Question

At shown instant a thin uniform rod $AB$ of length $L=1m$ and mass $m=1kg$ is coincident with y-axis such that centre of rod is at origin. The velocity of end $A$ and centre $O$ of rod at shown instant are $\overrightarrow{V_A}=-2im/s$ and $\overrightarrow{V_o}=10im/s$ respectively. Then the kinetic energy of rod at the shown instant is :

• A. $42J$
• B. $56J$
• C. $74J$
• D. None of these

Answer 1

The correct option is C $74J$

Total K.E = K.E of Centre of Mass + K.E about Centre of Mass

The centre of mass of the rod is the geometric centre of the rod, at O.

m = 1kg; $v_{CM}$ = 10 m/s

K.E of Centre of Mass = $\frac{1}{2}m{v_{CM}}^2$ = $\frac{1}{2}\times 1\times 10\times 10=50J$

K.E about Centre of Mass = $\frac{1}{2}I{\omega }^2$,

Where I = Moment of inertia of rod and $\omega$ = angular velocity of the rod

I = $\frac{m{L}^2}{12}$ = $\frac{1\times 1^2}{12}$ = $\frac{1}{12}$

We find $\omega$ by using the equation $\overrightarrow{v_{ij}}=\overrightarrow{\omega }\times \overrightarrow{r_{ij}}$

$\overrightarrow{v_{AO}}=\overrightarrow{v_A}-\overrightarrow{v_O}$ = -2i ${ms}^{-1}$ - 10i ${ms}^{-1}$ = -12i ${ms}^{-1}$

$\overrightarrow{r_{AO}}$ = $\frac{1}{2}$j m

=> $\omega$ = 24k $s^{-1}$

K.E about centre of mass = $\frac{1}{2}\times \frac{1}{12}\times {24}^2=24J$

Total K.E = 50 + 24 = 74 J