Question

Two particles of same mass $2\text{kg}$ are rigidly attached to the ends of a massless rod of length $10\text{m}$. The rod is clamped at the centre such that the system rotates about the perpendicular bisector of the rod at an angular speed $10\text{rad/s}$. Calculate the magnitude of angular momentum of the system about the axis of rotation [in ${\text{kg m}}^2\text{/s}$]

• A. $100$
• B. $1000$
• C. $500$
• D. $10$

Answer 1

The correct option is B $1000$

As we know,
${\overrightarrow{L}}_{\text{system}}=I_{\text{axis}}\overrightarrow{\omega }$
where, $I=$ MOI of the system about the axis of rotation
$I=m_1{r}_1^2+m_2{r}_2^2$
$I=2\times {\left(\frac{10}{2}\right)}^2+2\times {\left(\frac{10}{2}\right)}^2$
$I=100{\text{kg-m}}^2$
& $\omega =10\text{rad/s}$

From , ${\overrightarrow{L}}_{\text{system}}=I_{\text{axis}}\overrightarrow{\omega }$
${\overrightarrow{L}}_{\text{system}}=100\times 10$
${\overrightarrow{L}}_{\text{system}}=1000{\text{kg m}}^2\text{/s}$