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Circular Motion Problem

  1. Apr 15, 2007 #1
    Problem: Spiderman (mass = 90kg) is swinging from his web. His center of mass is 10.0m off of the ground and his web has a length of 8.0m. It is connected to a pivot point that is the same height as his center of mass (so the web is held horizontal to the ground as he begins his swing). While holding the web during his swing, his center of mass is an additional meter from the end of the web.
    a) What is the force on the web at the moment he reaches the bottom of his swing?
    b) The web has a breaking limit of 3300N. Will the web snap and if so, how can he change his starting height to prevent this from happening?

    Attempt:
    a) m=90kg, r=8.0m + 1.0m = 9.0m

    i.
    Eg=mgh
    =(90 kg)(9.8 m/s^2)(9.0m)
    =7938 J
    ii.
    Ek=0.5mv^2
    7938 J = 0.5(90 kg)v^2
    v=13.2816 m/s
    iii.
    Fac = mv^2/r
    = (90 kg)(13.2816 m/s)^2/(9.0 m)
    =1764 N
    iv.
    Fg = mg
    =(90 kg)(9.8 m/s^2)
    =882 N
    v.
    Ft = Fg + Fac
    =1764 N + 882 N
    =2646 N
    =2.6e3 N

    Therefore the force on the web at the moment he reaches the bottom of his swing is 2646 N.

    b) (Not really sure about my use of distance in this one; I used the radii of the circle and of the rope to find the force acting on the rope.)
    i.
    W=Fd
    W=(2646 N)(9.0m)
    W=23814 J
    ii.
    W=Fd
    23814 J = F(8.0m)
    F=2976.75 N
    =3.0e3 N

    Therefore the web will not break.


    I'm not really confident that any of this is correct, so please provide insight if you can. Thanks!!
     
  2. jcsd
  3. Apr 15, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Your answer to part a looks good to me, though you could have saved a bit of calculator arithmetic by solving it algebraically until the last step.

    I don't know what you are trying to do in part b, though. (Looks like you are trying to apply the concept of work, but it's not working! For one thing, that force is the tension in the web at the bottom of the swing; it's not constant.) Hint: You already calculated the maximum tension in the web in part a--use it to answer part b.
     
  4. Apr 15, 2007 #3
    Thank you for your help!

    I have one more question:

    A local circus act features a motorcycle rider inside a hollow sphere. He accelerates his bike to a certain velocity in order to allow himself to be able to drive along the equator of the sphere, parallel to the ground inside the sphere. The sphere has an inner diameter of 7.00m, the motorcycle and rider have a mass of 300kg and his tires have a [tex]\mu_s[/tex] value of 0.900 with the inside surface of the sphere. What is the minimum speed he must attain in order to keep from falling inside the sphere?

    Answer:
    i.
    [tex] \Sigma F_v = 0 [/tex]
    [tex]\Sigma F_v = F_s - F_g[/tex]
    [tex]0 = F_s - F_g[/tex]
    [tex]F_s = F_g [/tex]

    ii.
    [tex]F_a_c = F_N[/tex]
    [tex]F_s = F_N \mu_s [/tex]
    [tex]F_s = F_a_c \mu_s [/tex]

    iii.
    [tex] F_s = F_g [/tex]
    [tex] F_a_c \mu_s = F_g [/tex]
    [tex] F_a_c = \frac{F_g}{\mu_s} [/tex]

    iv.
    [tex]F_a_c = \frac{F_g}{\mu_s} [/tex]
    [tex]\frac{F_g}{\mu_s} = \frac{mv^2}{r} [/tex]
    [tex]\frac{mg}{\mu_s} = \frac{mv^2}{r} [/tex]
    [tex]v^2 = \frac{rg}{\mu_s}[/tex]
    [tex]v = \sqrt{\frac{rg}{\mu_s}} [/tex]

    v.
    [tex]v = \sqrt{\frac{rg}{\mu_s}}[/tex]
    [tex]v = \sqrt{\frac{3.50m * 9.8 m/s^2}{0.900}}[/tex]
    [tex]v = 6.173 m/s[/tex]
    [tex]v = 6.2 m/s [/tex]

    Therefore the minimum speed is 6.2 m/s.
     
  5. Apr 15, 2007 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Looks good to me! (Realize that you are modeling the motorcycle as a particle, which is OK.)
     
  6. Apr 15, 2007 #5
    Thanks again for your help!

    I assumed that the cyclist was a point, though I guess in reality it would be something more of a problem involving the man on the motorcycle with his density/mass as a function of height and involving rigidity and centers of mass, etc. That's a lot more complex than a point mass, though :|.
     
  7. Apr 15, 2007 #6
    I dare say. Even if modeled more simply than you suggest, the overall center of mass that gravity acts upon is is some distance from the sphere. In other words there is significant torque wanting to make this problems sol'n a bloody mess. In other words, imagine adding a stick figure sticking out from your point mass. Not asking for your thoughts, just adding some food for thought as was done for me a bit back on a similar problem.

    Good post.
     
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