Circular Motion / Newton's Laws

In summary: In the first situation with Friction up the plane, the cube would move with a velocity of W revolutions per second.
  • #1
Destrio
212
0
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
 
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  • #2
so Fn = cosA*Fg = cosA*mg
Ff = usFn = us*cosA*mg
correct.

Also don't forget the horizontal force due to rotation.
 
  • #3
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
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
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
never mind about mg for now... You have Fn. What is its horizontal component and vertical component (in terms of Fn and theta)?
 
  • #7
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
Destrio said:
I'm not sure I'm following you
How can I find Fn without mg?

yes, you will need that afterwards.

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

and horizontal:
x = Fn*cos(theta)

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
y = Fn*cosA
x = Fn*sinA
 
  • #10
Destrio said:
y = Fn*cosA
x = Fn*sinA

exactly. And how about the x and y components of friction using A?
 
  • #11
for friction wouldn't just be the us of Fn?
Ff = usFn
Fn = Ff/us

y = (Ff/us)*cosA
x = (Ff/us)*sinA
 
  • #12
Destrio said:
for friction wouldn't just be the us of Fn?
Ff = usFn
Fn = Ff/us

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

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

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

for Ff
y = Ff*sinA
x = Ff*cosA
 
  • #14
Destrio said:
for Fn
y = Fn*cosA
x = Fn*sinA

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

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 [tex]\Sigma{F_x} = mv^2/r[/tex]
 
  • #15
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
Destrio said:
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

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)

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

yes. get the velocity in terms of W.
 
  • #17
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
The W/s isn't right. W is the number of revolutions per second.

so [tex]2\pi{r}W[/tex] is distance traveled per second...

ie [tex]v = 2\pi{r}W[/tex]
 
  • #19
Destrio said:
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

don't forget the square.
 
  • #20
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:
  • #21
careful about v^2...

also you can get Fn, using the [tex]\Sigma{F_y}=0[/tex] equation. then substitute Fn into this equation.
 
  • #22
Destrio said:
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]

Looks good. Now you can get Fn from the Fy equation and plug it in here.
 
  • #23
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)

1st:
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]

Fnet, y = 0 = Ff*sinA - Fn*cosA
Ff*sinA = Fn*cosA
Fn = Ff*sinA / cosA

W = sqrt[Fn(-us*cosA + sinA) / mr4pi^2]
W = sqrt[(Ff*sinA / cosA)(-us*cosA + sinA) / mr4pi^2]
 
  • #24
Don't forget mg, the weight... also watch the directions... the y-components of both friction and the normal force are upward.

And use the relationship between normal force and friction... you want to solve for the Fn in terms of m and g and A... not in terms of friction.
 
  • #25
Fn = mg / cosA
W = sqrt[mg(-us*cosA + sinA) / cosAmr4pi^2]
 
  • #26
Destrio said:
Fn = mg / cosA

No. What are the forces acting in the y-direction?
 
  • #27
Ff , Fn , Fg

Ff*sinA = Fn*cosA
y = Ff*sinA (upward)
wont the y component also = Fg?
 
  • #28
Destrio said:
Ff , Fn , Fg

Ff*sinA = Fn*cosA
y = Ff*sinA (upward)
wont the y component also = Fg?

[tex]\Sigma{F_y} = ma_y[/tex], so

[tex]\Sigma{F_y} = 0[/tex]

insert the 3 forces into the left side of this equation. watch the signs. take upward as positive, downward as negative.
 
  • #29
Fnet, y = ma = 0
Fnet, y= Fg + Fn + Ff = 0
m(-a) + Fn + Ff = =
Fn = ma / -Ff

I don't know what I can plug in for Ff without using x or y variables
unless i want usFn

Fn = ma / -usFn
Fn^2 = -ma/us
Fn = sqrt(-ma/us)
 
  • #30
[tex]\Sigma{F_y} = 0[/tex]

[tex]F_ncosA + F_fsinA-mg = 0[/tex]

[tex]F_ncosA + \mu_s F_nsinA -mg = 0[/tex]

solve for Fn
 
  • #31
Fnet, y = 0
FncosA + FfsinA - mg = 0

I see where I was confused, I needed to take the enture expression containing Ff and Fn that = y to find the forces in the y direction, I think me using y instead of Fny confused me

FncosA + usFnsinA - mg = 0

Fn(cosA+us*sinA) = mg
Fn = mg / (cosA+us*sinA)

sooo

W = sqrt[Fn(-us*cosA + sinA) / mr4pi^2]
W = sqrt[(mg / (cosA+us*sinA))(-us*cosA + sinA) / mr4pi^2]
W = sqrt[(g)(-us*cosA + sinA) / (r4pi^2)(cosA+us*sinA)]
 
  • #32
cool. so now, you need to do the problem again but with friction directed down the plane instead of up... that gives the other W.
 
  • #33
ok, so from the beginning

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)

1st:
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^2 = (2pi*rW)^2= [rFn(-us*cosA + sinA) / m]
W^2 = Fn(-us*cosA + sinA) / mr4pi^2
W = sqrt[Fn(-us*cosA + sinA) / mr4pi^2]

FncosA + usFnsinA - mg = 0
Fn(cosA+us*sinA) = mg
Fn = mg / (cosA+us*sinA)

W = sqrt[Fn(-us*cosA + sinA) / mr4pi^2]
W = sqrt[(mg / (cosA+us*sinA))(-us*cosA + sinA) / mr4pi^2]
W = sqrt[(g)(-us*cosA + sinA) / (r4pi^2)(cosA+us*sinA)]

2nd:
y = -Ff*sinA (downward)
x = Ff*cosA (leftward)

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)
v^2 = (2pi*rW)^2 = [rFn(us*cosA - sinA) / m]
W^2 = [Fn(us*cosA - sinA) / mr4pi^2]
W = sqrt[Fn(us*cosA - sinA) / mr4pi^2]

FncosA + usFnsinA - mg = 0
Fn(cosA+us*sinA) = mg
Fn = mg / (cosA+us*sinA)

W = sqrt[Fn(us*cosA - sinA) / mr4pi^2]
W = sqrt[(mg / (cosA+us*sinA))(us*cosA - sinA) / mr4pi^2]
W = sqrt[(g)(us*cosA - sinA) / (r4pi^2)(cosA+us*sinA)]

Is this all correct?

Thanks
 
  • #34
Destrio said:
ok, so from the beginning

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)

1st:
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^2 = (2pi*rW)^2= [rFn(-us*cosA + sinA) / m]
W^2 = Fn(-us*cosA + sinA) / mr4pi^2
W = sqrt[Fn(-us*cosA + sinA) / mr4pi^2]

FncosA + usFnsinA - mg = 0
Fn(cosA+us*sinA) = mg
Fn = mg / (cosA+us*sinA)

W = sqrt[Fn(-us*cosA + sinA) / mr4pi^2]
W = sqrt[(mg / (cosA+us*sinA))(-us*cosA + sinA) / mr4pi^2]
W = sqrt[(g)(-us*cosA + sinA) / (r4pi^2)(cosA+us*sinA)]

Yeah, the above looks good.

2nd:
y = -Ff*sinA (downward)
x = Ff*cosA (leftward)

Fnet,x = mv^2/r = Ff*cosA - Fn*sinA

think about this part again.

FncosA + usFnsinA - mg = 0

this too.

careful about directions.
 
  • #35
I'm confused
if the 1st is
y = Ff*sinA (upward)
x = -Ff*cosA (rightward)

shouldnt downward be -
and leftward be +?
 

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