Question

a body of mass root 2 kg is projected with a speed V ata an angle if 45 to horizontal, when the body is at the highest point in its path , angular momentym of the body about the point of projection is

Answer 1

Answer:

2v

Explanation:

momentum = mvtan45 = 2vtan45= 2v

Answer 2

Therefore the angular momentum of the body at the maximum height is '2vr'.

Given:

The mass of the body = m = 2 kg

Speed of the body at which it is projected = v

The angle at which the body is projected = θ = 45°

To Find:

The angular momentum of the body at the highest position about the point of projection.

Solution:

The given question can be solved very easily as shown below.

The velocity of the object at any point in the line of action of projectile motion has 2 components.

⇒ Horizontal component = v Cos θ

⇒ Vertical component = v Sin θ

And the resultant becomes total velocity = V = √ [ ( v Cos θ )² + ( v Sin θ )²

Hence at the maximum point, the Final velocity will have only a horizontal component because the vertical component becomes zero.

So at the maximum height, Final velocity = V = v Cos θ

And the 'θ' becomes zero at the maximum height, So V = v Cos 0° = v

Hence the linear momentum of the body = Mass × Velocity = mv = 2v

⇒ Angular momentum = Moment of Inertia × Angular velocity = Iω

Moment of Inertia = I = mr² where r = radius

Angular momentum = ω = v/r

⇒ Angular momentum = Iω = ( mr² ) × ( v/r )

⇒ Angular momentum = mvr = 2vr

Therefore the angular momentum of the body at the maximum height is '2vr'.

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