1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Impulse = ∫Force(t)dt = m∫dv = Δmv via u-substitution?

  1. Dec 3, 2015 #1
    1. The problem statement, all variables and given/known data

    This is not an assigned hw problem but could be considered a hw-like problem. I do not have a physics book currently. I looked on Wikipedia, but did not find anything. I posted here in the hw section because I felt the problem may not be general enough for the general physics section.

    I'm trying to derive Δmv from a non-constant Force(t) function.

    If I have a Force vs. Time graph, the area under the curve is impulse.

    I would like to convert [itex]\int^{t_{2}}_{t_{1}}F(t)dt[/itex] into Δmv by u-substitution

    2. Relevant equations

    trying to find a u - substitution (or v-substitution) that makes sense.

    as in let v = v(t), so dv = v'(t)dt
    or let v = t, so dv = t'dt

    this is where I'm not making any progress I believe....

    3. The attempt at a solution


    I = [itex]\int^{t_{2}}_{t_{1}}F(t)dt[/itex] = [itex]\int^{t_{2}}_{t_{1}}m(a(t))dt[/itex] =

    [itex]\int^{t_{2}}_{t_{1}}m(v'(t))dt[/itex]

    Then for substitution, I decided to randomly let v = v(t)
    because that's the only way I can get dv = v'(t)dt
    so that from here, I can substitute and get:

    [itex]m\int^{t_{2}}_{t_{1}}(v'(t))dt[/itex] ==> [itex]m\int^{t_{2}}_{t_{1}}dv[/itex]

    But if I change dt to dv, then I have to change [itex]\int^{t_{2}}_{t_{1}}[/itex] to [itex]\int^{v_{2}}_{v_{1}}[/itex] somehow?

    The integral [itex]m\int^{v_{2}}_{v_{1}}dv[/itex] seems to equal mv_{2} - mv_{1} = Δmv

    However, I don't think I am choosing the right substitution from the start because suppose I have a function:

    Force(t) = F(t) = 60t^{2}

    [itex]\int^{t_{2}}_{t_{1}}F(t)dt[/itex] = [itex]\int^{4s}_{2s}60t^{2}dt[/itex]

    and I want to express the function in terms of velocity, v:

    [itex]\int^{v_{2}}_{v_{1}}(?)dv[/itex]

    I can't choose v = v(t) because there is no v's in the function to begin with, only t's.

    How can I express the Force in terms of time function as simply mv?

    P.S.

    I already know that [itex]\int F(t)dt[/itex] = [itex]m\int a(t)dt[/itex]
    [itex]a\int\frac{dv}{dt}dt[/itex] = [itex]m\int dv[/itex] = Δmv

    I am not looking to equate a(t) = dv/dt

    I am instead trying to find a substitution such that the limits of the integral reflect velocity values.

    So instead of choosing two different times and solving [itex]\int^{t_{2}}_{t_{1}}F(t)dt[/itex]

    I guess I'm trying to find be able to choose two different velocities and perhaps solve [itex]m\int^{v_{2}}_{v_{1}}dv[/itex]

    I just don't know how to arrive at [itex]m\int^{v_{2}}_{v_{1}}dv[/itex] via the u-substitution method.

    P.P.S.

    Hope my question makes some sense.
    Thank you.
     
    Last edited: Dec 3, 2015
  2. jcsd
  3. Dec 3, 2015 #2

    Ken G

    User Avatar
    Gold Member

    It sounds like what you need to do is simply let Delta v be your variable, that which you are solving. So take any given F(t) and carry out the integral for I, equate that to m times Delta v, and solve for Delta v. You seem to be wanting to plug something in for the endpoints of the v integration, but you cannot know those endpoints, you have to get them from F(t). So what you can figure out from F(t) is either Delta v, or else if you are given v1, you can use Delta v to figure out v2. But the problem would be overconstrained if you are told F(t) and also v1 and v2.

    Also, you seemed bothered by what to do with the limits of the integration when you change integration variables. But remember, the limits of the integration just label the start and end, so those labels can be expressed in any variable choice, but it has to be the integration variable if you want to substitute them in when integration is all done. So what this means is, you just manually replace the endpoints as needed to refer to the integration variable, they are just ways of labeling the endpoints so there's nothing to do when you shift from the t endpoints to the v endpoints except match them up appropriately.
     
  4. Dec 3, 2015 #3
    With the right substitution, I don't really understand why it would not be possible to get from:

    Impulse = [itex] \int^{t_{2}}_{t_{1}}F(t)dt [/itex] to.. [itex]m\int^{v(t_{2})}_{v(t_{1})}dv [/itex] = Δmv

    I believe it should be possible if a suitable substitution could be determined. I'm stuck on determining a substitution that both works and makes sense.

    Suppose I start with x = [itex] t^{4} [/itex],

    then v(t) = [itex] 4t^{3} [/itex]

    a(t) =[itex]12t^{2} [/itex]

    [itex]F(t) = m(a(t)) = 5(12t^{2}) = 60t^{2}[/itex]

    and I want to evaluate the Force from time [itex] t_{1} = 2s [/itex] to [itex] t_{2} = 4s [/itex]
    which means I want to evaluate the velocity from:
    [itex] v(t_{1}) = v(2s) = v_{1} = 4t^{3} = 4(2^{3}) = 32 \frac{m}{s} [/itex]
    to
    [itex] v(t_{2}) = v(4s) = v_{2} = 4t^{3} = 4(4^{3}) = 256 \frac{m}{s}[/itex]

    So to get from:

    [itex] \int^{t_{2}=4s}_{t_{1}=2} F(t)dt [/itex] = [itex] \int^{t_{2}=4s}_{t_{1}=2s} 60t^{2}dt [/itex]

    to:

    [itex] m\int^{v(t_{2})=256 \frac{m}{s}}_{v(t_{1})=32 \frac{m}{s}} dv[/itex] = [itex] m\int^{v(t_{2})=256 \frac{m}{s}}_{v(t_{1})=32 \frac{m}{s}} (????) dv[/itex]

    my assumption is that (????) must just equal a constant of 1 because that's the only way the formula for momentum would be produced. So, I'm wondering how I could arrive at such a function via the original function of force in terms of time, via substitution?

    Now, by your instruction:

    Carry out the integration for I:

    I = [itex] \int^{t_{2}=4s}_{t_{1}=2} F(t)dt [/itex] = [itex] \int^{t_{2}=4s}_{t_{1}=2s} 60t^{2}dt [/itex] = [itex]20t^{3} N*s[/itex]

    equate to mdv:

    [ I = mdv ] = [[itex] \frac{I}{m} = dv[/itex]] = [itex] \frac{20t^{3}}{5}[/itex] = [itex]4t^{3} = dv[/itex]

    hmm... both [itex]v(t) = 4t^{3}[/itex] and [itex]dv = 4t^{3} [/itex]? If [itex]v(t) = 4t^{3}[/itex] , then how can [itex]dv = 4t^{3}[/itex]? Unless somehow v(t) = dv?

    So, not exactly sure if I'm doing what you instructed correctly and also still wondering if it's possible to use a substitution, given the added information regarding the mass = 5kg and times = 2s and 4s and v(t) = 4t^3
     
    Last edited: Dec 3, 2015
  5. Dec 3, 2015 #4

    Ken G

    User Avatar
    Gold Member

    But you showed exactly the substitution above, and said that is not what you are asking. The substitution is v' dt = dv. The problem you had trouble making sense of is what to do with the limits of integration, but that's no problem, they just label the endpoints so you can manually shift from t labels to v labels, as you did.
    That holds if t1=0, which is what you implicitly assumed. If you want t1 = 2 s instead, you need to go back to your integration over F(t) and correctly insert that limit, which you didn't do above.
    v(t) = Delta v if you start at t=0. If you start at t=2 s, then v(t) = Delta v + v(2 s) = Delta v + 32.
    The only substitution that is ever needed is v' dt = dv. I think your problem is more around keeping track of how to use the limits of integration in the process of doing an integration, think about that part a bit more, it's the calculus there that is messing you up.
     
  6. Dec 3, 2015 #5
    For just a regular u-substitution,

    [itex]F(t) = 60t^{2} [/itex]

    [itex] \int^{t_{2}}_{t_{1}} F(t) dt [/itex] = [itex] \int^{4}_{2} 60(t)^{2} dt [/itex]

    in order to do u-substitution, don't we have to let the function inside the parentheses equal to a new variable?

    As in: Let u = t
    then, du = t'dt = dt
    that is, du = dt

    so, [itex] \int^{4}_{2} 60(t)^{2} dt [/itex] = [itex] \int^{u(4)}_{u(2)} 60(u)^{2} du [/itex] , right?

    So, in this case, we initially substituted the t, which already exists in the function, for a variable u and then did the derivative of u.

    But in the situation where I'm trying to substitute with a v (as opposed to a u).

    Shouldn't it be necessary to substitution with the variable that originally exists in the function, which is "t"? As in, let the inner function "t" equal to some other value, such that when I take the derivative of both sides, I end up with v' dt = dv.

    How come we are not setting t = to something?

    Instead, it looks as though you've made an initial substitution of:

    Let v = v(t)
    then, [itex] \frac{dv}{dt} = v'(t) [/itex]
    so that, dv = v'(t)dt

    Yes, v'(t)dt can be replaced by dv, but wouldn't the initial variable we started with, "v" have to exist somewhere in the function to begin with?
     
  7. Dec 3, 2015 #6
    Well maybe I'm starting to see it now.

    Perhaps it is not necessary to begin the substitution in any particular way.

    All that matters is that the substation be relevant to the initial function.

    If we set v = v(t)
    which is [itex] v = \int (a(t))dt = \int v'(t)dt = \int 12t^{2} = 4t^{3} [/itex]
    that is, [itex]v = 4t^{3}[/itex]

    so, [itex] dv = v'(t)dt = 12t^{2}dt[/itex]

    Therefore,

    [itex] m\int^{t_{2}}_{t_{1}} v'(t)dt = m\int^{t_{2}}_{t_{1}} 12t^{2}dt = m\int^{v(t_{2})}_{v(t_{1})}dv [/itex] = Δmv


    Thank you.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Impulse = ∫Force(t)dt = m∫dv = Δmv via u-substitution?
Loading...