What is the magnitude of the angular momentum relative to the origin of the 200 g particle in (figure 1)?.

The angular momentum relative to the origin of the 200 g particle is;

L = 0.601 kg.m²/s

Radius from the origin to the particle is gotten by;

R = √(2² + 1²)

R = √5

Now formula for moment of inertia here is;

I = mR²

Where;

m is mass = 200 g = 0.2 kg

R is radius

Thus;

I = 0.2 × (√5)²

I = 1 kg.m²

We are given v = 3 m/s. Thus;

Velocity component at 90° to R will be;

V = 3(sin 90 - θ)

Where;

θ = 90 - (tan^(-1) (1/2))

θ = 63.4°

Thus;

V = 3(sin (90 - 63.4))

V = 1.3433 m/s

Formula for angular velocity is;

ω = V/R

Thus;

ω = 1.3433/√5

ω = 0.601 rad/s

Formula for angular momentum is;

L = I × ω

L = 1 × 0.601

L = 0.601 kg.m²/s

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