Search any question & find its solution
Question:
Answered & Verified by Expert
If a particle of mass $m$ is moving with constant velocity $v$ parallel to $x$-axis in $x$-y plane as shown in fig. Its angular momentum with respect to origin at any time $t$ will be
Options:
Solution:
2591 Upvotes
Verified Answer
The correct answer is:
$-m v b \hat{k}$
We know that, Angular momentum
$\vec{L}=\vec{r} \times \vec{p}$ in terms of component becomes
$\vec{L}=\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ p_x & p_y & p_z\end{array}\right|$
As motion is in $x-y$ plane $\left(z=0\right.$ and $\left.P_z=0\right)$, so $\vec{L}=\vec{k}\left(x p_y-y p_x\right)$
Here $x=v t, y=b, p_x=m v$ and $p_y=0$
$\therefore \vec{L}=\vec{k}[v t \times 0-b m v]=-m v b \hat{k}$
$\vec{L}=\vec{r} \times \vec{p}$ in terms of component becomes
$\vec{L}=\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ p_x & p_y & p_z\end{array}\right|$
As motion is in $x-y$ plane $\left(z=0\right.$ and $\left.P_z=0\right)$, so $\vec{L}=\vec{k}\left(x p_y-y p_x\right)$
Here $x=v t, y=b, p_x=m v$ and $p_y=0$
$\therefore \vec{L}=\vec{k}[v t \times 0-b m v]=-m v b \hat{k}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.