Using Integration?

Question

Using Integration, Calculate the Rotational Inertia of a uniform hoop of mass M and radius R about any daimeter
Hint: equation; I = (1/2)MR^2

Answer 1

The moment of inertia of a point mass m at a radius r is given by:

I = mr²

So the moment of inertia contributed by a small section of the hoop of length dℓ is λr² dℓ, where λ is the linear mass density of the hoop and r is the distance from the axis of rotation. So all we have to do is integrate:

[hoop]∫λr² dℓ

Let us view the hoop from above, with the axis of rotation aligned horizontally. For each point on the hoop, we can draw a ray through that point which makes a given angle θ with the axis of rotation. Therefore, we can parameterize the path around the hoop using the angle that ray makes with the axis of rotation. In this parameterization, dℓ = R dθ and the limits of integration are 0≤θ≤2π. Thus we have:

[0, 2π]∫λr² R dθ

Now note that r is NOT a constant, since it represents the distance of the point from the axis of rotation (which runs through the hoop), not the distance from the center of the hoop. However, since we have aligned the axis of rotation horizontally, the distance r is simply the vertical distance from the x-axis, which is |R sin θ|. Therefore we have:

[0, 2π]∫λ|R sin θ|² R dθ

Simplifying and integrating:

[0, 2π]∫λR³ sin² θ dθ
λR³ [0, 2π]∫sin² θ dθ (R is constant, as is λ since the hoop is uniform)
λR³ [0, 2π]∫(1 - cos (2θ))/2 dθ
λR³ [0, 2π]∫1/2 - cos (2θ)/2 dθ
λR³ (θ/2 - sin (2θ)/4)|[0, 2π]
πλR³

So all that remains is to find λ. The total mass of the hoop is M, the total length is 2πR, since the hoop is of uniform density, the linear mass density is simply the mass divided by the length, which is M/(2πR). So we have:

π(M/(2πR))R³
MR²/2

Which is the answer provided.

Answer 2

I would use polar coordinates. See diagram: http://i13.tinypic.com/8e1puvk.gif

The linear mass density ρ is M/(2πR)

An element ds of the hoop has length R dθ

The mass dm of that small piece is

ρ ds = [M/(2πR)] R dθ = [M/(2π)] dθ

The element ds travels in a circle r of radius R sin(θ)

The moment of inertia I is the integral of r² dm

So

I = integral from 0 to 2π of [R sin(θ)]² [M/(2π)] dθ
= (MR²/π) integral from 0 to 2π of sin²(θ) dθ

If you use the identity sin²(θ) = ½(1 - cos(2θ)), you can show that the integral is ½, so

I = ½ MR²

Answer 3

Its been a while, but I think you just need to integrate the equation from 0 to 2R dR

Answer 4

Who cares!