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course MTH 279
10:01 am 3/2
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:
I'm using the book's question outline since just 'solve' doesn't completely make sense to me here. The book says to solve for v(x) meaning transform dv/dt to v dv/dx.
mv dv/dx = -kv/(1+x)
-m/k dv = dx/(1+x)
-m/k*v = ln(1+x) + C
v = -(k/m)(ln(1+x) + C)
It's given that v(0)= v_0
so C= -m/k*v_0
v = -k/m(ln(1+x) + v_0
Also, the object stops when v=0 which is at
x = e^(v_0k/m)-1
@& This would be
x = e^(m / k * v_0) - 1.
Dimensional analysis would be useful here:
k has units of force * position / velocity, or mass * acceleration * position / velocity.
m / k therefore has units of
velocity * mass / (mass * acceleration * position),
which you can verify come out in units of
position / time * mass / (mass * position / time^2 * position) = time / position
so when multiplied by v_0 in units of position / time (to get k /m * v_0) we get a unitless exponent for v.*@
confidence rating #$&*:
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Given Solution:
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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 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:
m dv/dt = -mg - kv^2
mv dv/dy = -mg-kv^2
Divide by mv:
dv/dy = -m/v - k/m v
Arrange to Bernoulli form:
dv/dy + k/m v= mv^-1
n = -1 --> m= 2 --> u = y^2
du/dy + 2k/mu = m
u = -mg/k + Ce^(-2k/my)
v^2 = u --> v= sqrt u
v= sqrt ( -mg/k + Ce^(-2k/my))
v(0)= v_0 = sqrt (-mg/k+C)
C= v_0^2 + mg/k
v = sqrt ( -mg/k + (v_0^2 + mg/k)e^(-2k/my))
y= max @ v=0
0 = sqrt ( -mg/k + (v_0^2 + mg/k)e^(-2k/my))
mg/k = (v_0^2 + mg/k)e^(-2k/my))
e^(-2k/my)) = (mg/k)/(v_0^2 + mg/k)
y = -2m/k
@& Very good. However note that
(mg/k)/(v_0^2 + mg/k) = 1 / (k / (m g) v_0^2 + 1) so that
(ln(mg/k)/(v_0^2 + mg/k)) =
ln(1 / (k / (m g) v_0^2 + 1) )
= - ln (k / (m g) v_0^2 + 1)
which gives you a simpler solutoin for y.
*@
v_0 = 80 m/s y_max = 40 m
Seems we need to know the mass as well, which is not given.
Plastic BB mass is about 0.12 g so I'll use that.
However, you still can't solve for k, which is frustrating since that's something you'd really want to be able to solve for.
@& If we know that x_max = 40 m when v_0 = 80 m/s and m = .12 grams (equal to .00012 kg) then our equation becomes
40 = .00012 / (2 k) * ln ((k / (.00012 * 9.8) * 80^2 + 1) ).
The zero(s) of the function
f(k) = .00012 / (2 k) * ln ((k / (.00012 * 9.8) * 80^2 + 1) ) - 40
will correspond to the value of k.
The zero of this function can be found by numerical methods (e.g., Newton's method) or other means to be
k = 10^-14, approximately.
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).
*@
confidence rating #$&*:65
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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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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:
I'm conjecturing here:
F= P/v
Fnet = P/v = m dv/dt
dv/dx = v dx/dt
P dx = mv^2 dv
Px +C= m/3v^3
v = (3P/m*x + C)^1/3
But particularly here the limits of int are v_1--> v_2 and 0--> x_2
m/3(v_2^3 - v_1^3) = P(x_2-0)
x_2 = (m/3(v_2^3 - v_1^3))/P
@& My solution agrees with your solution.
However it isn't necessary to assume a value for x_2.
P x + c = m / 3 * v^3
implies that
m / 3 * v_2 ^ 3 - m / 3 * v_1 ^ 3 = P x_2 + c - (P x_1 + c) = P (x_2 - x_1)
x_2 - x_1 is the distance traveled, which is therefore
dist traveled = m / (3 P) (v_2^3 - v_1^3)*@
confidence rating #$&*:
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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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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:
m dv/dt = -GMm/r^2 + kv^2
dv/dt = dv/dr*dr/dt = v dv/dr
mv dv/dr = -GMm/r^2 + kv^2
Setup as Bernoulli
dv/dr-k/m*v= -Gm/r^2(v^-1)
n= -1--> m= 2--> u = v^2
du/dr -2k/m*u = -GM/r^2
e^(-2k/mr)*u = -GM*Int(e^(-2rk/m)/r^2 dr)
u = (-GM/e^(-2k/mr))*Int(e^(-2rk/m)/r^2 dr)
v = sqrt ((-GM/e^(-2k/mr))*Int(e^(-2rk/m)/r^2 dr))
v_imp occurs when r = 0 so e^(-2k/mr)= 1 and we have
v_imp = sqrt (-GM*Int(e^(-2rk/m)/r^2 dr))
@& Very good, but I believe your integrand is incorrect, and r is distance from center and impact occurs when r = radius of Earth. See the note below.*@
@& We do get
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. At least I can't, and neither can my computer algebra system.
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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