Question

The position vector of a particle of mass 2 kg is given as a function of time by ~r = (5 m) ˆı + (5 m/s)t ˆ . Determine the magnitude of the angular momentum of the particle with respect to the origin at time 7 s

Answer 1

Answer:

$L = 50 \hat{k}\ kg.m^2/s$

Explanation:

given,

mass of the particle = 2 Kg

position vector of the particle is given as

$r = 5\hat{i} + 5 t \hat{j}$

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

$\dfrac{dr}{dt} = 5 \hat{j}$

angular momentum

L = m (r x v)

$L = m[(5\hat{i} + 5 t \hat{j}\ )\times (5 \hat{j})]$

cross product of vectors

i x j = k

j x j = 0

$L = m\times 25 \hat{k}$

$L = 2\times 25 \hat{k}$

$L = 50 \hat{k}\ kg.m^2/s$

Angular momentum is not dependent on the time function.