First slide

Rolling motion

Question

A ring is rolling on a surface without slipping. What is the ratio of its translational to rotational kinetic energies ?

Easy

A

5:7
B

2:5
C

2:7
D

1:1

Solution

Translational K.E. $= \frac{1}{2} {mv}^{2}$
Rotational K.E. $= \frac{1}{2} {Iω}^{2} = \frac{1}{2} ({mR}^{2}) ω^{2}$
$\begin{matrix} (∵ For a ring I = {mR}^{2}) \\ = \frac{1}{2} m (ωR)^{2} = \frac{1}{2} {mv}^{2} \end{matrix}$
So, translational K.E. = rotational K.E.
$∴$ translational K.E. : rotational K,E. = 1 :1

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