Separating angular and linear momentum

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  • #26
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I think I have some fundamental confusion here.

DocAl said:
Only two forces act on the rod: Gravity and the GRF. Since gravity acts vertically, only the horizontal component of the GRF creates the horizontal acceleration of the center of mass. That's why we're trying to find the acceleration of the center of mass.

I understand that net external forces are needed to produce a change in momentum (Newton's second). I run into difficulty with cause and effect. To me it seems that the falling rod's weight causes GRF so to count the effect of gravity and the effect of GRF seems like counting the same force twice.
To help me resolve this confusion (I know I am wrong), can you explain in words how the GRF comes into being?


DocAl said:
This won't work because you ignored one of the forces acting on the rod--the GRF also has a radial component. To use Newton's law you'd need the net radial force, not just the radial component of the weight.

OK, so we have the acceleration from weight, a = g cosθ and the force from GRF. The GRF is what I am trying to calculate, so I cannot find the net radial force like this.


DocAl said:
What I wanted you to do was calculate the radial acceleration kinematically. Are you familiar with the standard formulas for centripetal acceleration?

I've just had a look and I think F = m|v|^2 / R is the kinematic equation for centripetal force.

In the example, that gives F = |v|^2

Now we need to resolve v, so we can use the standard equation of motion v = u + at, but that would introduce the time variable into our equations which does not seem useful - I'm guessing there is a better approach?
 
  • #27
Doc Al
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I understand that net external forces are needed to produce a change in momentum (Newton's second). I run into difficulty with cause and effect. To me it seems that the falling rod's weight causes GRF so to count the effect of gravity and the effect of GRF seems like counting the same force twice.
To help me resolve this confusion (I know I am wrong), can you explain in words how the GRF comes into being?
It's certainly true that without gravity acting there would be no GRF. But they are two separate forces, and you must consider all forces when applying Newton's 2nd law.

Imagine yourself standing on the floor. If the only force on you were gravity, then you'd be accelerating downward at a = g. But the floor exerts a force on you to prevent you from pushing through it--that force is the GRF. If the floor is successful in stopping your potential motion, then your acceleration is zero and the GRF will equal your weight.

The case of the falling rod is a bit more complicated, since the rod is accelerating as it falls over. But we can calculate the acceleration and then use that to figure out the GRF via Newton's 2nd law.

OK, so we have the acceleration from weight, a = g cosθ and the force from GRF. The GRF is what I am trying to calculate, so I cannot find the net radial force like this.
The thing to do is calculate the acceleration directly, not via Newton's 2nd law.

I've just had a look and I think F = m|v|^2 / R is the kinematic equation for centripetal force.

In the example, that gives F = |v|^2
For this problem we need the centripetal acceleration, which is given by:
ac = v²/r = ω²r

Now we need to resolve v, so we can use the standard equation of motion v = u + at, but that would introduce the time variable into our equations which does not seem useful - I'm guessing there is a better approach?
Indeed there is. Use conservation of energy. See my hints in post #13.
 
  • #28
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Doc Al,

Applying conservation of energy, I suppose we can say that the energy of the system at the start is its gravitic potential energy. At any point during the fall, the total energy will still be the same magnitude, but will be made up of a component of reduced gravitic potential energy and a component of kinetic energy.

Can a solution for the net force be established using kinetic and remaining potential energy at a particular point during the fall?
 
  • #29
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Applying conservation of energy, I suppose we can say that the energy of the system at the start is its gravitic potential energy. At any point during the fall, the total energy will still be the same magnitude, but will be made up of a component of reduced gravitic potential energy and a component of kinetic energy.
That's right.
Can a solution for the net force be established using kinetic and remaining potential energy at a particular point during the fall?
Use conservation of energy to find the speed of the center of mass as a function of angle. (Use my earlier hints.) Then use that speed to figure out the radial acceleration. Since you already know how to find the tangential acceleration, you'll then have the full acceleration. Applying Newton's 2nd law will be the last step.
 
  • #30
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Finally got around to this again and I have worked out what may be a solution.

So, following your advice to find the velocity of the centre of mass as a function of angle:

We know that the energy E is constant such that the gravitic potential and kinetic are constant so E = mgh - 0.5mv^2

Before the rod falls, v is zero and h of COM = 1m, unit mass, so,

E = g

Therefore, g = mgh - 0.5mv^2
g = gh - 0.5v^2

Height can be expressed as a function of angle as h = rcosθ = cosθ, so,

g = gcosθ - 0.5v^2

and so the velocity as a function of angle is,

v^2 = 2g - 2gcosθ

If I am correct so far, the next step is to calculate the radial acceleration = v^2/r = v^2, so,

radial a = 2g - 2gcosθ

Tangential acceleration has already been calculated for the example as 0.75gsinθ

Total acceleration horizontally = tangential.cosθ - radial.cosθ

a = 0.75gsinθcosθ - cosθ(2g - 2gcosθ)

Unit mass so applying F= ma, horizontal force = 0.75gsinθcosθ - cosθ(2g - 2gcosθ)

Putting numbers into there gives results in line with what I would expect - is it right..?
 
  • #31
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Do I need to add more of my working to make it clearer to follow?
 
  • #32
Doc Al
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Finally got around to this again and I have worked out what may be a solution.

So, following your advice to find the velocity of the centre of mass as a function of angle:

We know that the energy E is constant such that the gravitic potential and kinetic are constant so E = mgh - 0.5mv^2

Before the rod falls, v is zero and h of COM = 1m, unit mass, so,

E = g

Therefore, g = mgh - 0.5mv^2
g = gh - 0.5v^2

Height can be expressed as a function of angle as h = rcosθ = cosθ, so,

g = gcosθ - 0.5v^2

and so the velocity as a function of angle is,

v^2 = 2g - 2gcosθ

If I am correct so far, the next step is to calculate the radial acceleration = v^2/r = v^2, so,

radial a = 2g - 2gcosθ

Tangential acceleration has already been calculated for the example as 0.75gsinθ
Looks good.

Total acceleration horizontally = tangential.cosθ - radial.cosθ
Almost. The horizontal component of the radial acceleration would be -radial.sinθ.


a = 0.75gsinθcosθ - cosθ(2g - 2gcosθ)

Unit mass so applying F= ma, horizontal force = 0.75gsinθcosθ - cosθ(2g - 2gcosθ)

Putting numbers into there gives results in line with what I would expect - is it right..?
Correct the above and you've got it.

Do I need to add more of my working to make it clearer to follow?
Nope. Just took me a while to get around to it. :smile:
 
  • #33
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Thanks Doc Al, appreciate your guidance.



horizontal force = 0.75gsinθcosθ - sinθ(2g - 2gcosθ)
 

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