- #1

PrudensOptimus

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Can someone distribute their knowledge of the above topics? I have a test on these tomorrow.

Thanks.

Thanks.

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- Thread starter PrudensOptimus
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- #1

PrudensOptimus

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Can someone distribute their knowledge of the above topics? I have a test on these tomorrow.

Thanks.

Thanks.

- #2

turin

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1) Integrate the dot product of the force with the position displacement over the path. This has a simplification that comes up in first semester physics:

2) Calculate the change in mechanical energy. Usually this will be simpler, so, if you are given a bunch of energy info, then you probably want to use this one.

To calculate

The rate of change of

Impluse can be calculated in two ways (notice the analogy to work):

1) It is the integral of the force (as a function of time) over time. This sometimes simplifies to

2) Calculate the change in momentum:

To help you organize:

as

That is not to say that whenever you have work or impulse, that energy and momentum are not conserved. If you consider work or impulse inside the system, then energy and momentum are conserved. If you consider work and energy as acting on the system from outside, then energy and momentum can break conservation in the system.

If we consider a mass falling due to gravity, then we can say that gravity is doing work on the object because it is applying a downward force to the object, and the object is moving downward. The force is mg, the distance it travels is y, therefore the work done on the object is mgy (Fd). This is manifest in the fact that the object is speeding up as it falls (and thus gaining kinetic energy). Now, consider the same situation, but include the gravitational potential in the system. In this view, the increase in kinetic energy is due to the decrease in gravitational potential energy. Thus, the energy is conserved. In the first point of view, we treated gravity as external to the system. In the second point of view, we included the potential energy of gravity as part of the system.

This one is a bit more tricky. I will use a bit more obscure example, because I can't think of a more reasonable one. A ball hits a wall and bounces back. Its velocity has changed, and therefore its momentum has changed. Does this violate the conservation of momentum. No. Actually, the wall recoiled ever so slightly, but, since it is rigidly connected to the earth, and the Earth is so huge, then the recoil was negligible. So, the change in mometum of the ball was mV, and the change in the momentum of the Earth was Mv, and mV must equal Mv, but, since M >> m, then v << V.

Another example for momentum would be the falling object again. Its downward momentum increases because the force of gravity puts an impulse on it of Ft = mgt. Thus, it seems like the impulse breaks the conservation of momentum. But actually, the Earth's momentum is again the compensation. The Earth moves upward to meet the object as it falls. The impulse on the Earth is -mgt. Since M >> m, then, again, the change in velocity isn't noticible.

- #3

turin

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I forgot to mention rotational stuff:

**angular moment of inertia** and **angular velocity** is to **angular momentum** is to **torque**

as

**mass** and **velocity** is to **momentum** is to **force**

(Usually, when you just hear "momentum" and "velocity" without a pre-qualifier, that refers to the linear version)

Symbolically:

Iw : L : τ :: mv : p : F

Angular momentum is conserved, separately from linear momentum (in the absence of a torque applied from outside the system).

Don't forget your centripetal acceleration equations:

**a_c = m (v^2) / r = m (w^2) r**

as

(Usually, when you just hear "momentum" and "velocity" without a pre-qualifier, that refers to the linear version)

Symbolically:

Iw : L : τ :: mv : p : F

Angular momentum is conserved, separately from linear momentum (in the absence of a torque applied from outside the system).

Don't forget your centripetal acceleration equations:

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- #4

PrudensOptimus

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I tried to integrate Work, ∫ΣF dx, where x is displacement, replacing F with ma, I end up getting δKE∫1/dx dx. Does ∫1/dx dx give you ln|x| + C, or something else? I'm not sure. And what is the bottom and top index of the Work Integral? Is it the initial position and the final posistion?

- #5

PrudensOptimus

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I saw some questions where W = ΔKE + ΔPE.

Or was that just for non conservative works? If it is for non conservative works only, then that means in cases when W = ΔKE, the W must mean for conservative?

- #6

PrudensOptimus

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"It must be emphasized that all the forces acting on a body must be included in equation 6-10 either in the potential energy term on the right (if it is a conservative force), or in the work term W

EQ 6-10 : W

- #7

turin

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I'm assuming that you mean you tried to use the integrate idea to find work. Don't forget that force is a vector, so it is a dot product of the force with the displacement. Also, one subtle issue, work is <i>usually</i> associated with a particular force (i.e. friction), not the resultant, so you usually will not have that summation in the integrand. I'm really sorry, but I don't quite follow what you are trying to do.Originally posted by PrudensOptimus

I tried to integrate Work, ∫ΣF dx, where x is displacement, replacing F with ma, I end up getting δKE∫1/dx dx.

Is this what you were trying to do?

∫Fdx = ∫madx = m∫(dv/dt)dx = m∫(dv/dx)(dx/dt)dx

= m∫(dv/dx)vdx = m∫vdv = (m/2)[v

- #8

turin

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You probably use this notion when you are considering theOriginally posted by PrudensOptimus

_{net}= ΔKE?

I saw some questions where W = ΔKE + ΔPE.

Or was that just for non conservative works? If it is for non conservative works only, then that means in cases when W = ΔKE, the W must mean for conservative?

Also, for instance,

- #9

turin

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It looks like they are using the subscript, "NC," to indicate that they are strictly talking about "Non-Conservative" work. That means that there is no potential energy that you can associate with it. Friction, I'm pretty sure, will contribute to the WOriginally posted by PrudensOptimus

"It must be emphasized that all the forces acting on a body must be included in equation 6-10 either in the potential energy term on the right (if it is a conservative force), or in the work term W_{NC}, on the left(but not in both!)"

EQ 6-10 : W_{NC}= ΔKE + Δ PE.

It's a little tricky, though, because it depends on how you define your system. I would say that it really doesn't matter, as long as you stick with the same defined system through your calculations.

The first law of thermodynamics says that ΔE = Q - W, where E is the energy in the system (accounting for ALL forms), Q is the heat transferred into the system, and W is the work done by the system. Maybe you will see the first law written as ΔE = W - Q. In this case, you just change the directions of W and Q. The sign just tells you which way the work and heat is going (into the system or out of the system). If you've seen this thermo stuff, then that W

I changed some other stupid statements as well. I hope I didn't screw you up on your test.

Oh hell, I don't really know what to say about this. You would probably do best to ignore the thermodynamics part; I'm confusing myself with it.

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- #10

PrudensOptimus

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Originally posted by turin

I'm assuming that you mean you tried to use the integrate idea to find work. Don't forget that force is a vector, so it is a dot product of the force with the displacement. Also, one subtle issue, work is <i>usually</i> associated with a particular force (i.e. friction), not the resultant, so you usually will not have that summation in the integrand. I'm really sorry, but I don't quite follow what you are trying to do.

Is this what you were trying to do?

∫Fdx = ∫madx = m∫(dv/dt)dx = m∫(dv/dx)(dx/dt)dx

= m∫(dv/dx)vdx = m∫vdv = (m/2)[v^{2}- v_{0}^{2}] = KE - KE_{0}= ΔKE

When I replaced F with ma, and a with (v^2 - v

- #11

PrudensOptimus

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mç(dv/dx)vdx --- how did you get to that part?

- #12

turin

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F is the integrand, and it can be a function of the position of the particle, in general. If you replace it by ma, then the a must be a function of the position of the particle, in general. You cannot replace this by a constant, in general.Originally posted by PrudensOptimus

When I replaced F with ma, and a with (v^2 - v_{0}^2)/2Δx, isn't m((v^2 - v_{0}^2)/2) suppose to be treated as constants?

- #13

turin

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This is the chain rule in action:Originally posted by PrudensOptimus

mç(dv/dx)vdx --- how did you get to that part?

Velocity, at a particular instant of time, has a value. You can make a convenient rule for this value which will accept, as its input, the time, and will give, as its output, the value of the velocity. This rule is the time derivative of the position of the particle, evaluated at the specific time input into the rule. In other words, this is the function, f(t):

v = f(t)

Velocity, at a particular point in space, has a value (because we're talking about a trajectory). I couldn't tell you off the top of my head what exactly that rule is, or where it comes from, without mentioning energy, but let's see what happens when we differentiate this rule (the function, g(x)) with respect to time:

v = g(x)

Now for the chain rule:

d[g(x)]/dt = d[g(x(t))]/dt

Here, I have just explicated the time dependence. Now, using the chain rule:

= g'(x(t))(dx/dt) = (dg/dx)(dx/dt)

But, going back to the first rule for assigning a value to v, you can see that:

(dx/dt) = v

so that:

d[g(x)]/dt = (dg/dx)v

But, g(x) also equals v at this point on the trajectory, so:

d[v]/dt = (dv/dx)v

But,

dv/dt = a

Thus,

a = v(dv/dx).

This is usually done a bit sloppily, referring to the velocity itself as an inherrent function of position and time. I think that doing so causes confusion. It is just the chain rule, but try to remember that we're talking about functions that return a value that is v, not v as a function in itself. After having said this, my notation is probably a bit sloppy. To try to make the point a bit more clear:

We usually talk about v as a function of t, and we denote it:

v = v(t)

This is the f(t).

What we want to end up with is an expression with a, v, and x, without t hanging around and bothering us. What we do in terms of the math is to start with an integral of a function of t over the variable x. This is messy. We could express x in terms of t, but we choose t in terms of x. We use the chain rule to get the time derivative of f(t) in terms of a function, g(x), and its derivative. So, I should probably give it to you like this:

h(x) = g(x)(dg/dx)

where, h(x) gives the acceleration at the point, x:

a = h(x)

and g(x) gives the velocity at the point x

v = g(x).

So:

df/dt = g(x)(dg/dx).

It should be emphasised that this does not generally happen with the chain rule that the function just kind of pops out of the derivative. This occurs in this case because of the relationship between the two choices of independent variable that we're playing around with to each other.

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