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
Read more at; https://brainly.com/question/3245307
The image of the particle in figure 1 is missing and so i have attached it.
