First slide

Dynamics of rotational motion about fixed axis

Question

A uniform disc of mass m is fitted (pivoted smoothly) with a rod of mass $\frac{m}{2}$ . If the bottom of the rod is pulled with a velocity v, it moves without changing its angle of orientation and the disc rolls without sliding. Thd kinetic energy of the system (rod + disc) is.

Moderate

A

$\frac{{mv}^{2}}{4}$
B

$\frac{{mv}^{2}}{2}$
C

$\frac{5 {mv}^{2}}{16}$
D

${mv}^{2}$

Solution

Rolling can be considered as pure rotation about point of contact
$K_{rolling} = \frac{1}{2} I_{P} ω^{2} = \frac{1}{2} (\frac{{mR}^{2}}{2} + {mR}^{2}) ω^{2} = \frac{3}{4} {mR}^{2} ω^{2}$ -----------(i)
The rod translates with the velocity v hence velocity of centre of disc will also be v
$v = ωR - - - - - - (ii)$
From (i) and (ii); $K_{rolling} = \frac{3}{4} {mv}^{2}$
Kinetic energy of the rod $K_{rod} = \frac{1}{2} (\frac{m}{2}) v^{2} = \frac{{mv}^{2}}{4}$ -------(iii)
$K_{total} = K_{rolling} + K_{rod} = {mv}^{2}$

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