Question

The rotational inertia of a collapsing spinning star changes to 1/6 its initial value. What is the ratio of the new rotational kinetic energy to the initial rotational kinetic energy

Answer 1

Answer:

Explanation:

Let the initial rotational inertia be I and final rotational inertia be I / 6 .

Let the initial angular velocity be ω₁ and final angular velocity be ω₂.

Applying conservation of angular momentum law

I x ω₁ = I / 6 x ω₂

6 ω₁  = ω₂

initial rotational kinetic energy = 1/2 I x ω₁ ²

Final  rotational kinetic energy = 1/2 ( I / 6 ) x ω₂ ²

Final  rotational kinetic energy / initial rotational kinetic energy

= ( 1 / 6 ) x ω₂ ² / ω₁ ²

=  ω₂ ² / 6ω₁ ²

= 36 ω₁ ² / 6ω₁ ²

= 6 .

Answer 2

The ratio will be "6". A further solution is below.

Let,

The initial rotational inertia,

• $\omega_1 = 1$

The final rotational inertia,

• $\omega_2 = \frac{1}{6}$

By applying the conservation of angular momentum law, we get

→ $1\times \omega_1 = \frac{1}{6}\times \omega_2$

→    $6 \ \omega_1 = \omega_2$

Now,

Initial rotational K.E = $\frac{1}{2}1\times \omega_1^2$

Final rotational K.E = $\frac{1}{2} (\frac{1}{6} )\times \omega_2^2$

hence,

→ $\frac{Final \ rotational \ K.E}{Initial \ rotational \ K.E} = \frac{(\frac{1}{6} )\times \omega_2^2}{\omega_1^2}$

                                  $= \frac{\omega_2^2}{6 \omega_1^2}$

                                  $= 6$

Thus the answer above is right.

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