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Cal II question with physics involved.

  1. Dec 14, 2009 #1
    hey. I got these three math questions that I can't figure out. These are for a Calculus II class. it's the "word problem" & physics part of it that confuses me, because if i was given an equation to figure out i'm certain i could do it. but i'll follow your guidelines and give it a shot.

    Question A)
    1. The problem statement, all variables and given/known data
    Find the total mass of a metal rod with length "l" if the density at every point is proportional to the cube of the distance from the left end of the rod with coefficient of proportionality "k".


    2. Relevant equations
    since its' a rod, we'll use disc method to find the total volume
    & p=m/v


    3. The attempt at a solution
    pi * integral from one to infinity of (k/(x^3))^2

    pi*k*(1/-5x^5) ] from one to infinity
    =pi*k/-5



    Question B)
    1. The problem statement, all variables and given/known data
    b) a chain 50 ft long and weighing 1 lb/ft is hanging vertically. How much work is done in raising the chain to a horizontal position at the level of the top of the chain?

    2. Relevant equations
    W=F*d

    3. The attempt at a solution
    i'll try a Sum attempt by breaking it into 25 parts of 1 feet ; where i'm raising the bottom of the chain by 2 foot.
    (1 foot of chain*2 feet * 1lb/foot)+(2 feet of chain*2 feet * 1lb/foot)+(3 foot of chain*2 feet * 1lb/foot)...+(25 foot of chain*2 feet * 1lb/foot)
    =25! feet of chain*2 feet * 1lb/foot

    work done = 25!*2

    I believe this problem might need me to use hyperbolic functions, which were not covered..


    Question C)
    1. The problem statement, all variables and given/known data
    c) Two particles repel each other with a force inversely proportional to the square of the distance between them. If one particle remains fixed at a point on the x-axis 2 units to the right of the origin, find the work done in moving the second particle along the x-axis from a point 3 units to the left of the origin to the origin if the coefficient of proportionality is k.

    2. Relevant equations
    ????

    3. The attempt at a solution
    integral from -5 to -2 of (1/k)*d
    =(25/2k)-(4/2k)
    =21/2k




    any help on any of these three problems will be greatly appreciated!
     
  2. jcsd
  3. Dec 14, 2009 #2

    Mark44

    Staff: Mentor

    Why do you have pi in the expressions above? And why are you using disks? What you need to do is to find an expression that gives the mass of a typical mass element, [itex]\Delta m[/itex]. You are given that the density (which I think you can assume is in units of mass/length rather than mass/ volume) is proportional to the cube of the distance from the left end.

    Let x = the distance your typical mass element is from the left end of the rod.
    So d = kx^3.
    Now, the mass of the typical mass element will be its density times its length. Your total mass will be the integral over the length of the rod of all of the typical mass elements.
     
  4. Dec 14, 2009 #3
    ok.

    let me know if this is correct:
    Question A)
    M=integral(pdV). p=kr^3 and dV=drda with da=cross section of rod. limits of integration are 0 and L.

    M=(k*l^4)/4 ??



    Question B)
    w=fd or PE=mgh. the center of mass is at the mid point of the chain, so h or d is 25 feet. m is the mass of the chain, and g is the gravitational constant, 32 ft/s^2

    25 feet * 50lb * 32 ft/s^2

    =40000lb/s^2

    should'nt there be calculus involved in this?


    Question c)
    w=fd. f=k/r^2. d=dr, integrating from infinite radius to r=5.

    =k/5
     
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