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Homework Help: Barrel of fun - Friction force

  1. Nov 8, 2013 #1
    1. The problem statement, all variables and given/known data

    A barrel of fun consists of a large vertical
    cylinder that spins about the vertical axis.
    When it spins fast enough, any person inside
    will be held up against the wall.



    If you double the angular speed (i.e., [itex]\omega _2=2\cdot \omega _1[/itex]) what is the new friction force [itex]f_2[/itex]?

    Answer choices: [itex]2\cdot F_1,1/3,1/4,1/2,3,1.5,1/3.5,1,2.5,4[/itex]

    2. Relevant equations

    [itex]F_f=\mu_k\cdot F_N[/itex]

    [itex]v=r\cdot \omega[/itex]


    3. The attempt at a solution

    The frictional force is defined as [itex]F_f=\mu_k\cdot F_N[/itex].

    The normal force of the object is the centripetal force that is pointing towards the center of the cylinder (perpendicular to surface).

    Thus frictional force is

    [itex]\begin{align*}F_f&=\mu_k\cdot (m\cdot a_c)\\&=\mu_k\cdot \left(m\cdot \dfrac{v^2}{r}\right)\\

    This means that [itex]\boxed{F_f\propto \omega^2}[/itex]

    so [itex]F_2=4\cdot F_1[/itex], but when I submitted it, it said that I was wrong.

    What am I doing wrong?
    Last edited: Nov 8, 2013
  2. jcsd
  3. Nov 8, 2013 #2


    User Avatar

    Staff: Mentor

    Hi vaio-911. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

    I don't think it is, actually. I think that is the maximum frictional force before relative motion can occur. Isn't that so?

    The actual frictional force to secure each person against gravity is their weight, and that doesn't change just because you spin them faster!

    Maybe this is a trick questionhttp://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon5.gif [Broken] Is your professor a barrel of fun? :smile:
    Last edited by a moderator: May 6, 2017
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