Calculating Max Angular Velocity for Rotation and Force

[hastings]

Homework Statement

\omega=constant

R=0.2m

\theta=30°

\mu _s=0.5 between the iron cube and the surface

\omega_{max}=?

Homework Equations

Fx=Fsin(α) or cos(α)
Friction=µN

The Attempt at a Solution

in pink I drew the gravity force and the normal force.
Which other force acts on the body? the friction force is directed along the surface, but in which direction going towards the center of the cone or outward? Why?

 

[Mentz114]

I can't make out the setup from your diagram. Centripetal force is directed towards the center of rotation.

 

[hastings]

the exercise is a little difficult to explain in words, it would be much easier with a picture.
An iron cube is on the inside rough surface ( \mu_s=0.5) of a cone which is upside down. this cone rotates around its centre axis with a \omega=constant. between the axis and the surface there's an angle of \theta=30°. the cube is at a distance R=20cm from the centre axis. find the \omega_{max}=?, maximum angular velocity, so that the cube remains still on the cone's surface.

I drew 2 imaginary axis x and y with origin in the cube. I am sure there are at least 2 forces:gravity,pointing down, and normal pointing away form the surface. which is the direction of the friction force? is it toward the centre of the cone or outward? why?

 

[Mentz114]

OK, I think I understand the setup now. You need the sliding forces to cancel out. The downward sliding is due to gravity, and the counterforce is the component of the centripetal force that is parallel to the surface. Remember that the normal force will be made up of two components - gravity and centripetal force.

 

[hastings]

m (\omega^2R) = N+G where N and G stand for Normal force and Gravity force.

omega² ∙R is the centripetal acceleration.
How do I fit in the friction force?

Is it right? I'm pretty much confused so please give me clear hints.

 

[Mentz114]

Ignoring friction for the moment, do you agree that

downward sliding force = mgcos(theta)

upward sliding force = mrw^2sin(theta)

?

 

[hastings]

upward sliding force = mrw^2sin(theta)?

Why?
I agree with the 1st statement, but I'm still having doubts on the direction of the centrifugal force. Is it parallel to the ground (not the cone's surface) point out?
Could you just show it with a picture post? This will clear out all my doubts. Thank you.

 

[Mentz114]

There is a component of the centripetal force that is parallel to the surface.
I suggest you draw a diagram.

 

[hastings]

Could you just tell me if the axis x and y are right as I drew them?
Did I draw all the forces acting on the object correctly ?

 

[Mentz114]

Yes. That's what I drew. Now, I could be wrong, but in equilibrium those sliding forces are equal. Also friction is irrelevant because nothing moves.

If you equate the forces, mass cancels out as it should, and you can solve for omega. Mass cancels out because gravity and centripetal 'force' are actually accelerations not forces.

 

[hastings]

ok, thanks a lot for your help
!

 

[D H]

Yes. That's what I drew. Now, I could be wrong, but in equilibrium those sliding forces are equal. Also friction is irrelevant because nothing moves.

If it weren't for static friction, there would be no range of rotation rates over which the iron cube is stationary. There would instead be one equilibrium rate. If the cone rotates slow than this rate the cube falls down the cone. For a fixed rotation rate w, the w cross r term decreases as the cube falls down the cone and increases as the cube works its way up the cone. The equilibrium rate thus represents an unstable equilibrium.

When you account for static friction (nothing is moving, so this is a static friction problem), the unstable equilibrium point becomes a range of rates.