An amusement park ride consists of a large vertical cylinder that spins about its axis fast enough that a person inside is stuck to the wall and does not slide down when the floor drops away. The acceleration of gravity is 9.8 m/s 2 . Given g = 9.8 m/s 2 , the coefficient Âµ = 0.52 of static friction between a person and the wall, and the radius of the cylinder R = 3.8 m. For simplicity, neglect the personâ€™s depth and assume he or she is just a physical point on the wall. The personâ€™s speed is v = 2Ď€R T 4 where T is the rotation period of the cylinder (the time to complete a full circle). Find the maximum rotation period T of the cylinder which would prevent a 50 kg person from falling down. Answer in units of s

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aristocles aristocles · Answer 1 · 2019-01-24T01:03:05+01:00

here since the speed of man while he is in contact with the wall will be given by

[tex]v = \frac{2\pi R}{T}[/tex]

here we know that

R = 3.8 m

Now when cylinder is rotating along with the man stick with the wall

So at this case normal force on the man will be the centripetal force on it

so here we can say

[tex]N = \frac{mv^2}{R}[/tex]

[tex]N = \frac{m(\frac{2\pi R}{T})^2}{R}[/tex]

[tex]N = \frac{4\pi^2mR}{T^2}[/tex]

now by the formula of friction force we know

[tex]F_f = \mu N[/tex]

now if man do not slip on the wall so friction force will balance the weight of the man

[tex]F_f = mg[/tex]

[tex]\mu\frac{4\pi^2mR}{T^2} = mg[/tex]

[tex]\mu \frac{4\pi^2 R}{T^2} = g[/tex]

[tex]T^2 = \frac{4\mu \pi^2 R}{g}[/tex]

[tex]T^2 = \frac{4(0.52)\pi^2 (3.8)}{9.8}[/tex]

[tex]T^2 = 7.96 s^2[/tex]

[tex]T = 2.82 s[/tex]

so time period is 2.82 s