A particle of mass 2, kg is on a smooth horiz | vedclass

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

Question

(B) The angular momentum $\vec{L}$ of a particle about a point $O$ is given by $\vec{L} = \vec{r} 	imes \vec{p} = \vec{r} 	imes (m\vec{v})$.Here,the

Vedclass · Accepted Answer

$8.64$

Vedclass · Answer

(B) The angular momentum $\vec{L}$ of a particle about a point $O$ is given by $\vec{L} = \vec{r} 	imes \vec{p} = \vec{r} 	imes (m\vec{v})$.Here,the particle moves in a horizontal circle of radius $r = 0.6\, m$ at a height $h = 0.8\, m$ above the ground.Let the point on the ground directly below the center of the circle be $O'$. The position vector $\vec{r}$ of the particle relative to $O'$ has a horizontal component $r = 0.6\, m$ and a vertical component $h = 0.8\, m$.The velocity of the particle is $\vec{v}$,which is tangential to the circular path,so its magnitude is $v = r\omega = 0.6 	imes 12 = 7.2\, m/s$.The angular momentum about $O'$ is $L = |\vec{r} 	imes m\vec{v}| = mvr_{\perp}$,where $r_{\perp}$ is the perpendicular distance from the axis of rotation passing through $O'$ to the velocity vector. Since the velocity is horizontal and the point $O'$ is on the vertical axis,the perpendicular distance from the line of velocity to the point $O'$ is the radius $r = 0.6\, m$.Thus,$L = mvr = (2\, kg) 	imes (7.2\, m/s) 	imes (0.6\, m) = 8.64\, kg\, m^2\, s^{-1}$.

Similar Questions

$A$ uniform rod $AB$ of mass $m$ and length $l$ is at rest on a smooth horizontal surface. An impulse $P$ is applied to the end $B$. The time taken by the rod to turn through a right angle is

$A$ particle with position vector $\overrightarrow{r}$ has a linear momentum $\overrightarrow{p}$. Which one of the following statements is true in respect of its angular momentum $\overrightarrow{L}$ about the origin?

$A$ mass '$M$' is moving with constant velocity parallel to the $X$-axis. Its angular momentum with respect to the origin is

Write the general formula of total angular momentum of rotational motion about a fixed axis.

$A$ particle of mass $5 \ kg$ moves along the straight line $y = 2x + 4$ with a velocity of $3\sqrt{5} \ m/s$. Find its angular momentum about the origin in $kg \ m^2 \ s^{-1}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module