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I have a doubt regarding finding Lagrangian of a body on an inclined plane.

Attached below are excerpts of 2 problems from a classical mechanics book that I am referring to.

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In the above question, while calculating potential energy V, the height of the centre of mass of sphere above the ground is taken into account.

However,

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In this case we see that potential energy V is simply height of body above horizontal level.

Why do we have different expression of V for these two problems?

I suppose I am lacking some concepts.

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Sharmi C
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  • Sharmi. If not mistaken: The first term is the height of the point of contact above ground level. You need the height of the centre of mass, centre of the sphere above ground level. See if you can.put it together(drawing). – Peter Szilas Mar 26 '20 at 08:43
  • Peter are you referring to the second question ? – Sharmi C Mar 26 '20 at 08:47
  • In the first question both height of body above ground plus height of Centre of mass above inclined is taken. But in 2nd question only height of body above ground is taken. Are you suggesting to add the height of center of mass in second question? – Sharmi C Mar 26 '20 at 08:50
  • Sharmi.Just looked at the first question only, seems ok. To repeat in the first question you need the height of the center of mass above ground level, which is the height of the point of contact plus a term. – Peter Szilas Mar 26 '20 at 09:00
  • Does That mean in the second question it is required to add height of centre of mass? – Sharmi C Mar 26 '20 at 09:16
  • They neglected the correction term in the second problem. I cannot see why (drawing). – Peter Szilas Mar 26 '20 at 09:21
  • This is precisely where I was confused. Thank you for clarifying. I guess this had nothing to do with it being a hoop instead of a solid sphere – Sharmi C Mar 26 '20 at 09:25
  • A good point. When neglecting one term against another term some quantities are smaller. The 2 terms here involve mg(l-x) sin and mga cos, cannot see why one is much smaller than the other(both for sphere and hoop). So I am with you, the hoop problem owes us an explanation. :) – Peter Szilas Mar 26 '20 at 09:41

1 Answers1

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The correct expression for the potential energy regarding the mass center is

$$ V(x) = \left((s-x)\sin\theta + a\cos\theta)\right)mg $$

but $a\cos\theta m g$ is constant along the movement then

$$ V(x) =m g (s-x)\sin\theta+C_1 = -m g x\sin\theta + C_2 $$

Cesareo
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  • Thank you for a cogent expression. It helps to explain both the problems – Sharmi C Mar 26 '20 at 09:51
  • I would like to clarify one more thing. Potential energy at top of incline means energy of body relative to that on ground. However on ground the mass centre of the body will be higher by its radius(since it never reaches the ground). So doesn't the vertical distance remain $((s-x)\sin\theta$ ? – Sharmi C Mar 26 '20 at 16:28