A particle of mass m is moving along a line y | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The equation of the line is $y = x + a$. Comparing this with the slope-intercept form $y = mx + c$,we get the slope $m_{slope} = 1$. Since $	an \

Vedclass · Accepted Answer

$\frac{mva}{\sqrt{2}}$

Vedclass · Answer

(C) The equation of the line is $y = x + a$. Comparing this with the slope-intercept form $y = mx + c$,we get the slope $m_{slope} = 1$. Since $	an 	heta = m_{slope} = 1$,the angle $	heta = 45^{\circ}$.The velocity vector $\vec{v}$ makes an angle of $45^{\circ}$ with the positive $x$-axis. Thus,$\vec{v} = v \cos 45^{\circ} \hat{i} + v \sin 45^{\circ} \hat{j} = \frac{v}{\sqrt{2}} (\hat{i} + \hat{j})$.The angular momentum $\vec{L}$ about the origin is given by $\vec{L} = \vec{r} 	imes \vec{p} = m (\vec{r} 	imes \vec{v})$.The perpendicular distance $d$ from the origin $(0, 0)$ to the line $x - y + a = 0$ is given by $d = \frac{|(1)(0) - (1)(0) + a|}{\sqrt{1^2 + (-1)^2}} = \frac{a}{\sqrt{2}}$.The magnitude of angular momentum is $L = mvd = m v \left( \frac{a}{\sqrt{2}} ight) = \frac{mva}{\sqrt{2}}$.

$A$ particle of mass $m$ is moving along a line $y = x + a$ with a constant velocity $v$. The angular momentum of the particle about the origin is

Similar Questions

$A$ particle of mass $m$ moves with a velocity $v$ along the line $PC$ as shown in the figure. What is the angular momentum of the particle about $O$?

$A$ ring of mass $10 \ kg$ and diameter $0.4 \ m$ is rotating about its axis. If it completes $2100$ revolutions per minute,what is its angular momentum in $kg \cdot m^2/s$?

$A$ particle starts from the point $(0, 8) \, m$ and moves with a uniform velocity of $\vec{v} = 3 \hat{i} \, m/s$. What is the angular momentum of the particle about the origin after $5 \, s$ (mass of the particle is $1 \, kg$)?

$A$ particle is rotating in a circle with uniform speed as shown. The angular momentum of the particle with respect to the origin is .........

Vedclass Test Series

Exam Paper Generator

Online Exam Module