) in highschool. What math courses are you taking in college? Surely you wouldn't be taking an Engineering Physics class (honors or not) without having taken calculus?
  5. Sep 23, 2003 #4
    alright i got the first one and the 3rd one. Apparently the integration for the 3rd one is something u have to just see as our TA explained today.

    Are you saying that with the given information, problem 2 cant be solved, as our TA showed that v approaches infinity as time does. Do you agree with this?

    As for my physics career - I took physics 1 during a summer course. We didnt cover much and the only thing I really learned was how to calculate trajectory of a projectile.

    Then I went to AP physics, where my teacher was completely incompetent, and didnt know any of the material, explaining all of it wrong. I got a 5 on the AP test, but not because I understood the material - only because I memorized the formulas and got lucky on the free response. I thought that taking physics 7a (intro to mechanics) would be too easy in college, so I took H7a, thinking itd be the same as 7a, just with more depth and analyzing more complex examples. As it turns out, I was wrong - the material is approached from a completely different angle ignoring the basics which I never fully understood. We use the book by Kleppner and Kolnokov, if you know what that is. They're both MIT professors. In summary, this class will probably destroy my gpa and I will gain nothing out of it, but a hatred of physics and therefore engineering.
  6. Sep 23, 2003 #5
    Maybe you should request a transfer to 7a before you waste the semester.
  7. Sep 24, 2003 #6
    i think its too late for that

    this is week 6 of the semester - midterm is in 2 weeks

    wait, also for problem 1, is that the motion of the center of mass?
    Last edited by a moderator: Sep 24, 2003
  8. Sep 24, 2003 #7
    Re #1:
    I don't think the center of mass is going anywhere -- doesn't conservation of momentum preclude that?

    It's really not clear to me exactly what they are asking you for. I would try to answer it in terms of an expression (two expressions, really) describing the length of the spring as a function of time.

    I haven't seen a problem like this before, with two masses at opposite ends of a spring, & I can't find anything similar on the web. But just thinking about it, it seems that you have to divide this into two problems. For the first half-period, m1 isn't going anywhere, so the motion of m2 can be represented the same as any simple harmonic motion problem with just one mass attached to a compressed spring, released with 0 initial velocity. So you can use x = A*cos(ωt) where A is the amplitude and ω is sqrt(k/m) to describe the position of m2 relative to its release point, and therefore, you can fiddle around with this a little to get an expression for the length of the spring, right?. I don't want to give the whole thing away. :)

    After the first half-period, when the spring starts contracting, it's pulling simultaneously on both masses, and both masses start oscillating, but I guess they would accelerate at different rates depending on their relative masses (unless they are equal). Even then, I don't think the center of mass of the system can move. Need to think some more about how to express the motion of each mass individually. But I think the LENGTH of the spring would vary the same as it would if one end were fixed to the wall, and both masses were attached together to the other end. If so, the same expression you got for the first half-period, modified for the new mass, should work.

    Anybody else have any opinions about that?
  9. Sep 24, 2003 #8
    ok, i tried that solution, but im having trouble fitting the bottle in my mouth. Can you explain a bit more on how I should swallow them
  10. Sep 24, 2003 #9
    Uh, where are you having trouble with that?

    If the length at rest is L and the initial displacement is 1/2 L and displacement as a function of time is x = Acos(ωt), then what's an expression for length as a function of time? Pick another letter, or lowercase l, for the variable length, since your L is a constant.

    That gets you through the first half-period. I assume you can figure out the length of that half-period, given ω. If not, you'd better review harmonic motion in your text to find the formula.

    Then follow the same reasoning for the motion after the first half-period, except ω is different because the moving mass is different.
  11. Sep 25, 2003 #10
    lol sorry some guy posted on how i should go kill myself earlier
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