Question

It is given that M = 185 kg and R = 1.65 m. What if an inner solid cylinder is cut out from it, with a radius of (1/2)R?

a) What is the moment of inertia around the central axis (as in the example) of this inner cylinder?

b) What is the moment of inertia of the remaining hollow cylinder around the same axis? (Notice that all answers are independent of the length L.)

Answer 1

Answer:

Let us go to the basics first.

Dimensions of inner solid cylinder:

Since density(ρ) of cylinders will be same,

so, ρ_{original cylinder} = ρ_{inner solid cylinder}

=> M / V = m / v (since, density = mass / volume)

=> M / m = V / v

=> m / M = v / V

=> m / M = πr²l / πR²L

(l = L , lengths of the cylinders will be same as inner cylinder is cut from the Original cylinder)

=> m / M = r² / R²

=> m = Mr² / R²

=> m = 185* (0.5R)² / R² [since, r = (1/2)R = 0.5R given]

=> m = 185* (0.5)²

=> m = 46.25 kg (mass of inner solid cylinder)

Thus, the moment of inertia around the central axis of this inner cylinder = (1/2) mr²

= (1/2) * 46.25 *0.825² (since, r = 0.5R = 0.5*1.65 = 0.825 m )

= 15.7394 kgm² (Answer a)

Now, moment of inertia of the remaining hollow cylinder around the same axis:

Remaining mass, M^' = 185 - 46.25 = 138.75 kg

R = 1.65 m

r = 0.825 m

We knw that moment of inertia of hollow cylinder is given by:

I = (1/2)M' (R² + r²)

=> I = (1/2)*138.75*(1.65² + 0.825²) = 236.0917 kgm² (Answer b)