A particle is projected at time t = 0 from a point P on the ground with a speed v0, at an angle of 45° to the horizontal. The angular momentum of the particle about P at time t = v0/g is (A) (mv03/2√2g) (B) (mv03/√2g) (C) 3(mv03/√2g) (D) (√2mv03/g)

Let L = angular momentum of the particle about P at time t = v0/g At time t = v0/g: Let px = x-component of momentum of particle vx = x - component of velocity of particle x = displacement along x-axis. Let py , vy and y denote the quantities along y-axis <br? Angular momentum, L = x py - y px or L = x (mvy) - y(mvx) or L = m [x vy - yvx] .....(i) To find vx and vy: vx = Horizontal velocity of the projectile or vx = v0 cos 45° = (v0/√2), .....(ii) It remains constant all along. vy = vertical velocity of particle or vy = (v0 sin 45°) - g × ((v0/g)) [∵ v= u+at] or vy = ((v0/√2) - v0) ......(iii) To find x and y: x= vx× time or x= (v0/√2) ×(v0/g) = (v02/√2g) ∴ x= (v02/√2g) .....(iv) y =v0 sin θ t - (gt2/2) y = (v0/√2) ×(v0/g) -(g/2)((v0/g))2 or y = ((v02/√2g) - (v02/2g)) .....(v) ∴ Substitute (ii), (iii), (iv) and (v) in (i) L= m[(v02/√2 g)×((v0/√2) -v0) - ((v02/√2g) - (v02/2g))(v0/√2)] or L =(mv03/g)[((1/2)-(1/√2))-((1/2)-(1/2√2))] or L =(mv03/g)×((-1/2√2)) = -(mv03/2√2g) L is ⊥ to plane of motion and is directed away from the reader i.e., along -ve z direction.

Q. A particle is projected at time $t = 0$ from a point $P$ on the ground with a speed $v_{0}$ , at an angle of $4 5^{\circ}$ to the horizontal. The angular momentum of the particle about $P$ at time $t = v_{0} / g$ is

3524 235 System of Particles and Rotational Motion Report Error

$\frac{m v _{0}^{3}}{2 2 g}$

$\frac{m v _{0}^{3}}{2 g}$

$3 \frac{m v _{0}^{3}}{2 g}$

$\frac{2 m v _{0}^{3}}{g}$

Solution:

Let $L =$ angular momentum of the particle about $P$ at time $t = v_{0} / g$
At time $t = v_{0} / g$ :
Let $p_{x} = x$ -component of momentum of particle
$v_{x} = x -$ component of velocity of particle
$x =$ displacement along $x$ -axis.
Let $p_{y}, v_{y}$ and $y$ denote the quantities along $y$ -axis $L = x p_{y} - y p_{x} or L = x (m v_{y}) - y (m v_{x}) or L = m [x v_{y} - y v_{x}] ..... (i) To find v_{x} and v_{y} : v_{x} = Horizontal velocity of the projectile or v_{x} = v_{0} cos 4 5^{\circ} = \frac{v _{0}}{2}, ..... (ii) It remains constant all along. v_{y} = vertical velocity of particle or v_{y} = (v_{0} s in 4 5^{\circ}) - g \times (\frac{v _{0}}{g}) [∵ v = u + a t] or v_{y} = (\frac{v _{0}}{2} - v_{0}) ...... (iii) To find x and y : x = v_{x} \times time or x = \frac{v _{0}}{2} \times \frac{v _{0}}{g} = \frac{v _{0}^{2}}{2 g} ∴ x = \frac{v _{0}^{2}}{2 g} ..... (i v) y = v_{0} s in θt - \frac{g t ^{2}}{2} y = \frac{v _{0}}{2} \times \frac{v _{0}}{g} - \frac{g}{2} (\frac{v _{0}}{g})^{2} or y = (\frac{v _{0}^{2}}{2 g} - \frac{v _{0}^{2}}{2 g}) ..... (v) ∴ Substitute (ii), (iii), (i v) and (v) in (i) L = m [\frac{v _{0}^{2}}{2 g} \times (\frac{v _{0}}{2} - v_{0}) - (\frac{v _{0}^{2}}{2 g} - \frac{v _{0}^{2}}{2 g}) \frac{v _{0}}{2}] or L = \frac{m v _{0}^{3}}{g} [(\frac{1}{2} - \frac{1}{2}) - (\frac{1}{2} - \frac{1}{2 2})] or L = \frac{m v _{0}^{3}}{g} \times (\frac{- 1}{2 2}) = - \frac{m v _{0}^{3}}{2 2 g} L is ⊥ to plane of motion and is directed away from the reader i.e., along -ve z direction.$

Q. A particle is projected at time t=0 from a point P on the ground with a speed v0​, at an angle of 45∘ to the horizontal. The angular momentum of the particle about P at time t=v0​/g is

22​gmv03​​

2​gmv03​​

32​gmv03​​

g2​mv03​​

Q. A particle is projected at time $t = 0$ from a point $P$ on the ground with a speed $v_{0}$ , at an angle of $4 5^{\circ}$ to the horizontal. The angular momentum of the particle about $P$ at time $t = v_{0} / g$ is

$\frac{m v _{0}^{3}}{2 2 g}$

$\frac{m v _{0}^{3}}{2 g}$

$3 \frac{m v _{0}^{3}}{2 g}$

$\frac{2 m v _{0}^{3}}{g}$