- #1

- 28

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For a specific question I came up with this

F=ma

Integral (m * da)

da=dv/t

Integral m * dv/t

(m/t) * dv

(m/t) * v

So

F = (m/t) * v

Does this seem right? Or I'm I making illegal moves. Help.

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- I
- Thread starter Dynamics101
- Start date

- #1

- 28

- 4

For a specific question I came up with this

F=ma

Integral (m * da)

da=dv/t

Integral m * dv/t

(m/t) * dv

(m/t) * v

So

F = (m/t) * v

Does this seem right? Or I'm I making illegal moves. Help.

- #2

Chandra Prayaga

Science Advisor

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So I do not know what is meant by the second line: integral (m * da).

If you don't know the acceleration as a function of time, but know the velocity as a function of time, then the acceleration can be obtained by

- #3

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- 4

- #4

- 28

- 4

Sorry I'm not very good at this.

So I do not know what is meant by the second line: integral (m * da).

If you don't know the acceleration as a function of time, but know the velocity as a function of time, then the acceleration can be obtained bydifferentiatingthe velocity, since a = dv/dt. So the third line, da = dv/t is not correct, and integrating that has no significance.

- #5

jbriggs444

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The manipulations you are making are both invalid and poorly motivated. Instead of writing a sequence of formulas down, you need to write a sequence of equations down and explain how each one follows from the ones.If force isn't constant and you want to find the force of something at a specific time, you'd integrate right?

For a specific question I came up with this.

This is a good starting point. An equation that is valid for constant acceleration.F=ma

This expression comes out of the blue and is incorrect. What is it that you are trying to write down here? Do you, for instance, want to write down the formula for the momentum transferred by a non-constant force acting over a period of time? If so, where is the time variable that should appear in the formula?Integral (m * da)

- #6

Chandra Prayaga

Science Advisor

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That will give you the acceleration. Multiplying that with mass gives the sum of all the forces acting on the body.

- #7

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I wrote the da to indicate the quantity that is changing. [emoji26]The manipulations you are making are both invalid and poorly motivated. Instead of writing a sequence of formulas down, you need to write a sequence of equations down and explain how each one follows from the ones.

This is a good starting point. An equation that is valid for constant acceleration.

This expression comes out of the blue and is incorrect. What is it that you are trying to write down here? Do you, for instance, want to write down the formula for the momentum transferred by a non-constant force acting over a period of time? If so, where is the time variable that should appear in the formula?

- #8

jbriggs444

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But that's not how the "d" symbol is conventionally used. Normally you would consider acceleration as a function of some other parameter such as time. Then you would denote the integral of "acceleration over time" as ##\int a(t) \, dt##I wrote the da to indicate the quantity that is changing. [emoji26]

- #9

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So writing acceleration as that tell me that acceleration is changing over time? If I want to integrate something which is not constant I need to consider that something as a function of another thing?But that's not how the "d" symbol is conventionally used. Normally you would consider acceleration as a function of some other parameter such as time. Then you would denote the integral of "acceleration over time" as ##\int a(t) \, dt##

Exammple, if I want to find the kinetic energy of something (a spinning disk), would we write velocity in terms of distance from the center of the disk? Since eery small area on the disk has a different velocity.

Lol sorry I'm new to calculus infused physics

- #10

nasu

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Yes, you will need the velocity as a function of radius and then integrate over the radius.

- #11

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- #12

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So when you specify the variable of integration, you can then integrate. Thank You [emoji5]. Last question, just so I understand this, let's say I want to find the area of a disk, I would integrate area with respect to radius? A(r) ?

- #13

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So when you specify the variable of integration, you can then integrate. Thank You [emoji5]. Last question, just so I understand this, let's say I want to find the area of a disk, I would integrate area with respect to radius? A(r) ?

In that case the variable of integration is implied, but for the force, a force can be a function of time , or a function of spatial coordinates (as for example ##F=-kx##) or force is always function of acceleration, so you have to say if you want to integrate force w.r.t to acceleration to get the integral ##\int mada=\frac{1}{2}ma^2##, though not sure if this last integral has some physical meaning.

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- #15

jbriggs444

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let's say I want to find the area of a disk, I would integrate area with respect to radius? A(r) ?

I disagree with this. The variable of integration is not implied. There is no rule that requires us to find the area of the disk as the sum of the infinitesimal areas of a bunch of concentric rings. We could equally well find it as the sum of the infinitesimal areas of a bunch of side-by-side isosceles triangles with each of their tips at the center of the disk. Or we could find it as the sum of the area of a bunch of side by side vertical strips running from top to bottom of the disk.In that case the variable of integration is implied

Let us work the problem. We have a disk of radius r and we want to compute its total area.

First the concentric ring approach. We want to add up the areas of a bunch of thin rings. Each ring is at a radius x from the center of the disk and is of infinitesimal width, "dx". If we unroll one of these rings to a rectangle, its area is its length (##2 \pi x##) times its width (##dx##). So our formula is

[tex]A = \int_0^r 2 \pi x \, dx[/tex]

The ##2pi## term is a constant. We can take that out of the integrand and get

[tex]A = 2 \pi \int_0^r x \, dx[/tex]

That integral is easy to evaluate. ##\int x \, dx = \frac{x^2}{2}##. We are evaluating a definite integral (the bottom and top of the evaluation range are specified). The limits of integration are 0 and r, so the formula becomes:

[tex]A = 2 \pi ( \frac{r^2}{2} - \frac{0^2}{2} )[/tex]

This obviously reduces to

[tex]A = \pi r^2[/tex]

Now the triangle approach. We want to add up the areas of a bunch of narrow isosceles triangles. Each triangle has one vertex at the center of the disk and its other two vertices on the rim of the disk separated by an infinitesimal distance dx and at a distance x from an arbitrary starting point. [Think about measuring x along the rim of the disk with a flexible measuring tape]. Each of these triangles has an area of one half base (##dx##) times height (r). We want to add up triangles until the distance measured on the rim gets back to the starting point, a distance ##2 \pi r## away. So our formula is

[tex]A = \int_0^{2 \pi r}\frac{r}{2} \, dx[/tex]

Our variable of integration is x. ##\frac{r}{2}## does not depend on x. It behaves like a constant. We can pull it out of the integrand and get

[tex]A = \frac{r}{2} \int_0^{2 \pi r}1 \, dx[/tex]

That's a trivial integral. It reduces to the difference between the upper limit of integration and the lower. So our formula becomes

[tex]A = \frac{r}{2} ( 2 \pi r - 0 )[/tex]

Which simplifies to

[tex]A = \pi r^2[/tex]

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