Power difference when calculated from Force and Energy

[AGNuke]

Problem Statement:
We have a chain lying on the ground, with uniform linear density λ. It is held by one of its end and is pulled vertically upwards with a constant velocity, v₀. So calculate the following :-
1.)The Force in terms of the length (l) of the end pulled.
2.)Change in energy up to that point.
3.)Power


Solution. 1) By differentiating momentum with respect to time, we get the force applied.
$$F = \frac{dp}{dt} = \frac{d(mv)}{dt} = m\frac{dv}{dt} + v\frac{dm}{dt}$$
$$F = \lambda lg + \lambda v_0^2$$

Power, as calculated by the Force,
$$P = Fv_0$$
$$P = \lambda lgv_0 + \lambda v_0^3$$

2.) Kinetic Energy - Every infinitesimally small element of the chain off the ground is moving with same velocity v₀. ∴
$$K = \frac{1}{2}\lambda lv_0^2$$

Potential Energy, calculating the PE for infinitesimally small element of length dx at height x.
$$dU = \lambda dx.g.x$$
$$U = \int_{0}^{l}\lambda dx.g.x = \frac{1}{2} \lambda gl^2$$

Power from Energy,
$$P = \frac{dE}{dt} = \frac{dK}{dt} + \frac{dU}{dt}$$
$$P = \frac{1}{2}\lambda v_0^2\frac{dl}{dt} + \frac{1}{2}\lambda g\frac{dl^2}{dt}$$
$$P = \frac{1}{2}\lambda v_0^3 + \lambda glv_0$$

My question is, the power calculated from the energy is less than the power from Force. Any ideas what that imply?

 

[gneill]

In your calculation of the power via rate of change in energy, what does the $dl^2/dt$ represent? How does it manage to become $l v_o$ ?

For the rate of change of potential energy, consider that at some instant the weight of the hanging chain is $\lambda L g$ and its center of mass is rising at velocity $v_o$.

 

[AGNuke]

$$\frac{dl^2}{dt} = 2l\frac{dl}{dt} = 2lv_0$$

 

[rude man]

Problem Statement:
We have a chain lying on the ground, with uniform linear density λ. It is held by one of its end and is pulled vertically upwards with a constant velocity, v₀. So calculate the following :-
1.)The Force in terms of the length (l) of the end pulled.
2.)Change in energy up to that point.
3.)Power


Solution. 1) By differentiating momentum with respect to time, we get the force applied.
$$F = \frac{dp}{dt} = \frac{d(mv)}{dt} = m\frac{dv}{dt} + v\frac{dm}{dt}$$

The v in the second term has to be that of the c.g. What is the velocity of the c.g. in relation to v₀?

I don't know how you got your equation but with that correction it's right. I would have started with
∑forces = F - mg = dp/dt ... etc.

Using F = dp/dt for variable-mass systems is very tricky. I had to revert to my textbook to find the error myself.

 

[AGNuke]

$$C.M. = \frac{\int_0^l \lambda dx. x}{\int_0^l \lambda dx}$$
$$CM = \frac{\lambda \frac{l^2}{2}}{\lambda l}$$
$$CM = \frac{l}{2}$$
$$v = \frac{d(l/2)}{dt} = \frac{v_0}{2}$$

So, I see. The Power was coming incorrect from the Force. And regarding the use of F - mg, I'd remember it for the future.

Thanks.

 

[vela]

You might want to check out this thread about a similar problem:

https://www.physicsforums.com/showthread.php?t=729207

 

[rude man]

$$\frac{dl^2}{dt} = 2l\frac{dl}{dt} = 2lv_0$$

I believe the OP meant (dl/dt)^2 = v^2.

 

[rude man]

$$C.M. = \frac{\int_0^l \lambda dx. x}{\int_0^l \lambda dx}$$
$$CM = \frac{\lambda \frac{l^2}{2}}{\lambda l}$$
$$CM = \frac{l}{2}$$
$$v = \frac{d(l/2)}{dt} = \frac{v_0}{2}$$

So, I see. The Power was coming incorrect from the Force. And regarding the use of F - mg, I'd remember it for the future.

Thanks.

OK. BTW the basic equation for the variable-mass problem is

F_ext = m dv/dt + v dm/dt - u dm/dt
v = velocity of c.g. of "large" mass
u = velocity of c.g. of "expelled" mass

Typically, v is the velocity of a rocket and u is the velocity of the expelled matertial Of course, u and v have opposite signs in this case. Both u and v are w/r/t a reference inertial coordinate system. In your case, u = 0 for the part of the chain still on the ground, obviously.

You can kind of think of this problem in reverse: start with say a 2m long vertically suspended cable and drop it slowly to the ground. The vertical part of the chain is the "large" mass and the grounded part is the "expelled" mass. Of course, signs get changed appropriately.

 

[haruspex]

You might want to check out this thread about a similar problem:

https://www.physicsforums.com/showthread.php?t=729207

That's not for constant velocity. But the constant velocity problem has been posted on these forums a couple of times.
The trap is to assume that work is conserved. There's no basis for supposing that. Indeed, if you consider a small element of the chain as it lifts off the table it seemingly goes from rest to v0 instantly. This is a clue that we are dealing with inelastic impulses.
Conservation of momentum is the way to go, but the equation dp/dt = d(mv)/dt = m dv/dt + v dm/dt is not as solid as it looks. In the real world, closed systems do not change mass. The portion of the chain not on the table does not constitute a closed system.
Here, it happens to give the right answer because the portion of the chain that is on the table has no momentum in your reference frame. If you were to try using this equation from a reference frame moving upwards at speed v0 you would get the wrong answer.