# Circular Motion / Newton's Laws

1. Oct 1, 2007

### Destrio

A very small cube of mass m is places on the inside of a funnel rotating about a vertical axis at a constant rate of W revolutions per second. The wall of the funnel makes an angle A with the horizontal. The coefficient of static friction between the cube and funnel is us and the center of the cube is at a distance r from the axis of rotation. Find the largest and smallest values of W for which the cube will not move with respect to the funnel.

I'm not really sure how to even begin this one
The Fnet in the y direction must = 0 for it to not move
and the fnet in the x direction must move with the funnel
we have the force of gravity acting on the mass
Fg = mg
we need force of friction
normal force will be perpindicular to the funnel
so Fn = cosA*Fg = cosA*mg
Ff = usFn = us*cosA*mg

is this correct so far?

Thanks

2. Oct 1, 2007

### Astronuc

Staff Emeritus
correct.

Also don't forget the horizontal force due to rotation.

3. Oct 2, 2007

### Destrio

Fn = cosA*Fg = cosA*mg
Ff = usFn = us*cosA*mg

Horizontal force due to rotation would be a centripedal force towards the centre of the circle?
Fc = mv^2/r

So I have all the forces acting on the mass (I believe)

I need to find a way to incorporate W revolutions per second into this.
Could I use W revolutions per second instead of velocity in the Fc equation?
Fc = m(W/s)^2/r
Or would that cause the units to go bonkers?

4. Oct 2, 2007

### learningphysics

In this particular case I don't think Fn = mgcos(theta).

You do know that in the vertical direction net force = 0. And you know that in the horizontal direction, net force = mv^2/r.

What are the components of the normal force and the frictional force in the vertical diretion?

What are their components in the horizontal direction?

5. Oct 2, 2007

### Destrio

for normal force,
would Fn in the vertical direction be Fn = sinA/mg
and the the horizontal Fn = cosA/Fc = cosA/(mv^2/r)

and frictional force being usFn
vert Ff = us(sinA/mg)
hori Ff = us[cosA/(mv^2/r)]
?

6. Oct 2, 2007

### learningphysics

never mind about mg for now... You have Fn. What is its horizontal component and vertical component (in terms of Fn and theta)?

7. Oct 2, 2007

### Destrio

I'm not sure I'm following you
How can I find Fn without mg?

would I want the verticle component to be
y = Fn*sin(theta)

and horizontal:
x = Fn*cos(theta)

8. Oct 2, 2007

### learningphysics

yes, you will need that afterwards.

yes, sorry but can you write y and x in terms of the angle A? If you were using theta as A, then the components aren't right.

9. Oct 2, 2007

### Destrio

y = Fn*cosA
x = Fn*sinA

10. Oct 2, 2007

### learningphysics

exactly. And how about the x and y components of friction using A?

11. Oct 2, 2007

### Destrio

for friction wouldnt just be the us of Fn?
Ff = usFn
Fn = Ff/us

y = (Ff/us)*cosA
x = (Ff/us)*sinA

12. Oct 2, 2007

### learningphysics

Yes, this is true at the limit, when it is just about to slip.

ok. but these aren't the vertical and horizontal components of friction... these are still the vertical and horizontal components of the normal force.

what are the vertical and horizontal components of friction... do it the same way you did it with the normal force.

13. Oct 2, 2007

### Destrio

for Fn
y = Fn*cosA
x = Fn*sinA

for Ff
y = Ff*sinA
x = Ff*cosA

14. Oct 2, 2007

### learningphysics

exactly. now here's the main trick to the problem... the friction can act either way... either up the plane or down the plane... depending on the rate of W. meaning the components for friction could be:

y = Ff*sinA (upward)
x = Ff*cosA (rightward)

or

y = -Ff*sinA (downward)
x = -Ff*cosA (leftward)

in the moment when the cube is about to slip Ff = us*Fn.

you need to solve the problem for the two situations... when the friction is up the plane with a magnitude of us*Fn... that gives one W. And then when the friction is down the plane with a magnitude of us*Fn. that gives the other W.

you know that the y-components of friction, normal force cancel with gravity...

And you know that the x-components of friction and normal force give rise to centripetal motion... hence $$\Sigma{F_x} = mv^2/r$$

15. Oct 2, 2007

### Destrio

for Fn
y = Fn*cosA
x = Fn*sinA

for Ff
y = Ff*sinA (upward)
x = Ff*cosA (rightward)
or
y = -Ff*sinA (downward)
x = -Ff*cosA (leftward)

in the moment when the cube is about to slip Ff = us*Fn.

So first situation with Friction up the plane
y = Ff*sinA (upward)
x = Ff*cosA (rightward)

Fnet,x = mv^2/r = Ff*cosA + Fn*sinA
mv^2/r = us*Fn*cosA + Fn*sinA
mv^2/r = Fn(us*cosA + sinA)

I'm not sure how to incorporate the W into the expression, can I use W/s to replace velocity?

16. Oct 2, 2007

### learningphysics

careful here. we're looking for the centripetal force... the x-component of the normal force is acting inward... so it should be positive. But the x-component of the frictional force acts outward so it should be negative. (unless I'm misunderstanding the problem)

yes. get the velocity in terms of W.

17. Oct 2, 2007

### Destrio

ah yes you're right

So first situation with Friction up the plane
y = Ff*sinA (upward)
x = -Ff*cosA (rightward)

Fnet,x = mv^2/r = -Ff*cosA + Fn*sinA
mv^2/r = -us*Fn*cosA + Fn*sinA
mv^2/r = Fn(-us*cosA + sinA)
v = rFn(-us*cosA + sinA) / m
v = W/s = rFn(-us*cosA + sinA) / m

18. Oct 2, 2007

### learningphysics

The W/s isn't right. W is the number of revolutions per second.

so $$2\pi{r}W$$ is distance travelled per second....

ie $$v = 2\pi{r}W$$

19. Oct 2, 2007

### learningphysics

don't forget the square.

20. Oct 2, 2007

### Destrio

ah, alright, let me try again

y = Ff*sinA (upward)
x = -Ff*cosA (rightward)

Fnet,x = mv^2/r = -Ff*cosA + Fn*sinA
mv^2/r = -us*Fn*cosA + Fn*sinA
mv^2/r = Fn(-us*cosA + sinA)
v^2 = [rFn(-us*cosA + sinA) / m]
v = (2pi*rW)^2= [rFn(-us*cosA + sinA) / m]
W^2 = Fn(-us*cosA + sinA) / mr4pi^2
W = sqrt[Fn(-us*cosA + sinA) / mr4pi^2]

Last edited: Oct 2, 2007