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Spinning around a cylinder with a person inside 
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#1
Apr1405, 08:56 AM

P: 255

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 forcediagram 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 :) Sincerley Fred 


#2
Apr1405, 09:24 AM

P: 453

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.



#3
Apr1405, 09:48 AM

P: 255

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 Zaxis 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]?? Sincerely Fred 


#4
Apr1405, 10:05 AM

P: 453

Spinning around a cylinder with a person inside
[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. Therefore, [tex]f_s = mg[/tex] [tex]\mu_{s}m(\frac{2\pi}{T})^2r = mg[/tex] or, [tex]\mu_{s}\frac{4\pi ^2}{T^2}r = g[/tex] or, [tex]T = \frac{\sqrt{4\pi ^2\mu_{s}r}}{g}[/tex] Regards, Ervin 


#5
Apr1405, 10:24 AM

Sci Advisor
HW Helper
P: 2,002

Just a minor comment: the g should go under the root sign:
[tex]T=2\pi \sqrt{\frac{\mu_s R}{g}}[/tex] 


#6
Apr1405, 10:32 AM

P: 453




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