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course MTH 279
Query 10 Differential Equations*********************************************
Question: 3.7.4. Solve the equation m dv/dt = - k v / (1 + x), where x is position as a function of t and v is velocity as a function of t.
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Your solution:
mv(dv/dt) = -kv / 1+x. Simplifying we get dv = -k/m(1+x). Next we can integrate to get v = -k/m (|1+x|) + C. We can assume v(0) = v_0, so we can get v_0 = -k/m ln(|1+0|) + C.
The ln(1) = 0, we C = v_0, and our solution is: v(x) = -k/m ln(|1+x|) + v_0. If we wanted to know where an object would stop we would solve for v(x) = 0. This would be: v(x) = -k/m(ln|1+x|)+v_0 = 0.
so -k/m(ln|1+x|) = -v_0.
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Given Solution:
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The question also asks how far the object travels before stopping, but that seems to have been left off the document you were using (not sure how that happened, but it was on this end, not on yours).
Your solution is right on target as far as it goes.
You do still need to solve for
To find where the object stops, solve for v(x) = 0.
v(x) = -k / m ln | 1 + x | + v_0 = 0
when
-k/m ln | 1 + x | = -v_0,
which occurs when
ln | 1 + x | = m / k * v_0
| 1 + x | = e^(m / k * v_0)
If x > -1 then | 1 + x | = 1 + x and we get
x = e^(m / k * v_0) - 1.
If x < -1 then | 1 + x | = -1 - x and we get
x = -e^(m / k * v_0) - 1.
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Question:
3.7.6. A vertical projectile has initial velocity v_0, and experiences drag force k v^2 in the direction opposite its motion. Assume that acceleration g of gravity is constant. Find the maximum height to which the projectile rises.
If the initial velocity is 80 m/s and the projectile, whose mass is .12 grams, rises to a height of 40 meters (estimated quantities for a plastic BB), what is the value of k?
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Your solution:
Our equation is m(dv/dt) = -k(v^2) - mg. We can rewrite this so that v (dv) / (k/m (v^2) + g) = (-dx). Now we can integrate that to get m / (2k) (ln |(k/m)(v^2)+g|) = -x + c. We can simpify to get
ln(|(k/m)(v^2) + g)| = -2k / m * x +C. Now we can use a coefficient A, to get: k / m (v^2) + g = e^(2 k / m * x + c) = A e^(-2 k / m * x). Solving for v we get v = (sqrt of ((m/k) * A * e^(-2k/m*0) - mg / k) )= (sqrt of (A * (m/k) - mg/k)).
Lets assume that v(0) = v_0. If this is the case, we can plug in v_0 to get: v_0 = sqrt(m / k * A e^(-2 k / m * 0) - m g / k ) = sqrt(A * m / k - m g / k). Next we will solve for A, getting A = (v_0^2+ m g / k) / (m / k) = k / m ( v_0^2 + m g / k). Plugging this
back into our equation, we will get v(t) = sqrt( (v_0^2 + m g / k) e^(-k/m * x) - m g / k). Now we are looking for the maximum height, which we know occurs when v(t) = 0. Therefore, plugging in, we get v(t) = sqrt( (v_0^2 + m g / k) e^(-2 k / m * x) - m g / k) = 0, where x is our maximum height.
solving for x, we get x = m / (2 k) * ln ((k / (m g) * v_0^2 + 1) ). We are told that x = 40 meters when v_0 = 80 m/s and we also know that the mass is equal to .12 grams, or .00012 kg. Plugging these facts into our equation, we get: 40 = .00012 / (2 k) * ln ((k / (.00012 * 9.8) * 80^2 + 1) ).
Solving for k we get that k is almost equal to 10^(-14).
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Very good.
A little more information on the units:
The SI units of k are Newtons / (m/s)^2 = kg / m.
The drag force on the BB at 80 m/s would be
(80 m/s)^2 * 10^-14 N / (m/s)^2 = 6.4 * 10^-11 N.
This force is very small compared to m g (about .001 N for this mass), and this is clearly erroneous.
The solution to x_max = m / (2 k) * ln ((k / (m g) * v_0^2 + 1) ) is dimensionally consistent (k / (m g) has units of velocity^2 and m / (2 k) has units of position).
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Given Solution:
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Self-critique (if necessary): This was so hard to work out even with my notes...not sure that I will be able to pull this off on a test!
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Question:
3.7.8. Mass m accelerates from velocity v_1 to velocity v_2, while constant power is exerted by the net force.
At any instant power = force * velocity (physics explanation: power = dw / dt, the rate at which work is being done; w = F * dx so dw/dt = F * dx/dt = F * v).
How far does the mass travel as it accelerates?
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Your solution:
Since we know that power = force * velocity, or F = P / v, we also know that F = P/v = m dv/dt, or dv/dt = P / v. We know that dv/dt = (dv/dx)(dx/dt) = v (dv/dx),
so m(v)(dv/dx) = P/v, and therefore m(v^2) dv = P dx. Plugging in we can see that P x + C = m (v^3)/3 = m/3 * (v^3). If we have velocity 1 and 2 (v_1, v_2), we know that
m / 3 * (v_2 )^ 3 - m / 3 *( v_1) ^ 3 = P (x_2) + C - (P( x_1) + C) = P ((x_2) -( x_1)), which allows us to see that (x_2) - (x_1) is the travel distance, and so the mass travels
m / 3P * ((v_2)^3 - (v_1)^3).
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Given Solution:
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Question:
3.7.11. An object falls to the surface of the Earth from a great height h, and as it falls experiences a drag force proportional to the square of its velocity.
Assume that the gravitational force is - G M m / r^2.
What will be its impact velocity?
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Your solution:
Impact velocity is calculated by kniwing the time of impact. First we would find the equation for x, displacement, which is dx/dt = v(t). Then we integrate. Knowing the intitial displacement, we would
substitute in t and x and find out what C was. Then we would plug C back into our equation. Then we would find T, which would be the time when we hit the ground. Once we found that value, we would plug that into the equation for velocity, and we would have the impact velocity.
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Here's a solution, which makes use of the fact that dv/dt = dv/dr * dr/dt = v dv/dr:
m dv/dt = -G M m / r^2 + k v^2
dv/dt = dv/dr*dr/dt = v dv/dr
mv dv/dr = -GMm/r^2 + kv^2
m v v ' = -GMm/r^2 + kv^2
where v ' means dv / dr.
v ' = - G M / r^2 * ( 1 / v ) + k / m * v
v ' - k / m * v = - G M / r^2 * (1 / v)
This is a Bernoulli equation
v ' - k / m * v = - G M / r^2 * v^(-1),
of form
v ' - p * v = q * v^n
with p = - k / m, q = - G M / r^2 and n = -1.
Letting u = v^m, with m = 2, we will get the form
u ' - k / m * u = - G M / r^2
with integrating factor e^(-k / m * r), giving us
(u e^(-k / m * r) ) ' = - G M / r^2 * e^(-k / m * r).
Integrating the right-hand side, we find that we can't integrate the right-hand side.
If we could integrate, we would get an integration constant, which we could use to account for the initial position, which would be h + r_Earth.
Another comment:
This model might not be very realistic, because if h is great the atmosphere will thin. However the thinning is exponential in nature, and this could to an extent balance the fact that the air resistance at high speeds involves turbulence and an effective proportionality to a power of v greater than the power 2 assumed here. This might be interesting to investigate, but in the interest of time we'll probably have to leave it at that.
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
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Check my notes and let me know if you have questions.
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