Question
A disk of radius R and mass m has a hole in the center of radius r. It is spun around its center point, much like a CD. Use calculus to derive an expression for the moment of inertia of the disk/ring.
Answer
Consider a thin ring of width dx at a distance x from the center of the ring, r < x < R.
The moment of inertia of this thin ring is
dI = (2πx*t*ρ)*x^2 dx, where, t = thickness and ρ = density
Integrating,
Moment of inertia, I
= (2π*t*ρ)* [x^4/4] (x = r to R)
= 0.5πρt(R^4 - r4).
Answer in terms of mass, m = πρt(R^2 - r^2)
=> moment of inertia
I = 0.5 m (R^4 - r^4) / (R^2 - r^2)
= 0.5m (R^2 + r^2)
Alternate method:
Using the formula for moment of inertia of the disc,
I = (1/2)m1*R^2 - (1/2)m2*r^2
where
m1 = mass of the disc having radius R = πρt*R^2 and
m2 = mass of the disc having radius r = πρt*r^2
=> m = m1 - m2 = 0.5πρt(R^2 - r^2)
=> I = 0.5 πρt (R^4 - r^4), same as above
This is then coverted in terms of mass of the annular ring, m as
I = 0.5 m (R^2 + r^2).
A disk of radius R and mass m has a hole in the center of radius r. It is spun around its center point, much like a CD. Use calculus to derive an expression for the moment of inertia of the disk/ring.
Answer
Consider a thin ring of width dx at a distance x from the center of the ring, r < x < R.
The moment of inertia of this thin ring is
dI = (2πx*t*ρ)*x^2 dx, where, t = thickness and ρ = density
Integrating,
Moment of inertia, I
= (2π*t*ρ)* [x^4/4] (x = r to R)
= 0.5πρt(R^4 - r4).
Answer in terms of mass, m = πρt(R^2 - r^2)
=> moment of inertia
I = 0.5 m (R^4 - r^4) / (R^2 - r^2)
= 0.5m (R^2 + r^2)
Alternate method:
Using the formula for moment of inertia of the disc,
I = (1/2)m1*R^2 - (1/2)m2*r^2
where
m1 = mass of the disc having radius R = πρt*R^2 and
m2 = mass of the disc having radius r = πρt*r^2
=> m = m1 - m2 = 0.5πρt(R^2 - r^2)
=> I = 0.5 πρt (R^4 - r^4), same as above
This is then coverted in terms of mass of the annular ring, m as
I = 0.5 m (R^2 + r^2).
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