First slide

Rolling motion

Question

A thin-walled pipe rolls along the floor. What is the ratio of its translational kinetic energy to its rotational kinetic energy about the central axis parallel to its length?

Easy

A

1
B

$\frac{3}{2}$
C

$\frac{2}{3}$
D

$\frac{4}{3}$

Solution

The rotational kinetic energy is K = $\frac{1}{2} {Iω}^{2}$ , where I = ${mR}^{2}$ is its rotational inertia about the centre of mass the ratio is
$\frac{K_{transl}}{K_{rot}} = \frac{\frac{1}{2} {mv}_{com}^{2}}{\frac{1}{2} ({mR}^{2}) {(\frac{v_{com}}{R})}^{2}} = 1.00 .$

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