Question

A particle of mass 2.0 kg is moving in the xy plane at a constant speed of 0.80 m s in the x direction along the line y 4 m As the particle travels from x 3 m to x 3 m the magnitude of its angular momentum with respect to the origin is?

Answer 1

Answer:

6.4 kgm²/s

Explanation:

the formula for angular momentum is

Angular momentum, L= radius/distance, r * linear momentum, p

i.e L = r * p

and linear momentum = mass * velocity  i.e p = m*v

Hence, L = m*v*r

we are given the following:

mass of particle, m = 2 kg

velocity/speed of particle, v = 0.8 m/s

radius/distance, r = the value in y axis = 4m

substituting the values, we have

L = 2 * 0.8 * 4 = 6.4 kgm²/s

Answer 2

The magnitude of its angular momentum with respect to the origin is 6.4 $kgm^{2}/s$

The angular momentum of an object is given by

L = r × p

L = rpsinθ

where r is the distance and p is the linear momentum of the object.

Now for a particle moving on the line y = 4, the value of rsinθ remains constant from the origin and equal to 4

L = mvrsinθ  

since p = mv, where m is the mass of the particle and v is the velocity of the particle.

L = 2.0 × 0.8 × 4

L = 6.4 $kgm^{2}/s$ is the angular momentum of the particle.

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