Question

A particle is moving along line y = 4x with a speed 10 m/s. If the mass of the particle is 2 kg then its angular momentum about the point (1, 4) is

Answer 1

Dear Student,

$$\begin{gathered} \overrightarrow{L}=m\left(\overrightarrow{r}\times \overrightarrow{v}\right) \\ L=mrv\sin \theta =mv\left(r\sin \theta \right)=mv{r}_{\perp } \\ {r}_{\perp }=perpendiculardis\tan cebetweenvelocityvectorandpoint. \\ Perpendiculardis\tan cebetweenlineax+by+c=0andpoint\left(x_1,y_1\right)=\left|\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}\right| \\ {r}_{\perp }=\left|\frac{4\times 1-1\times 4}{\sqrt{4^2+1^2}}\right|=0 \\ L=0[asris0] \end{gathered}$$


Regards,