Physics, 27.07.2020 01:01 9Time

An amusement park ride has a vertical cylinder with an inner radius of 3.4 m, which rotates about its vertical axis. Riders stand inside against the carpeted surface and rotate with the cylinder while it accelerates to its full angular velocity. At that point the floor drops away and friction between the riders and the cylinder prevents them from sliding downward. The coefficient of static friction between the riders and the cylinder is 0.87. What minimum angular velocity in radians/second is necessary to assure that the riders will not slide down the wall?

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Answer from: zhellyyyyy

The minimum angular velocity necessary so that the riders will not slide down the wall is 1.58 rad/s

The riders will experience a centripetal force from the cylinder

F = mrw^2

where m is the mass of the rider

r is the inner radius of the cylinder = 3.4 m

ω is the angular speed of the rider

For the riders not to slide downwards, this centripetal force must be balanced by friction. The frictional force is given as

f = μN

where

μ = coefficient of friction = 0.87

N is the normal force = mg

f = μmg

Equating centripetal force of and frictional force of we'll get

mrw^2 = umg

rw^2 = ug

w^2 = ug/r

$w= \sqrt{ug/r}$

$w= \sqrt{0.87*9.8/3.4}$

ω = 1.58 rad/s is the minimum angular velocity needed to prevent the rider from sliding.

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Answer from: hijrjdkfjd

The minimum angular velocity necessary to assure that the riders will not slide down the wall is 1.58 rad/second.

Explanation:

The riders will experience a centripetal force from the cylinder

$F_{C}$ = mrω^2 .... equ 1

where

m is the mass of the rider

r is the inner radius of the cylinder = 3.4 m

ω is the angular speed of of the rider

For the riders not to slide downwards, this centripetal force is balanced by the friction between the riders and the cylinder. The frictional force is given as

$F_{f}$ = μR ....equ 2

where

μ = coefficient of friction = 0.87

R is the normal force from the rider = mg

where

m is the rider's mass

g is the acceleration due to gravity = 9.81 m/s

substitute mg for R in equ 2, we'll have

$F_{f}$ = μmg ....equ 3

Equating centripetal force of equ 1 and frictional force of equ 3, we'll get

mrω^2 = μmg

the mass of the rider cancels out, and we are left with

rω^2 = μg

ω^2 = μg/r

ω = $\sqrt{\frac{ug}{r} }$

ω = $\sqrt{\frac{0.87*9.81}{3.4} }$

ω = 1.58 rad/second

Answer from: Quest

Ithink the answer is c but then again i'm probably wrong

Answer from: Quest

D. a bar magnet with its south pole on the left

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An amusement park ride has a vertical cylinder with an inner radius of 3.4 m, which rotates about it...

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