Question

Two bodies have moment of Inertia ‘I’ and ‘2I’ respectively about their axis of rotation. If their kinetic energies of rotation are equal, what will be the ratio of their angular momentum? ​

Answer 1

Answer:1/$\sqrt 2$

Explanation:

L = Iω

where L is angular momentum,I is moment of inertia,ω angular velocity

given,$I_{1}$=i  and  $I_{2}$=2i

kinetic energy=(1/2)Iω²

since kinetic energies are equal.

(1/2) i ω₁²=(1/2) 2i ω₂²

ω₁/ω₂ = $\sqrt{2}$

ratio of angular momentum =  $I_{1}$ ω₁ / $I_{2}$ ω₂

                                                = i ω₁ /2i ω₂

                                               =$\sqrt{2}$/2

                                                = 1/$\sqrt{2}$

Answer 2

Answer:-

1 : √2

Explanation:-

We know that :-

K.E. = 0.5Iω²

Where :-

• K.E. is the rotational kinetic energy.

• I is the moment of inertia.

• ω is the angular velocity.

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∵ ω = L/I

So, we can write the above formula as :-

=> K.E. = 0.5I(L/I)²

∴ K.E. = 0.5L²/I

Case (1) :-

=> K.E. = 0.5L₁²/1 -----(1)

Case (2) :-

=> K.E.' = 0.5L₂²/2I

=> K.E.' = 0.25L₂²/I

Since, it is given that the rotational kinetic energy of the bodies are equal .

∴ K.E. = K.E.'

=> K.E. = 0.25L₂²/I ------(2)

On dividing eq.1 by eq.2 we get :-

=> K.E./K.E. = [0.5L₁²/I][0.25L₂²/I]

=> 1 = 2L₁²/L₂²

=> L₁²/L₂² = 1/2

=> √(L₁²/L₂²) = √(1/2)

=> L₁/L₂ = 1/√2

=> L₁ : L₂ = 1 : √2

Thus, the ratio of angular momentum of the bodies is 1 : √2 .