The moment of inertia is a measure of how much torque is needed for a desired angular acceleration about a rotational axis. For a hollow cylinder, we can derive the moment of inertia step by step as follows:
Step 1: Understanding the Geometry
Consider a hollow cylinder with inner radius r1, outer radius r2, height h, and uniform mass distribution.
Step 2: Element of Mass
To derive the moment of inertia, we first consider a thin cylindrical shell at a radius r with thickness dr. The height of this element is h.
The volume of a thin shell is given by:
dV = 2πr dr h
The mass dm of this element can be expressed in terms of its volume and density ρ as:
dm = ρ dV = ρ (2πr dr h)
Step 3: Moment of Inertia of the Element
The moment of inertia of this thin shell about the central axis is given by:
dI = r2 dm
Substituting the expression for dm, we have:
dI = r2 (ρ (2πr dr h)) = 2πρhr3 dr
Step 4: Integrating to Find Total Moment of Inertia
The total moment of inertia I for the hollow cylinder can be found by integrating dI from r1 to r2:
I = ∫(from r1 to r2) 2πρhr3 dr
Step 5: Solving the Integral
Calculating the integral:
I = 2πρh [ (r4/4) ] (from r1 to r2)
This becomes:
I = 2πρh [ (r24/4) - (r14/4) ]
So, we can factor out the 2πh/4:
I = (πh/2) ρ (r24 - r14)
Step 6: Relating Mass to Density
Next, we relate the density and the total mass of the hollow cylinder. The mass M can be expressed as:
M = ρ (Volume) = ρ (πh (r22 - r12))
Thus, we can find ρ as:
ρ = M / (πh (r22 - r12))
Step 7: Final Moment of Inertia Expression
Substituting this back into the equation for I:
I = (M / (πh (r22 - r12))) * (πh / 2) (r24 - r14)
The πh cancels out, leading to:
I = (1/2) M ( (r22 + r12) )
Conclusion
Thus, the moment of inertia of a hollow cylinder about its central axis is given by:
I = (1/2) M (r22 + r12)