 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2017/prelims/mathematics-iv-resit/6/c/solution

a = 0

\overset{x}{¨}_{solid} = \frac{2}{3} g sin φ .

For the hollow cylinder,

a > 0

makes the denominator larger:

\overset{x}{¨}_{hollow} = \frac{2 b ^{2}}{3 b ^{2} + a ^{2}} g sin φ < \frac{2}{3} g sin φ

on a downward incline with

sin φ > 0

. Thus the solid cylinder accelerates faster, assuming rolling without slipping for both. More of the hollow cylinder's mass lies far from the rotation axis, increasing its moment of inertia relative to

M b^{2}

 New to topics? Read the docs here!