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.

Answer 2

The angular momentum of the particle with respect to the origin is 50 kgm²/s.

What is angular momentum?

The angular momentum of an object is the product of moment of inertia and angular velocity.

L = mvr

where;

• m is the mass
• v is the velocity
• r is the radius

r = 5i + 5tj

v = dr/dt

v = 5 m/s

L = m(v x r)

v x r = 5j x (5i + 5(7)j)

v x r = 5j x (5i + 35j)

v x r = -25k

|v x r| = 25

L = m(v x r)

L = 2 x 25

L = 50 kgm²/s

Thus, the angular momentum of the particle with respect to the origin is 50 kgm²/s.

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