 Solution

ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/section-b/8/a/solution

The central rod occupies the holes in the discs, so the disc masses must be annular rather than counted as solid discs overlapping the rod. Define

C = L_{1} ρ_{1} = 2 L_{2} ρ_{2}

. Then

M = π r_{1}^{2} L_{1} ρ_{1} + 2 π (r_{2}^{2} - r_{1}^{2}) L_{2} ρ_{2} = π C r_{2}^{2} .

I = \frac{π}{2} L_{1} ρ_{1} r_{1}^{4} + π L_{2} ρ_{2} (r_{2}^{4} - r_{1}^{4}) = \frac{π}{2} C r_{2}^{4} .

Thus $I = \frac{1}{2} M r_{2}^{2}$ , exactly the axial inertia of a uniform disc of outer radius

r_{2}

and mass

M

. The cancellation uses the given density–length relation.

 New to topics? Read the docs here!