Circular Motion / Newton's Laws

[Destrio]

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)]

 

[learningphysics]

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.

 

[Destrio]

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

 

[learningphysics]

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.

 

[Destrio]

I'm confused
if the 1st is
y = Ff*sinA (upward)
x = -Ff*cosA (rightward)

shouldnt downward be -
and leftward be +?

 

[learningphysics]

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.

 

[Destrio]

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?

 

[learningphysics]

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.

 

[Destrio]

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

 

[learningphysics]

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!

 

[Destrio]

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)]

 

[learningphysics]

Looks good!

 

[Destrio]

Excellent,
thanks very much for your help
This problem was terribly difficult to me, but I definitely understand the concepts of it better now :)

 

[learningphysics]

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.

 

[TEAL62]

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?