Question

A rod of mass $2\text{kg}$ and length $1\text{m}$ is lying in the horizontal plane and pivoted about its one end. Initially, it is rotating about its pivoted end (axis perpendicular to horizontal plane) with an angular velocity $2\text{rad/s}$. Suddenly, an angular impulse $\overrightarrow{J}$ is given to the rod, because of which its angular velocity becomes $10\text{rad/s}$. Find the magnitude of angular impulse $\overrightarrow{J}$.

• A. zero
• B. $\frac{4}{3}{\text{kg m}}^2\text{/s}$
• C. $\frac{20}{3}{\text{kg m}}^2\text{/s}$
• D. $\frac{16}{3}{\text{kg m}}^2\text{/s}$

Answer 1

The correct option is D $\frac{16}{3}{\text{kg m}}^2\text{/s}$

Given,
Mass of rod $=M=2\text{kg}$
Lemgth of rod $=L=1\text{m}$
MOI of rod about its end $I=\frac{M{L}^2}{3}=\frac{2\times 1^2}{3}$
$I=2/3{\text{kg m}}^2$
As we know from impulse-angular momentum theorem,
$\overrightarrow{J}=\underset{t_1}{\overset{t_2}{\int }}d\overrightarrow{L}=\overrightarrow{\Delta L}={\overrightarrow{L}}_2-{\overrightarrow{L}}_1$
where, $L_2=$ final angular momentum
$L_1=$ initial angular momentum $=I{\omega }_1$
$L_1=\frac{2}{3}\times 2=\frac{4}{3}{\text{kg m}}^2\text{/s}$
$L_2=$ Final angular momentum $=I{\omega }_2=\frac{2}{3}\times 10=\frac{20}{3}{\text{kg m}}^2\text{/s}$

Therefore,
$|\overrightarrow{J}|=|{\overrightarrow{L}}_2-{\overrightarrow{L}}_1|=\frac{20}{3}-\frac{4}{3}=\frac{16}{3}{\text{kg m}}^2\text{/s}$