In a carnival ride, passengers stand with their backs against the wall of a cylinder. The cylinder is set into rotation and the floor is lowered away from the passengers, but they remain stuck against the wall of the cylinder. For a cylinder with a 2.0-m radius, what is the minimum speed that the passengers can have so they do not fall if the coefficient of static friction between the passengers and the wall is 0.25?

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aristeus aristeus · Answer 1 · 2019-12-24T07:01:05+01:00

Answer:

Minimum speed will be equal to 2.213 m/sec

Explanation:

We have given radius of the r = 2 m

Coefficient of friction [tex]\mu =0.25[/tex]

At minimum speed frictional force will be equal to centripetal force

So [tex]\mu mg=\frac{mv^2}{r}[/tex]

[tex]\frac{v^2}{r}=\mu g[/tex]

[tex]v=\sqrt{\mu rg}=\sqrt{0.25\times 2\times 9.8}=2.213m/sec[/tex]

So the minimum speed will be equal to 2.213 m/sec

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jessrsmith48 jessrsmith48 · Answer 2 · 2021-04-06T21:54:09+02:00

Answer:

v = 8.9 m/s

Explanation:

1. f = mg

2. f=цn

3. mg=цn=цmv²/r

v=√(gr/ц)

v=√[(9.8 x 2) ÷ 0.25]

v=8.9 m/s

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