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In an amusement park ride, a small child stands against the wall of a cylindrical room that is then made to rotate. The floor drops down and the child remains pinned against the wall. If the radius of the device is 2.15 m and the relevant coefficient of friction between the child and the wall is 0.400, with what minimum speed is the child moving if he is to remain pinned against the wall?

User Erico
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1 Answer

4 votes

Answer:7.25 m/s

Step-by-step explanation:

Given

radius of device
r=2.15 m

coefficient of Friction
\mu =0.4

here centripetal Force will be equal to Normal reaction


F_C=N=(mv^2)/(r)

and frictional Force
f_r=\mu N

Friction force will balance the weight of child


\mu (mv^2)/(r)=mg


v=\sqrt{(rg)/(\mu )}


v=\sqrt{(2.15* 9.8)/(0.4)}


v=7.25 m/s

User Neysha
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