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Math Help - Work/energy

  1. #1
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    Work/energy

    heres the problem

    http://i14.photobucket.com/albums/a3...ristx/adsf.jpg

    heres my work

    http://i14.photobucket.com/albums/a3...istx/faaaa.jpg


    The speed is supposed to be 5.18 but im not getting that...

    any ideas?
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  2. #2
    Member Danshader's Avatar
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    ok first i am going to convert all the units given in tat question to SI units (i was taught in SI units)

    mass of the collar, m = 0.17 kg (1 oz = 0.0283495231 kg)
    spring constant, k =875.83 N/m (1 lb/in = 175.166 N/m)
    distance spring compressed, x = 0.0381 m (1 inch = 0.0254 m)

    Now total elastic potential energy,EPE stored in the spring before release is given by the formula,
    <br />
EPE = \frac{1}{2}kx^2<br />
    <br />
EPE = \frac{1}{2}(875.83)(0.0381)^2<br />
    <br />
 EPE = 0.635681793J<br />

    The collar is then released. At point B, the collar has already traveled 10 inches (0.254 m) vertically upwards.

    Hence gravitational potential energy,GPE gained by the collar is:
    <br />
GPE = mgh<br />
    <br />
GPE = (0.17)(9.81)(0.254)<br />
    <br />
 GPE = 0.4235958J<br />

    using principle of conservation of energy the total energy of the system is conserved:

    Initial energy = final energy
    EPE = Kinetic energy + GPE
    0.635681793J = KE + 0.4235958J
    KE = 0.21085993 J

    from the formula of kinetic energy, KE:
    <br />
\frac{1}{2}mv^2 = 0.21085993<br />
    <br />
v^2 = \frac{2(0.21085993)}{0.17} <br />
    <br />
v = 1.579597838 ms^{-1} = 5.1824 fts^{-1} \longrightarrow (1 ms^{-1} = 3.2808399fts^{-1})<br />



    well as for the force part take note that:

    elastic potential energy, EPE = \frac{1}{2}kx^2 = \frac{1}{2}Fx
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  3. #3
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    Thanks!....that really helped.....i realized one of my big problems was converting the spring constant which is what I forgot......I like working in metric better as well.....U.S. Customary is a pain.....


    What is the strategy for finding the force the collar is putting on the rod?
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  4. #4
    Member Danshader's Avatar
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    the motion of the collar now is in a circle, hence we can consider centrifugal and centripetal force acting on the collar.
    <br />
F = ma<br />
    <br />
F = mr\omega ^2<br />
    <br />
F = mr{(\frac{v}{r})}^2 \longrightarrow v = r\omega <br />
    <br />
F = m(\frac{v^2}{r})\longrightarrow (1)<br />

    well you have already calculated the linear speed,v (do not confuse with angular speed \omega) of the collar previously also r and m are given in the question. Hence just substitute those values into equation (1) and solve for F remember to state the direction of the force on the collar as well which in this case at point B is inwards (towards the left/ towarsd the center of the circle)
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  5. #5
    Forum Admin topsquark's Avatar
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    Two comments:
    Quote Originally Posted by Danshader View Post
    elastic potential energy, EPE = \frac{1}{2}kx^2 = \frac{1}{2}Fx
    Be VERY careful here. \vec{F} = -k \vec{x}. If you aren't careful your spring potential energy will be the negative of what you are looking for!

    Quote Originally Posted by Danshader View Post
    the motion of the collar now is in a circle, hence we can consider centrifugal and centripetal force acting on the collar.
    Never consider the centrifugal force unless you are working with a relative motion problem.

    -Dan
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