# Chapter 8 - Section 8.8 - Multiple-Choice Questions - Page 309: 4

The correct answer is: $(c) ~\frac{K_{rot}'}{K_{tr}'} \gt \frac{K_{rot}}{K_{tr}}$

#### Work Step by Step

Since the hole was drilled along the axis, the new shape has less mass located near the axis. For a cylinder, the rotational inertia is $\frac{1}{2}MR^2$ The new shape will have a rotational inertia $I'$ such that $I' \gt \frac{1}{2}MR^2$ Since the expression for the rotational inertia increases, a higher fraction of the energy at the bottom of the incline will be in the form of rotational kinetic energy. The correct answer is: $(c) ~\frac{K_{rot}'}{K_{tr}'} \gt \frac{K_{rot}}{K_{tr}}$

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