query 10

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course Phy 279

3/6 3Had a little trouble with some of the problems

Query 10 Differential Equations

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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:

m dv/dt = - k v / (1 + x), I believe we would first get this into the form dv/dx

m dv/dt = - k v / (1 + x), because dv/dt = dv/dx*dx/dt. We know dv/dt = dv/dx*v

mv dv/dx = - k v / (1 + x)

dv/dx = -kv/vm(1 + x)

v/-kv dv = 1/(m(1 + x)) dx

Int(v/-kv dv) = -k*Int(v/v dv) = -kInt(1 dv) = -kv

lnt(1/(m(1 + x)) dx) = 1/mInt(1/(x+1) dx)

= m*[Int(1/x dx )+ Int(1 dx)]=m*(ln|x| + x) + C

-kv = m*(ln|x| + x) + C

v = m*(ln|x| + x)/-k

????I think I did this completely wrong, instead of solving for change in v wrt x I think we were simply to consider that x is a fn of t when calculating lnt(1/(m(1 + x)) dx)??????

@& Up to this point your solution looks good.

dv/dt = dv/dx * dx/dt = v dv so the equation becomes

m v dv/dt = - k v / ( 1 + x ),

which rearranges to

dv = -k / m ( 1 + x ).

Integrating we get

v = -k / m ln | 1 + x | + c.

With the condition v(0) = v_0 we get

v_0 = - k / m ln | 1 + 0 | + c.

Since ln(1) = 0 we have c = v_0, giving us solution

v(x) = - k / m ln | 1 + x | + v_0.

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.

NOTE ON ERRONEOUS ANSWER:

Erroneous answer from careless algebra in last step (k/m instead of m/k, a very easy mistake to make since the quantity k / m occurs frequently and we're used to writing it):

x = e^(k / m * v_0)-1

This answer is not dimensionally consistent. The exponent must be unitless.

The correct answer is

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 m / k is multiplied by v_0 in units of position / time (to get m / k * v_0) we get a unitless exponent for v.

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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, v(0) = 0

We can write v’ in place of dv/dt to make solution easier to read

m*v’ = -mg - kv^2, divid through by -mg

mv’/-mg = 1 + k/(mg)v^2, m’s cancel out on right

v’/( 1 + k/(mg)v^2) = -g

(dv/dt)/ ( 1 + k/(mg)v^2) = -g

dv/( 1 + k/(mg)v^2) = -g dt

Int(1/( 1 + k/(mg)v^2) dv), letting u = tan^-1(`sqrt(k/mg)*v), du = 1/(1 + (`sqrt(k/mg)*v)^2) dv = (1 + (k/mg)*v^2) du

Int( 1/( 1 + (k/mg)v^2)*( 1 + (k/mg)*v^2) du) = Int(1 du) = u

Int(1/( 1 + k/(mg)v^2) dv) = tan^-1(`sqrt(k/mg)*v)

tan^-1(`sqrt(k/mg)*v) = -gt + C

For V(0) = V_0

C = tan^-1(`sqrt(k/mg)*v_0)

So…

v(t) = ((`sqrt(k/mg)*v_0) - gt) * `sqrt(k/mg)

???I’ve not getting the same answer as the one shown in the book, I’m messing up somewhere during my u sub to trying find the integral of (1/( 1 + k/(mg)v^2) dv)????????

@& It looks like you've got a chain rule problem. The derivative of tan^-1(`sqrt(k/mg)v) would have a chain rule factor of sqrt(k/mg). Otherwise your work looks pretty goo.

However the question asks about maximum height, which is easier to get to if you use x as your independent variable rather than t.

The equation is

m dv/dt = - k v^2 - m g,

which we can write as

m v dv/dx = - k v^2 - m g

so that

v dv/dx = -k/m v^2 - g

v dv / (k/m v^2 + g) = -dx.

Integrating we get

m / (2 k) ln | k/m v^2 + g | = -x + c

ln | k / m v^2 + g | = -2 k / m * x + c

k / m v^2 + g = e^(2 k / m * x + c) = A e^(-2 k / m * x), A > 0

v = sqrt(m / k * A e^(- 2 k / m * x) - m g / k ).

If v(0) = v_0 then

v_0 = sqrt(m / k * A e^(-2 k / m * 0) - m g / k ) = sqrt(A * m / k - m g / k).

Solving for A we get

A = (v_0^2+ m g / k) / (m / k) = k / m ( v_0^2 + m g / k).

Substituting this into our equation for v we obtain

v(t) = sqrt( m / k * k / m ( v_0^2 + m g / k) e^(-k/m * x) - m g / k) )

v(t) = sqrt( (v_0^2 + m g / k) e^(-k/m * x) - m g / k).

Maximum height occurs when v(t) = 0:

v(t) = sqrt( (v_0^2 + m g / k) e^(-2 k / m * x_max) - m g / k) = 0

when

(v_0^2 + m g / k) e^(-2 k / m * x_max) - m g / k = 0

e^(-2 k / m * x_max) = (m g / k) / ((v_0^2 + m g / k) ) = 1 / (k / (m g) * v_0^2 + 1)

-2 k / m * x_max = ln | 1 / (k / (m g) * v_0^2 + 1) |

x_max = m / (2 k) * ln ((k / (m g) * v_0^2 + 1) )

(note: on the last step we used the fact that ln(1 / a) = ln(1) - ln(a) = 0 - ln(a) = - ln(a) )

If we know x_max, m and v_0 then k is given implicitly by this result. However we can't explicitly solve this equation for k, so we need to use graphical or numerical techniques.

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).

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Given Solution:

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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:

F_net = m*a

F_1 * dx/dt =m*a, where a = dv/dt and dx/dt = v

F_1*v = mdv/dt,

dv/dt = vdv/dx

F_1*v = m*v*dv/dx, v is non zero over interval so we can cancel it out

dv/dx = F_1/m, v(0) =v_1

v(x) = (F_1/m)x + C, where C = v_1

????I’ve gotten confused with what I’m doing??????

@& Easy to do.

F= P/v

Fnet = P/v = m dv/dt

so dv/dt = P / v

dv/dt = dv/dx * dx/dt = v dv/dx

m v dv/dx = P / v

and

m v^2 dv/dx = P,

m v^2 dv = P dx

P x +c = m v^3 / 3 = m/3 * v^3

This 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)
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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:

Diff equ is

mdv/dt = -(GMm/r^2) + kv^2, dv/dt = d^2r/dt^2 because v = dr/dt or change in radius of earth wrt to time.

m(d^2r/dt^2) = -( GMm/r^2) + kv^2

= mv(dv/dr) = -( GMm/r^2) + kv^2

Using Bernoulli Equation

m*v*v’ = -( GMm/r^2) + kv^2

v’ = -(GM(1/vr^2) + kv/m

??????I’m stuck????

@& m dv/dt = -G M m / r^2 + k v^2

dv/dt = dv/drdr/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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@& You're doing fine; you're making some errors in detail, and in some cases you're using t as the independent variable when x would make things easier.

Check my notes.*@

query 10

@& You're doing fine; you're making some errors in detail, and in some cases you're using t as the independent variable when x would make things easier. Check my notes.*@

@& You're doing fine; you're making some errors in detail, and in some cases you're using t as the independent variable when x would make things easier.

Check my notes.*@