Moment of inertia of a disc using rods as differential elements

In summary: The OP got the right answer as did you so the algebra is correct.In summary, the conversation discusses different methods for computing the moment of inertia of a disc using rods as differential elements. The correct method involves using rings as differential elements and leads to a simpler integral. However, the original homework specifically requires the use of rods, making the exercise more challenging.
  • #1
Hamiltonian
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Homework Statement
find the moment of inertia of a disc using rods of variable lengths as differential element.(axis perpendicular to plane of disc and through its COM)
Relevant Equations
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I know there are more convenient differential elements that can be chosen to compute the moment of inertia of a disc(like rings).

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the mass of the differential element:
$$dm = (M/\pi R^2) (dA) = (M/ \pi R^2) (2\sqrt{R^2 - y^2})(dy)$$
the moment of inertia of a rod through its COM is ##(1/12)ML^2## hency by applying the parallel axis theorem :
$$dI = (1/12)(dm)(2\sqrt{R^2 - y^2})^2 + (dm)y^2$$
$$I = (\frac{2M}{3\pi R^2})\int_{-R}^{+R} (R^2 + 2y^2)\sqrt{R^2 - y^2}(dy) $$
$$I = (5/12)MR^2$$
where am I going wrong?
 
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  • #2
Looks like you set it up correctly. Are you sure you evaluated the integral correctly?
 
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  • #3
I agree with @TSny , the integral is not correctly performed.
 
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  • #4
TSny said:
Looks like you set it up correctly. Are you sure you evaluated the integral correctly?
$$I = (\frac{2M}{3\pi R^2})\int_{-R}^{+R} (R^2 + 2y^2)\sqrt{R^2 - y^2}(dy) $$
$$I = (\frac{2M}{3\pi R^2})\int_{-R}^{+R} (R^2\sqrt{R^2 - y^2} + 2y^2\sqrt{R^2 - y^2} ) (dy) $$

$$I = \frac{2M}{3\pi R^2} [R^2 (\frac{\pi R^2}{2}) + 2(\frac{\pi R^4}{8})] = (1/2)MR^2 $$
i used wolfram alpha to do it now i am getting the correct answer thanks!
 
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  • #5
Hamiltonian299792458 said:
$$I = (\frac{2M}{3\pi R^2})\int_{-R}^{+R} (R^2 + 2y^2)\sqrt{R^2 - y^2}(dy) $$
$$I = (\frac{2M}{3\pi R^2})\int_{-R}^{+R} (R^2\sqrt{R^2 - y^2} + 2y^2\sqrt{R^2 - y^2} ) (dy) $$

$$I = \frac{2M}{3\pi R^2} [R^2 (\frac{\pi R^2}{2}) + 2(\frac{\pi R^4}{8})] = (1/2)MR^2 $$
i used wolfram alpha to do it now i am getting the correct answer thanks!
Note that solving the integral involves a trig substitution which effectively converts it to having chosen the more sensible differential element in the first place. Is that a useful exercise?
Perhaps it is a lesson that when a trig substitution helps with an integral one should consider what it means in terms of the original analysis.
 
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  • #6
haruspex said:
Note that solving the integral involves a trig substitution which effectively converts it to having chosen the more sensible differential element in the first place. Is that a useful exercise?
Perhaps it is a lesson that when a trig substitution helps with an integral one should consider what it means in terms of the original analysis.
The original homework statement specifically requires use of "rods of variable lengths as differential element".

Is there a better approach than that used by the OP? Am I missing something obvious?

Edited - typo'.
 
  • #7
Steve4Physics said:
The original homework statement specifically requires use of "rods of variable lengths as differential element".

Is there a better approach than that used by the OP? Am I missing something obvious?

Edited - typo'.
you could use the differentiable elements as rings.

$$dm = \frac{M}{\pi R^2} 2\pi x dx$$
the moment of inertia of a ring is ##= MR^2##
$$dI = (dm)x^2$$
$$I = \int_0^R \frac{M}{\pi R^2} 2\pi x^3 dx$$
$$I = (1/2)MR^2$$
the resulting integral is much simpler in this case
 
  • #8
Hamiltonian299792458 said:
you could use the differentiable elements as rings.

$$dm = \frac{M}{\pi R^2} 2\pi x dx$$
the moment of inertia of a ring is ##= MR^2##
$$dI = (dm)x^2$$
$$I = \int_0^R \frac{M}{\pi R^2} 2\pi x^3 dx$$
$$I = (1/2)MR^2$$
the resulting integral is much simpler in this case
Yes, I know! (And in Post #1, you'd already mentioned that you realized this.)

But the original homework asks for an analysis in terms of 'rods'. Whoever set the homework specifically didn't want you to use rings - presumably to make the exercise a little more demanding.

I was pointing out that your method is the only one (as faras I can see) that properly answers the original question. So don't be tempted to hand-in a solution based on the ring method, except perhaps as an addendum to demonstrate that you know that this is better approach.
 
  • #9
Steve4Physics said:
The original homework statement specifically requires use of "rods of variable lengths as differential element".

Is there a better approach than that used by the OP? Am I missing something obvious?

Edited - typo'.
I'm sure the OP approach satisfies the given requirement in the way the setter expected. Just pointing out that the algebra can't help but switch to the standard route under the covers. I don't think there's a way to avoid it.
 
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FAQ: Moment of inertia of a disc using rods as differential elements

What is the moment of inertia of a disc using rods as differential elements?

The moment of inertia of a disc using rods as differential elements is a measure of the disc's resistance to changes in its rotational motion. It is calculated by adding up the contributions of each tiny rod that makes up the disc, taking into account their mass, distance from the axis of rotation, and orientation.

How is the moment of inertia of a disc using rods as differential elements calculated?

The moment of inertia of a disc using rods as differential elements is calculated using the integral of the square of the distance from the axis of rotation to each differential element (in this case, the rods) multiplied by the mass of each element. This integral is then summed up over the entire disc to get the total moment of inertia.

What is the difference between using rods as differential elements and using thin circular rings?

Using rods as differential elements is a more accurate method of calculating the moment of inertia for a disc compared to using thin circular rings. This is because rods take into account the varying mass distribution of the disc, whereas thin rings assume a uniform mass distribution.

Can the moment of inertia of a disc using rods as differential elements be negative?

No, the moment of inertia of a disc using rods as differential elements cannot be negative. This is because the distance from the axis of rotation is squared in the calculation, so even if the mass of the rod is negative, the moment of inertia will always be positive.

What is the physical significance of the moment of inertia of a disc using rods as differential elements?

The moment of inertia of a disc using rods as differential elements has physical significance in understanding the disc's rotational motion. It determines how much torque is needed to accelerate or decelerate the disc, and also affects the disc's stability and ability to maintain its rotational motion.

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