2011-02-16, 04:17 AM
(This post was last modified: 2011-02-16, 06:17 AM by 2147483647.)
The problem I am having is conceptual. Say that a particle of mass m starts from rest, and then moves along the path r(t) = <x(t), y(t)>. Does the forcing function always have to equal F(t) = m <x''(t), y''(t)>? (Just asking for confirmation on this part.) I'm going to proceed assuming that F(t) must equal the second derivatives, because it seems to be correct at the end of my derivation.
To calculate the work done by a force field and a path, the line integral is given by :
W= ∫ (F•T) ds
T is the unit tangent vector, r'(t)/||r'(t)||, and ds is the length, ||r'(t)|| dt. Therefore,
W= ∫ F • r'(t) dt
Because of the way I defined F(t), I can plug it in directly:
W= ∫ m <x''(t), y''(t)> • <x'(t), y'(t)> dt
Additionally, accounting for the force supplied against gravity (in this part, I am not sure whether the sign is correct. Someone PLEASE PLEASE check this):
F= m <0, g>
I can combine to obtain this:
W= ∫ m <x''(t), y''(t) +g> • <x'(t), y'(t)> dt
W= ∫ m *x'(t) *x''(t) + m *y'(t) *y''(t) + mg y'(t) dt
W = 1/2 m x'(t)^2 + 1/2 m y'(t)^2 + mg y(t) [a,b]
This seems to be in agreement with the work-energy theorem, which states that:
W = K = 1/2 mv^2
Now comes my first question. What if I have the same conditions, but instead of assuming that the input force is F = m <x''(t), y''(t)>, I assume that F = <g(x,y), h(x,y)>? If I go through the procedure starting from here:
W= ∫ F • r'(t) dt
Does this integral only tell me the work done by the field? What if I wanted to find the field that supplied the rest of the force?
My second question is, what if I'm given that the initial position of a particle is (0,0), and that the forcing function is F = m <x''(t), y''(t)>? Then do I assume that the particle always travels the path r(t) = <x(t), y(t)>? What if my path isn't time dependent?
Third, what if there is friction along this path? How do I account for this?
Fourth, my book (Giancoli, 4th ed.) states that the "work done by a varying force is":
W= ∫ F • dl = ∫ F cos(θ
dl
This is drastically different from my derivation. Can someone please explain this difference? I'm leaning toward the impression that my book is completely wrong, and the author tried to cater to dummies who have no real idea how to operate mathematics.
Fifth, what if the path isn't time dependent?
Someone please help me. I'm pretty desperate, because I have a midterm on this stuff on Thursday (tomorrow). To be precise, in 36 hours.
To calculate the work done by a force field and a path, the line integral is given by :
W= ∫ (F•T) ds
T is the unit tangent vector, r'(t)/||r'(t)||, and ds is the length, ||r'(t)|| dt. Therefore,
W= ∫ F • r'(t) dt
Because of the way I defined F(t), I can plug it in directly:
W= ∫ m <x''(t), y''(t)> • <x'(t), y'(t)> dt
Additionally, accounting for the force supplied against gravity (in this part, I am not sure whether the sign is correct. Someone PLEASE PLEASE check this):
F= m <0, g>
I can combine to obtain this:
W= ∫ m <x''(t), y''(t) +g> • <x'(t), y'(t)> dt
W= ∫ m *x'(t) *x''(t) + m *y'(t) *y''(t) + mg y'(t) dt
W = 1/2 m x'(t)^2 + 1/2 m y'(t)^2 + mg y(t) [a,b]
This seems to be in agreement with the work-energy theorem, which states that:
W = K = 1/2 mv^2
Now comes my first question. What if I have the same conditions, but instead of assuming that the input force is F = m <x''(t), y''(t)>, I assume that F = <g(x,y), h(x,y)>? If I go through the procedure starting from here:
W= ∫ F • r'(t) dt
Does this integral only tell me the work done by the field? What if I wanted to find the field that supplied the rest of the force?
My second question is, what if I'm given that the initial position of a particle is (0,0), and that the forcing function is F = m <x''(t), y''(t)>? Then do I assume that the particle always travels the path r(t) = <x(t), y(t)>? What if my path isn't time dependent?
Third, what if there is friction along this path? How do I account for this?
Fourth, my book (Giancoli, 4th ed.) states that the "work done by a varying force is":
W= ∫ F • dl = ∫ F cos(θ
dlThis is drastically different from my derivation. Can someone please explain this difference? I'm leaning toward the impression that my book is completely wrong, and the author tried to cater to dummies who have no real idea how to operate mathematics.
Fifth, what if the path isn't time dependent?
Someone please help me. I'm pretty desperate, because I have a midterm on this stuff on Thursday (tomorrow). To be precise, in 36 hours.