Movement caused by wind (changing force)

  • Thread starter BitterX
  • Start date
  • #1
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the problem is with b. and c. but I would like verification on a if possible, thanks!

Homework Statement


A cabin lays on a smooth track (so friction can be disregarded)
the wind exerts a pushing force [itex]F=\gamma u[/itex] u is the relative velocity between
the cabin and the wind. the wind's velocity is [itex]v_0[/itex]

a. what is the maximum velocity?
b. how much time will it take to the cabin to reach 86% of this velocity?
c. sand is spread on the track and the friction coefficient is [itex]\mu_k[/itex]
what is the maximum velocity in this situation?


Homework Equations


[itex]F=ma[/itex]
[itex]a=\frac{dv}{dt}[/itex]

The Attempt at a Solution


a. let's assume the velocity of the cabin is [itex]v_c[/itex] then:
[itex]u=v_0 - v_c[/itex]
the maximum velocity will be reached when there is no more acceleration.
[itex]ma=\gamma u = 0[/itex]
so [itex] u=0 , v_c=v_0[/itex]
so the maximum velocity is [itex]v_0[/itex]

b.
[itex]F=ma=m\frac{du}{dt} , m\frac{du}{dt}=\gamma u[/itex]
multiplying by dt dividing by m and u we get:
[itex]\frac{1}{u}du=\frac{\gamma}{m}dt[/itex]
integeral on both sides from final to initial:
[itex]u(t)=v_0 - v_c (t)[/itex]
[itex]\int_{v_0}^{u(t)}\frac{1}{u}du=\int_{0}^{t}\frac{\gamma}{m}dt[/itex]
this gives:
[itex]ln\frac{u(t)}{v_0}=\frac{\gamma}{m}t[/itex]
[itex]\frac{v_0 - v_c (t)}{v_0}=e^{\frac{\gamma}{m}t}[/itex]
so the velocity(time):
[itex]v_c (t)=v_0 (1- e^{\frac{\gamma}{m}t})[/itex]
this can't be right - more logical is [itex]v_c (t)=v_0 (1- e^{-\frac{\gamma}{m}t})[/itex]
but I don't see where my error is.
let's assume I have the right velocity, we need to find when it's 86% so
[itex]v_c (t)= 0.86v_0[/itex]
[itex]e^{-\frac{\gamma}{m}t}=0.14[/itex]
[itex]t=-\frac{mln0.14}{\gamma}[/itex]

c. now [itex]F= \gamma u -mg\mu_k[/itex]
but I'm not sure how I should proceed
 

Answers and Replies

  • #2
1,198
5
Going back to the start, change your variable to

V = V0 - Vc

where V0 is wind velocity and Vc cabin velocity. The result you think is correct will drop right out after integration and evaluation of constant of integration.

For part c, do another appropriate change of variable.
 
Last edited:
  • #3
36
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I'm terribly sorry but I didn't understand,
I did use [itex]u= v_0 - v_c[/itex]
and I tried using it from the start but ended up in the same place,

and as for c. I'm not sure what I should change "u" to,
because u has to have units of speed [itex]\frac{[L]}{[T]}[/itex] where can I add the [itex]g\mu_k[/itex]?
maybe
[itex]u= v_0 +g\mu_k t - v_c[/itex]? but I don't think the t fits there logically
 
  • #4
1,198
5
mdV/dt = F = gamma(Vw-V)

Define a new variable that is Vw-V. Then solve the differential equation using only indefinite integration where you pick up a constant of integration. You evaluate the constant by setting V=0 at time=0.

The part with the friction is very similar.
 
  • #5
1,198
5
Here is some more help:

m*dV/dt = F = gamma*(Vw-V)

Let h = Vw-V

so V = Vw-h

dV/dt = -dh/dt

-m*dh/dt = gamma * h

dh/h = -(gamma/m) * dt

Now do an indefinite integration. Evaluate the constant of integration after returning to original variable V.

For part with friction:

m*dV/dt = gamma*(Vw-V) - m*g*mu

Define new variable h in a similar way. I cannot tell you exactly what because I would essentially be doing the problem for you. You want to define it so you can separate the differential equation and do an indefinite integration. Evaluate constant of integration the same as before.

After you derive all this, do a dimensional analysis to make sure your units agree.
 
  • #6
36
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Thanks!
I now see what I did wrong: I used [itex]a= \frac{du}{dt}[/itex] ( the relative velocity)
instead of [itex]a= \frac{dv_c}{dt}[/itex] (the cabin's velocity)
 

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