Question

The position vector of a particle of mass 1.70 kg as a function of time is given by r with arrow = (6.00 î + 5.70 t ĵ), where r with arrow is in meters and t is in seconds. Determine the angular momentum of the particle about the origin as a function of time.

Answer 1

Answer:

The angular momentum of the particle is 58.14 kg m²/s along positive z- axis and is independent of time .

Explanation:

Given that,

Mass = 1.70 kg

Position vector $r=(6.00\hat{i}+5.70 t \hat{j})$

We need to calculate the angular velocity

The velocity is the rate of change of the position of the particle.

$v = \dfrac{dr}{dt}$

$v=\dfrac{d}{dt}(6.00\hat{i}+5.70 t \hat{j})$

$v=5.70\hat{j}$

We need to calculate the angular momentum of the particle

Using formula of angular momentum

$L=r\cdot p$

Where, p = mv

Put the value of p into the formula

$L=m(r\times v)$

Substitute the value into the formula

$L=1.70(6.00\hat{i}+5.70 t \hat{j}\times5.70\hat{j})$

$L=1.70\times34.2$

$L=58.14\ kgm^2/s$

Hence, The angular momentum of the particle is 58.14 kg m²/s along positive z- axis and is independent of time .