Centripetal force and friction

[sheepy]

Homework Statement

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 2m radius, what is the minimum speed of that passengers can have fr this to happen if the coefficient of friction between the passengers and the wall is 0.25.

Homework Equations

\mu= Friction/normal force
Centripetal force= (mv^2)/r

The Attempt at a Solution

I don't understand how I would go about the equation if I'm not given the mass of the passenger. Because I can't find the friction force through the normal force (because I don't have mass). Is there another way to approach it?
Thanks!

 

[Loopy]

Mass cancels out once you equate the two forces (frictional and gravitational).

 

[pooface]

centripetal acceleration = mu*g

you can equate that to V^2/R I think.

this is a new topic for me.

 

[Loopy]

No. That is incorrect. mu*g does not describe any force in this problem, while V^2/R is the definition of centripetal acceleration.

I already provided the answer to the original question.

 

[sheepy]

i still don't get what the equation should be.
at a certain point i thought it was
w=centripetal force
but where does the friction come into play?
or.. is friction = centripetal?

 

[salman213]

m(9.81) = (mv^2/2)(0.25)

?

like 8.86

 

[sheepy]

yeah that's what i thought? except why did the question mention friction if we didn't need it?

 

[salman213]

you did need it...

mu = Friction/normal force
Centripetal force= (mv^2)/r

gravitational force = frictional force therefore

mg =\mu*normal force

centripetal must be equal to normal therefore rearranging the equations u get

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

in which the masses cancel
g = \mu*(v^2)/r

root(9.81*2)/0.25= v

 

[sheepy]

sorry to be slow but can you explain why its weight is equal to the friction force, and why the centripetal must equal to its normal force?

 

[PhanthomJay]

sorry to be slow but can you explain why its weight is equal to the friction force,

to understand what is going on, it is essential to draw a free body diagram of the passenger and examine all the forces (contact forces and gravity forces) acting on the passenger, then apply Newton's laws. What are the forces acting in the vertical direction? What is the net force in the vertical direction ?(think Newton 1).

.and why the centripetal must equal to its normal force?

what are the forces acting in the horizontal (centripetal) direction? What is the net force in that direction? (think Newton 2).

 

[rl.bhat]

When a car takes sudden turn you are pushed in the outward direction due to centrifugal force. This force is perpendicular to the walls of the cylinder.Reaction to this force by the wall is the normal reaction. If the speed of the cylinder is not sufficient passengers slide downwards due to gravitational force. This produces the frictional force in the upward direction. To keep the passengers glued to the cylinder Fk = mu*mv^2/R = mg.

 

[Loopy]

Think of other models of centripetal acceleration that you have learned. If a ball on a string is swung, its centripetal acceleration is provided by the tension force towards the centre of the orbit. If a satellite is orbiting the earth, its centripetal acceleration is provided by gravitational pull to the earth. In this problem, the only force that can provide the needed centripetal acceleration is the cylinder's normal force on the passenger. Thus, the centripetal force equals the normal force.

 

[sheepy]

thanks guys, i get it :]

 

[rl.bhat]

Thus, the centripetal acceleration equals the normal force.
Forces always exist in pairs. Here the normal reaction experianced by the passenger is due to the centifugal force acting on the passenger in side a rotating cylinder.

 

[salman213]

isnt centrifugal force different then centripetal

 

[rl.bhat]

Yes.It is different. It exists only in the accelerated frame of reference, like rotating frame of reference. There is no physical origin for this force. That is why it is called fictitious force.