Moment of inertia of hollow cylinder, axis orthogonal to length

In summary, The speaker is working through the Feynman lectures on physics and trying to calculate the moment of inertia of a cylinder using the parallel axis theorem and the result for a rod. They initially make a mistake by considering the cylinder as a hollow one instead of a solid one, but after adjusting their approach and considering a partition into concentric hollow cylinders, they are able to arrive at the correct answer listed in table 19-2.
  • #1
jds17
5
0
Hi, I am working through the Feynman lectures on physics and trying to calculate the moment of inertia stated in the title.
(the taxis of rotation going through c.m., orthogonal to length).
My approach is to slice the cylinder into thin rods along the length, using the parallel taxis theorem and the result for a rod.
Unfortunately, I get as result: I = M ( L^2 / 12 + r^2 / 2). I.e. the last numerator comes out as 2 instead of 4, as stated in section
19-2. The corresponding expression comes from summing up dm sum( z_i ^ 2), where dm is the mass of a single rod and z_i
the height of the rod's center of inertia. Perhaps my mistake lies in handling the 2-dim slices as 3-dim rods?
 
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  • #2
I would slice the cylinder into rings instead. It makes integration far easier.
 
  • #3
jds17 said:
My approach is to slice the cylinder into thin rods along the length, using the parallel taxis theorem and the resultat for a rod.
Unfortunately, I get as result. I = M ( L^2 / 12 + r^2 / 2). I.e. the last numerator comes out as 2 instead of 4, as stated in section
19-4.
Your method looks OK to me. Feynman lists (in table 19-2) the moment of inertia of a solid cylinder.
 
  • #4
@Doc Al: Thank you for your reply, I took the cylinder as a hollow one, and this seems to be my mistake. I will try
to do the calculation again for the solid cylinder as soon as I get back home.

@K^2: thank you, too, but I wanted to find out what was wrong with my thinking instead of doing a different
calculation. I will try yours, too, although it seemed more complicated when I first considered it
 
  • #5
Hi, everything turned out nicely, considering a partition into concentric hollow cylinders, adding their M.I.s (calculated as before) up and going to the limit gives the answer in table 19-2!
 
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1. What is the moment of inertia of a hollow cylinder with its axis perpendicular to its length?

The moment of inertia of a hollow cylinder with its axis perpendicular to its length is given by the formula: I = MR2, where M is the mass of the cylinder and R is the radius of the cylinder.

2. How does the moment of inertia change if the radius of the hollow cylinder is doubled?

If the radius of the hollow cylinder is doubled, the moment of inertia also doubles. This is because the moment of inertia is directly proportional to the square of the radius.

3. What is the difference between the moment of inertia of a solid cylinder and a hollow cylinder with the same mass and dimensions?

The moment of inertia of a solid cylinder is greater than that of a hollow cylinder with the same mass and dimensions. This is because the mass of a solid cylinder is distributed further from the axis of rotation, resulting in a larger moment of inertia.

4. How does the moment of inertia of a hollow cylinder change if its mass is tripled?

If the mass of a hollow cylinder is tripled, the moment of inertia also triples. This is because the moment of inertia is directly proportional to the mass of the object.

5. Can the moment of inertia of a hollow cylinder be negative?

No, the moment of inertia of a hollow cylinder cannot be negative. It is always a positive quantity as it is a measure of an object's resistance to changes in its rotational motion.

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