 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2012/h/h2/d/solution

Choose the

z

-axis along the cylinder and the origin at its midpoint, so

x^{2} + y^{2} \leq b^{2}

and

- a \leq z \leq a

. Reflection symmetry puts the centroid at the origin and makes all products of inertia zero. The mass is

m = 2 π ρ a b^{2} .

Cylindrical integration gives

\int_{A} ρ x^{2} d V = \int_{A} ρ y^{2} d V = m b^{2} /4

, while the uniform height coordinate has

\int_{A} ρ z^{2} d V = m a^{2} /3

. Consequently

J_{G} = m diag (\frac{b ^{2}}{4} + \frac{a ^{2}}{3}, \frac{b ^{2}}{4} + \frac{a ^{2}}{3}, \frac{b ^{2}}{2}) .

The two equal transverse entries reflect the rotational symmetry about the cylinder axis.

 New to topics? Read the docs here!