Circular Motion / Newton's Laws

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Homework Help Overview

The problem involves a small cube of mass m placed inside a funnel that rotates about a vertical axis at a constant rate of W revolutions per second. The funnel's wall makes an angle A with the horizontal, and the coefficient of static friction between the cube and the funnel is us. The objective is to determine the largest and smallest values of W for which the cube remains stationary relative to the funnel.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the forces acting on the cube, including gravitational force, normal force, and frictional force. There are attempts to express these forces in terms of the angle A and the rotation rate W. Some participants question how to incorporate W into the equations for centripetal force and the relationship between linear velocity and angular velocity.

Discussion Status

The discussion is ongoing, with participants exploring different components of forces and their relationships. Some have provided insights into the directionality of forces and the need to consider the effects of friction in both upward and downward scenarios. There is a recognition of the need to derive expressions for W based on the forces involved, but no consensus has been reached on the final formulation.

Contextual Notes

Participants note the importance of correctly identifying the components of forces in both vertical and horizontal directions. There is also a reminder to consider the gravitational force acting on the cube, which is essential for determining the normal force.

  • #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)]
 
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  • #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 +?
 
  • #36
Destrio said:
I'm confused
if the 1st is
y = Ff*sinA (upward)
x = -Ff*cosA (rightward)

shouldnt downward be -
and leftward be +?

Yes. it should be downward. You didn't use downward in the equation though:

FncosA + usFnsinA - mg = 0

Why are you using the y-component of friction as upward here when it should be downward just as you said?

Also the Fnet,x equation is wrong.
 
  • #37
2nd:
y = -Ff*sinA (downward)
x = Ff*cosA (leftward)

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

FncosA - usFnsinA - mg = 0

is this better?
 
  • #38
Destrio said:
2nd:
y = -Ff*sinA (downward)
x = Ff*cosA (leftward)

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

The above equation isn't right. Describe how you are getting it.

FncosA - usFnsinA - mg = 0

is this better?

yes, this one's right.
 
  • #39
ooh, I see what I was doing, I was reading the y component for Ff instead of the Fn component

Fnet,x = mv^2/r = Ff*cosA + Fn*sinA
 
  • #40
Destrio said:
ooh, I see what I was doing, I was reading the y component for Ff instead of the Fn component

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

There you go! :smile:
 
  • #41
Hurray!

So for part 2:

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^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)]
 
  • #42
Looks good!
 
  • #43
Excellent,
thanks very much for your help
This problem was terribly difficult to me, but I definitely understand the concepts of it better now :)
 
  • #44
Destrio said:
Excellent,
thanks very much for your help
This problem was terribly difficult to me, but I definitely understand the concepts of it better now :)

No prob. It was definitely a tough problem.
 
  • #45
learningphysics said:
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. [/tex]

why can friction act either up or down the plane?
 

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