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* Write down the kinetic energy for each particle individually and add: T = 0.5*m*v_A^2 + 0.5*m*v_B^2 * Choose your gravitational datum and write down potential. If you choose the datum lines at the horizontal line passing through B: V1 = m*g*y1 V2 = m*g*y2 (Note that B does not change in potential as it moves.) * No work is done on the system except for the weight of block A (which is included in V). * Use x^2 + y^2 = L^2 to relate positions of A and B. * Differentiate this once with respect to time to get (for position 2): x2*v_B2 + y2*v_A2 = 0 which gives: v_A2=-(x2/y2)*v_B2 * Substitute into kinetic energy expression. * Write down work energy equation: V1 = T2 + V2 * Differentiate work-energy equation once with respect to time. Set dv_B2/dt = 0 to find the value of y2 at which v_B is a maximum. * Substitute y2 into work energy equation to find the maximum speed for B.
I worked this question out and got the correct answer but I believe the book is misleading. It asks you to find the max velocity of B but it doesn't specify that it means in the upper half of the triangle.
Block A would be moving much faster after it falls below B. Hence block B would move faster as well. Unfortunately, the book only considered the upper interval and got 0.962 m/s. I think this is because they differentiated the velocity equation and set it equal to zero and assumed that would be the max. I instead graphed it to see the whole velocity versus position chart.
I ran my same equation that I hope explains the both intervals and got 0.962 m/s for the upper part and 3.532 m/s for the lower part. This second number comes from the instant right before B would slam into the wall.
Let me know if I this makes sense or if I made a mistake.
if you set dv_B2/dt = 0 wouldn't you just have mgy_1 = mgy_2? I guess i don't know how to differentiate the work energy equation with respect to time. What happens to the x^2/y^2?
Your final equation should be just Vb in terms of y. So, you don't differentiate with respect to time. You differentiate with respect to y. If you need to see what's happening graph it on your calculator or computer.
Yeah, I guess that doing the differentiation with respect to y makes more sense that with respect to time, t, since y_B is appearing as a function of y.
However, either way should work since (by the chain rule):
dv_B/dt = (dv_B/dy)*(dy/dt)
we see that dv_B/dt = 0 is equivalent to dv_B/dy = 0 for dy/dt not equal to zero.
7 comments:
Here are some basic steps for the solution:
* Write down the kinetic energy for each particle individually and add:
T = 0.5*m*v_A^2 + 0.5*m*v_B^2
* Choose your gravitational datum and write down potential. If you choose the datum lines at the horizontal line passing through B:
V1 = m*g*y1
V2 = m*g*y2
(Note that B does not change in potential as it moves.)
* No work is done on the system except for the weight of block A (which is included in V).
* Use x^2 + y^2 = L^2 to relate positions of A and B.
* Differentiate this once with respect to time to get (for position 2):
x2*v_B2 + y2*v_A2 = 0
which gives:
v_A2=-(x2/y2)*v_B2
* Substitute into kinetic energy expression.
* Write down work energy equation:
V1 = T2 + V2
* Differentiate work-energy equation once with respect to time. Set dv_B2/dt = 0 to find the value of y2 at which v_B is a maximum.
* Substitute y2 into work energy equation to find the maximum speed for B.
Does your solution look something like that?
I worked this question out and got the correct answer but I believe the book is misleading. It asks you to find the max velocity of B but it doesn't specify that it means in the upper half of the triangle.
Block A would be moving much faster after it falls below B. Hence block B would move faster as well. Unfortunately, the book only considered the upper interval and got 0.962 m/s. I think this is because they differentiated the velocity equation and set it equal to zero and assumed that would be the max. I instead graphed it to see the whole velocity versus position chart.
I ran my same equation that I hope explains the both intervals and got 0.962 m/s for the upper part and 3.532 m/s for the lower part. This second number comes from the instant right before B would slam into the wall.
Let me know if I this makes sense or if I made a mistake.
i got the same thing chris b got
i got the 3.532 because i did not realize it meant upper triangle. thanks.
if you set dv_B2/dt = 0 wouldn't you just have mgy_1 = mgy_2? I guess i don't know how to differentiate the work energy equation with respect to time. What happens to the x^2/y^2?
Your final equation should be just Vb in terms of y. So, you don't differentiate with respect to time. You differentiate with respect to y. If you need to see what's happening graph it on your calculator or computer.
Yeah, I guess that doing the differentiation with respect to y makes more sense that with respect to time, t, since y_B is appearing as a function of y.
However, either way should work since (by the chain rule):
dv_B/dt = (dv_B/dy)*(dy/dt)
we see that dv_B/dt = 0 is equivalent to dv_B/dy = 0 for dy/dt not equal to zero.
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