Circular Motion / Newton's Laws

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SUMMARY

The discussion focuses on analyzing the forces acting on a small cube of mass m placed inside a funnel rotating about a vertical axis. The participants derive the equations for the normal force (Fn) and frictional force (Ff) in terms of the angle A, mass m, gravitational force (mg), and the coefficient of static friction (us). They establish the conditions under which the cube will not slip, leading to the formulation of the largest and smallest values of angular velocity (W) for the cube's stability. The final equations for W are expressed as W = sqrt[(g)(-us*cosA + sinA) / (r4pi^2)(cosA+us*sinA)].

PREREQUISITES
  • Understanding of Newton's Laws of Motion
  • Knowledge of centripetal force and its application
  • Familiarity with static friction and its coefficient
  • Basic trigonometry, particularly with angles and components
NEXT STEPS
  • Study the derivation of centripetal force in rotating systems
  • Learn about the implications of static friction in inclined planes
  • Explore the relationship between angular velocity and linear velocity
  • Investigate the effects of varying angles on stability in rotational dynamics
USEFUL FOR

This discussion is beneficial for physics students, mechanical engineers, and anyone interested in the dynamics of rotating systems and the application of Newton's Laws in real-world scenarios.

  • #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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