Question

A ring of mass $m$ is rolling without slipping with linear speed $v$ as shown in figure. Four particles each of mass $m$ are also attached at points $\text{A, B, C}$ and $\text{D}$. Find total kinetic energy of the system.

• A. $6m{v}^2$
  
• B. $3m{v}^2$
  
• C. $m{v}^2$
  
• D. $5m{v}^2$
  

Answer 1

The correct option is D $5m{v}^2$


Let the radius of the ring be $R$.

We know that in case of pure rolling,
$\omega =\frac{v}{R}$ and $v_A=0$

Since the angular velocity $\omega$ is constant, so we can write,

$\omega =\frac{v}{AO}=\frac{v_B}{AB}=\frac{v_D}{AD}=\frac{v_C}{AC}...\left(1\right)$

From,
$AO=R$ ; $AB=\sqrt{R^2+R^2}=R\sqrt{2}$ ; $AD=\sqrt{R^2+R^2}=R\sqrt{2}$ ; $AC=2R$

From equation $1$,

$\omega =\frac{v}{R}=\frac{v_B}{R\sqrt{2}}=\frac{v_D}{R\sqrt{2}}=\frac{v_C}{2R}$

Thus,
$v_B=v_D=\sqrt{2}v$ and $v_C=2v$

Now,
Total kinetic energy $=[$translational kinetic energy of four particles$]+[$translational kinetic energy of ring $+$ rotational kinetic energy of ring$]$

Substituting the values, we get
$(KE{)}_{total}=\left[\frac{1}{2}m\right(0{)}^2+\frac{1}{2}m(\sqrt{2}v{)}^2+\frac{1}{2}m(\sqrt{2}v{)}^2+\frac{1}{2}m\left(2v{)}^2\right]+[\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2]$

Substituing $I=m{R}^2$ and $\omega =\frac{v}{R}$, we get
$(KE{)}_{total}=[0+m(v{)}^2+m(v{)}^2+2m(v{)}^2]+[\frac{1}{2}m{v}^2+\frac{1}{2}\times m{R}^2{\left(\frac{v}{R}\right)}^2]$

$\therefore (KE{)}_{total}=5m{v}^2$

Hence, option (d) is the correct answer.