Question

Find angular momentum along point O when it reaches at point P
Please answer the question.​

Answer 1

Answer:

Angular Momentum is an axial vector represented as the cross product between position vector and linear momentum vector.

Mathematically, we can say :

Angular momentum = Linear Momentum × ⊥r

Calculation:

Linear momentum at highest point be v

$\bigstar \: \: P =m \{ u \cos( \theta) \}$

Perpendicular distance :

$\bigstar \: \: h_{max} = \dfrac{ {u}^{2} { \sin }^{2} ( \theta) }{2g}$

So, angular momentum at highest point shall be :

$\bigstar \: \: L = P \times (r\perp)$

$\implies \: L = m \{u \cos( \theta) \} \times \{ \dfrac{ {u}^{2} { \sin}^{2}( \theta) }{2g} \}$

$\implies \: L = \dfrac{m {u}^{3} { \sin}^{2}( \theta) \cos( \theta) }{2g}$

So final answer is :

$\boxed{ \red{ \huge{ \bold{ \: L = \dfrac{m {u}^{3} { \sin}^{2}( \theta) \cos( \theta) }{2g} }}}}$

Answer 2

Answer:

Answer:

Angular Momentum is an axial vector represented as the cross product between position vector and linear momentum vector.

Mathematically, we can say :

Angular momentum = Linear Momentum × ⊥r

Calculation:

Linear momentum at highest point be v

\bigstar \: \: P =m \{ u \cos( \theta) \}★P=m{ucos(θ)}

Perpendicular distance :

\bigstar \: \: h_{max} = \dfrac{ {u}^{2} { \sin }^{2} ( \theta) }{2g}★h

max

=

2g

u

2

sin

2

(θ)

So, angular momentum at highest point shall be :

\bigstar \: \: L = P \times (r\perp)★L=P×(r⊥)

\implies \: L = m \{u \cos( \theta) \} \times \{ \dfrac{ {u}^{2} { \sin}^{2}( \theta) }{2g} \}⟹L=m{ucos(θ)}×{

2g

u

2

sin

2

(θ)

}

\implies \: L = \dfrac{m {u}^{3} { \sin}^{2}( \theta) \cos( \theta) }{2g}⟹L=

2g

mu

3

sin

2

(θ)cos(θ)

So final answer is :

\boxed{ \red{ \huge{ \bold{ \: L = \dfrac{m {u}^{3} { \sin}^{2}( \theta) \cos( \theta) }{2g} }}}}

L=

2g

mu

3

sin

2

(θ)cos(θ)