Question

A particle of mass m=5 kg is moving with a uniform speed v=3√2 in the XOY plane along the line Y=x + 4.The magnitude of angular momentum of the particle about the origin is?​

Answer 1

Answer :-

Angular momentum L = 60 unit

Explanation :-

Method I

The equation of the line is Y= X+4

or X- Y +4 =0

Length of perpendicular from origin on this line is

R = 0-0+4/ √1^2+ 1^2 = 4/√2

∴ Angular momentum

L = mvR = 5 ×3√2 × 4/√2 = 60 units

Method II

Y= X +4 line is shown in the figure.

When X=0, Y=4,

So, OS = 4

To find slope of this line comparing this with equation of line

y = m'x + c

∴ Slope, m' = tanθ = 1

⇒θ= 45°

Length of perpendicular = OP

In ∆PSO, OP/OS = sin 45°

∴ OP = OS sin 45°

OP = 4 × 1/√2 = 4√2

∴ Angular momentum of particle going this line

= mvR

= 5 × 3 √2 × 4√2

= 60 unit

[ Refer attachment for better understanding ]