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Spinning around a cylinder with a person inside

  1. Apr 14, 2005 #1
    Hi I have physics problem which I'm stuck with.

    It goes like this.

    An amusement park ride consists of a large vertical cylinder that spins around its axis fast enough for any person inside is held up against the wall then the floor drops away. The coefficient of static friction is [itex]\mu_{s}[/itex] and the radius of the cylinder is R.

    I'm suppose to show the following: The maximum period of revolution necssary to keep the person from falling is [itex]T = \frac{(4 \pi^{2} R \mu_{s})}{g}^{1/2} [/itex]

    I know that in order to understand the situation I first need to draw a force-diagram displaying the forces acting on both the cylinder and the person inside.

    Do I add these forces together then?

    Any help/hint(s) will be appreciated :-)


  2. jcsd
  3. Apr 14, 2005 #2
    No, you don't need to add all forces together. Just break into vertical and horizontal (radial) components. Identify the three forces that act on the person and solve for the radial force, then for the angular velocity, and hence for period.
  4. Apr 14, 2005 #3
    Hi and thanks for Your answer.

    Then the forces acting on the person and cylinder must be [itex]f_{s} = \mu_{s} \cdot n[/itex] , [itex]F_{g} = m \cdot g[/itex] , [itex]F_{z} = m \cdot a[/itex]

    Since it turns around the Z-axis I guess [itex]F_{x} = 0[/itex] [itex]F_{y} = 0[/itex].

    I know both the cylinder and person are affect by acceleration too [itex]a_{c} = m \frac{v^{2}}{r}[/itex]

    The component forces are then:

    [itex]F_{\textrm{person} }} = m \cdot g + \mu_{s} \cdot n[/itex]??

    [itex]F_{\textrm{cylinder}} = m \cdot g + m \cdot \frac{v^2}{r}[/itex]??



  5. Apr 14, 2005 #4
    You didn't observe that the normal force the wall applies on the person is the only radial force, so n = ma, and thus

    [tex]f_s = \mu_{s}ma = \mu_{s}m\omega ^2r[/tex]

    On the other hand, f_s must balance the person's weight G.


    [tex]f_s = mg[/tex]

    [tex]\mu_{s}m(\frac{2\pi}{T})^2r = mg[/tex]


    [tex]\mu_{s}\frac{4\pi ^2}{T^2}r = g[/tex]


    [tex]T = \frac{\sqrt{4\pi ^2\mu_{s}r}}{g}[/tex]

    That is not the case. The centre of mass of the cylinder is not moving in the x or y direction, but the person is accelerating in the horizontal (radial) direction, so that F_x is not zero. F_y, yes is zero.

  6. Apr 14, 2005 #5


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    Just a minor comment: the g should go under the root sign:

    [tex]T=2\pi \sqrt{\frac{\mu_s R}{g}}[/tex]
  7. Apr 14, 2005 #6
    That looks much better, but I tried to reach the expected answer. :smile:

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