LCSphysicist
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Analyzing Net Forces and Equations in a Driven Mass on a Circular Path System
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It would help if you could post this using LaTeX (see the LaTeX Guide link at the lower left of the Edit window. Thanks.LCSphysicist said:What you think about this system:?
F*cosw - k(x1+x2) - k(x1-x2) = mx1''
-k(x1+x2) - k(x1-x2) = mx2''
Also, could you please explain the equations you are trying to write? It looks like you are trying to write F=ma type equations, but your terms are not clear to me (especially since some parts seem to be missing). Also, at some point fairly soon you will need to include the variables ##\theta_1## and ##\theta_2## to denote the positions of the two masses as functions of time...
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Some sign errors.LCSphysicist said:What you think about this system:?
F*cosw - k(x1+x2) - k(x1-x2) = mx1''
-k(x1+x2) - k(x1-x2) = mx2''
The x1 and x2 can be taken as angles, or arc lengths, whatever.berkeman said:you will need to include the variables θ1 and θ2
LCSphysicist
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Maybe the problem is adopt one clockwise and another counterclockwise?
This came to my mind when i attack the problem, but i went on just to see if i could try by this another way as well as adopt just clockwise [or counterclokwise]. But what i can't refut is why would it be wrong, that is:
## F*cos(wt) - [k(x1+x2)] - k(x1-x2) = m \frac{d^2 x1}{dt^2} ##
## [-k(x1+x2)] - k(x1-x2) = m \frac{d^2 x2}{dt^2} ##
the bracket being to the left spring:
If x2 = 0 and x1 > 0, will be a force on m1 in its negative direction, as to x2. Works as well to x1<0
Without bracket to the right spring:
If x2 = 0 and x1 > 0, will be a force on m1 in its negative direction, as to x2. Also works to x1<0
About the Latex, i will try ;)
This came to my mind when i attack the problem, but i went on just to see if i could try by this another way as well as adopt just clockwise [or counterclokwise]. But what i can't refut is why would it be wrong, that is:
## F*cos(wt) - [k(x1+x2)] - k(x1-x2) = m \frac{d^2 x1}{dt^2} ##
## [-k(x1+x2)] - k(x1-x2) = m \frac{d^2 x2}{dt^2} ##
the bracket being to the left spring:
If x2 = 0 and x1 > 0, will be a force on m1 in its negative direction, as to x2. Works as well to x1<0
Without bracket to the right spring:
If x2 = 0 and x1 > 0, will be a force on m1 in its negative direction, as to x2. Also works to x1<0
About the Latex, i will try ;)
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Consider the case x1=-x2, so both move the same direction around the hoop. What net forces will spring exert on them? What do your equations give?LCSphysicist said:Maybe the problem is adopt one clockwise and another counterclockwise?
This came to my mind when i attack the problem, but i went on just to see if i could try by this another way as well as adopt just clockwise [or counterclokwise]. But what i can't refut is why would it be wrong, that is:
## F*cos(wt) - [k(x1+x2)] - k(x1-x2) = m \frac{d^2 x1}{dt^2} ##
## [-k(x1+x2)] - k(x1-x2) = m \frac{d^2 x2}{dt^2} ##
the bracket being to the left spring:
If x2 = 0 and x1 > 0, will be a force on m1 in its negative direction, as to x2. Works as well to x1<0
Without bracket to the right spring:
If x2 = 0 and x1 > 0, will be a force on m1 in its negative direction, as to x2. Also works to x1<0
About the Latex, i will try ;)