- #1
tamtam402
- 201
- 0
Hey guys.
We were given a problem, which was to consider bv^2 as the force acting upwards on a body falling in water. We were asked to find the depth at which a body entering the water at 5 * ve (terminal speed) would end up with a speed of 1.1 * ve.
Starting from -bv^2 + mg = ma, the first step is to linearize the non-linear expression bv^2 at the equilibrium point (ma = 0).
The equation at the chosen equilibrium point becomes b(ve)^2 = mg
The linearization gives:
b(ve)^2 + 2b(ve)(Δv).Substitution into the original equation gives -b(ve)^2 - 2b(ve)(Δv) + mg = ma.
We know that at the chosen equilibrium point, b(ve)^2 = mg. Thus, mg - b(ve)^2 = 0. I can remove these 2 expressions in the above equation to get the linearized equation at the equilibrium point ve:
Linearized upward force at terminal speed = 2b(ve)(Δv) = ma
Also our teacher showed us that the 2b(ve) expression must be used to find the value of the linearized ve point. Thus, using the equilibrium equation above but substituting b(ve)^2 by the new constant coefficient 2b(ve) which multiplies ve, we get 2b(ve)*ve = mg -> ve = sqrt(mg/2b)
Given the above equation in bold, I was able to solve for height h by using the following substitution:
a = v * dv / dh
Using this substitution I could integrate on both sides of the above equation and solve for h pretty easily. I confirmed with my teacher that our answer was correct.
Here's my question:
I would like to solve this problem by using the conservation of energy equation instead of solving the F = ma equation. Before my teacher confirmed that we had the right answer, I wanted to solve the problem using this other method because I could've verified my answer by myself. However, I don't know how to solve this. I put the initial energy + the work on the left hand side, and the final energy on the right hand side.
Assuming the object falls into the water at a height h at an initial speed of 5ve,
V1 = mgh
T1 = m/2 * (5ve)^2
Work = 2b(ve)(Δv) * h
V2 = 0
T2 = m/2 * (1.1ve)^2
V1 + T1 - Work = V2 + T2
I don't know what to do with the expression in bold, for the work of the linearized force. Obviously an integral is needed, but if I integrate as I specified above, I have to divide both sides by Δv since dv is on the right side of the equation. I end up with 2b(ve)h = stuff. The left side of this equation is NOT equal to the Force times acceleration because the force has a Δv on the left side.
So, how can I find the "correct" work to put into the equation?
We were given a problem, which was to consider bv^2 as the force acting upwards on a body falling in water. We were asked to find the depth at which a body entering the water at 5 * ve (terminal speed) would end up with a speed of 1.1 * ve.
Starting from -bv^2 + mg = ma, the first step is to linearize the non-linear expression bv^2 at the equilibrium point (ma = 0).
The equation at the chosen equilibrium point becomes b(ve)^2 = mg
The linearization gives:
b(ve)^2 + 2b(ve)(Δv).Substitution into the original equation gives -b(ve)^2 - 2b(ve)(Δv) + mg = ma.
We know that at the chosen equilibrium point, b(ve)^2 = mg. Thus, mg - b(ve)^2 = 0. I can remove these 2 expressions in the above equation to get the linearized equation at the equilibrium point ve:
Linearized upward force at terminal speed = 2b(ve)(Δv) = ma
Also our teacher showed us that the 2b(ve) expression must be used to find the value of the linearized ve point. Thus, using the equilibrium equation above but substituting b(ve)^2 by the new constant coefficient 2b(ve) which multiplies ve, we get 2b(ve)*ve = mg -> ve = sqrt(mg/2b)
Given the above equation in bold, I was able to solve for height h by using the following substitution:
a = v * dv / dh
Using this substitution I could integrate on both sides of the above equation and solve for h pretty easily. I confirmed with my teacher that our answer was correct.
Here's my question:
I would like to solve this problem by using the conservation of energy equation instead of solving the F = ma equation. Before my teacher confirmed that we had the right answer, I wanted to solve the problem using this other method because I could've verified my answer by myself. However, I don't know how to solve this. I put the initial energy + the work on the left hand side, and the final energy on the right hand side.
Assuming the object falls into the water at a height h at an initial speed of 5ve,
V1 = mgh
T1 = m/2 * (5ve)^2
Work = 2b(ve)(Δv) * h
V2 = 0
T2 = m/2 * (1.1ve)^2
V1 + T1 - Work = V2 + T2
I don't know what to do with the expression in bold, for the work of the linearized force. Obviously an integral is needed, but if I integrate as I specified above, I have to divide both sides by Δv since dv is on the right side of the equation. I end up with 2b(ve)h = stuff. The left side of this equation is NOT equal to the Force times acceleration because the force has a Δv on the left side.
So, how can I find the "correct" work to put into the equation?