Question

An annular disc has a mass m, inner radius R and outer radius 2 R. The disc rolls on a flat surface without slipping. If the velocity of the center of mass is v, the kinetic energy of the disc is

• A. $\frac{9}{16}m{v}^2$
• B. $\frac{11}{16}m{v}^2$
• C. $\frac{13}{16}m{v}^2$
• D. $\frac{15}{16}m{v}^2$

Answer 1

The correct option is C $\frac{13}{16}m{v}^2$


m is mass in area $3\pi R^2[4\pi R^2-\pi R^2]$
mass in area $2\pi rdr=dm$
$dm=\frac{2mrdr}{3R^2}$
$I={\int }_R^{2R}dm{r}^2=\frac{2m}{3R^2}{\int }_R^{2R}{r}^{3}dr$
$=\frac{2m}{3R^2}{\left[\frac{r^4}{4}\right]}_R^{2R}$
$\frac{m}{6R^2}\left[15R^4\right]=\frac{5m{R}^2}{2}$
$K.E=(K.E{)}_{translation}+(K.E{)}_{rotation}$
$=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$
Where $V=2R\omega$
$\Rightarrow \omega =\frac{v}{2R}$
$\therefore K.E=\frac{1}{2}m{v}^2+\frac{1}{2}\times \frac{5m{R}^2}{2}\times \frac{v^2}{4R^2}$
$=\frac{1}{2}m{v}^2+\frac{5m{v}^2}{16}$
$K.E=\frac{8m{v}^2+5m{v}^2}{16}=\frac{13m{v}^2}{16}$
Method II:
Also solved above,
$I=\frac{5}{2}m{R}^2$
this is the moment of inertia about the centre of mass of the disc.
$\therefore$ Again, the moment of inertia about the instantaneous axis of rotaion.
$I_0=I+m(2R{)}^2$
$=\frac{5}{2}m{R}^2+4m{R}^2=\frac{13}{2}m{R}^2$
$\therefore$ Kinetic energy of the disc,
$KE=\frac{1}{2}{I}_0{\omega }^2$
$=\frac{1}{2}\left(\frac{13}{2}\right)m{R}^2\times {\left(\frac{V}{2R}\right)}^2$
$=\frac{13}{4}m{R}^2\times \frac{V^2}{4R^2}=\frac{13}{16}m{V}^2$