First slide

Dynamics of rotational motion about fixed axis

Question

A circular disc rolls down an inclined plane without slipping. What fraction of its total energy is translational ?

Moderate

A

$\frac{1}{2}$
B

$\frac{1}{3}$
C

$\frac{2}{3}$
D

$\frac{1}{4}$

Solution

The kinetic energy of translational motion is given by
$(K.E.)_{t} = \frac{1}{2} {mv}^{2} = \frac{1}{2} {mr}^{2} ω^{2}$
The ratational kinetic energy is given by
$\begin{array}{l} (K . E .)_{r} = \frac{1}{2} {Iω}^{2} = \frac{1}{2} (\frac{1}{2} {mr}^{2}) ω^{2} \\ = \frac{1}{4} {mr}^{2} ω^{2} \end{array}$
$\begin{array}{l} ∴ {Total (K.E.)}_{r} = \frac{1}{2} {mr}^{2} ω^{2} + \frac{1}{4} {mr}^{2} ω^{2} \\ = \frac{3}{4} {mr}^{2} ω^{2} \end{array}$
The ratio of translational kinetic energy to total kinetic energy is given by
$\frac{(K.E.)_{t}}{(K.E.)_{T}} = \frac{(1 / 2) {mr}^{2} ω^{2}}{(3 / 4) {mr}^{2} ω^{2}} = \frac{2}{3}$

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