Question

A particle of mass 2 kg is located at the position (i+j)m has a velocity 2(i+j+k)m/s. Its sngular momentum about z-axis is

Answer 1

Answer: The angular momentum is $L= 8 kg-m/s^{2}$.

Explanation:

Given that,

Mass m = 2 kg

Position  $r = (i+j)$

Velocity $v = 2(i+j+k) m/s$

We know that,

The angular moment is the product of the moment of inertia and angular velocity.

$L = I\omega$

$L = mr^{2}\times\dfrac{v}{r}$

$L = m(\vec r\times \vec v)$...(I)

Put the value in equation (I)

$L = 2[(i+j)\times(2i+2j+2k)]$

$L =2(2+2) kg-m/s^{2}$

$L= 8 kg-m/s^{2}$

Hence, the angular momentum is $L= 8 kg-m/s^{2}$.

Answer 2

"Given,

Mass 'm' = 2Kg        

Position $\\ 'r'\quad =\quad \left( i+j \right)$

Velocity $\\ 'v'\quad =\quad 2\left( i+j+k \right) \frac { m }{ s }$

The product of moment of inertia and angular velocity is called the angular moment. The angular moment around axis Z is given by,

$\\ L\quad =\quad -I\omega$

$\\ L\quad =\quad -m{ r }^{ 2 }x\frac { v }{ r }$

$\\ L\quad =\quad -m\left( r\times v \right)$

We know that$\\ v'\quad =\quad 2\left( i+j+k \right) \frac { m }{ s }.$

By substituting in above equation we get,

$\\ L\quad =\quad -2\left[ \left( i+j \right) x\left( 2i+2j+2k \right) \right]$

$\\ L\quad =\quad -2\left( 2+2 \right) \frac { Kgm }{ { s }^{ 2 } }$

$\\ L\quad =\quad -8\frac { Kgm }{ { s }^{ 2 } }$"