Physics, asked by ruchanshi873, 10 months ago

The angular velocity of the body changes from to without applying torque but by changing moment of inertia. The ratio of initial radius of gyration to the final radius of gyration is

Answers

Answered by ShivamKashyap08
13

Answer:

  • Ratio of initial radius of gyration to final radius gyration is √ (ω₂ / ω₁)

Assumptions:

  1. Let initial angular velocity be ω₁
  2. Let final radius of gyration be ω₂
  3. Let initial radius of gyration be K₁
  4. Let final radius of gyration be K₂

Explanation:

____________________

Since, the question says that no external torques act on the body to change its angular velocity, Then we need to apply law of conservation of angular momentum.

From law of conservation of momentum,

I ω = Constant

Where,

  • I denotes Moment of Inertia.
  • ω denotes angular velocity.

Now,

⇒ I ω = constant

⇒ I₁ ω₁ = I₂ ω₂

⇒ (M K²₁) × ω₁ = (M K²₂) × ω₂     ∵ [ I = M K²]

Now,

⇒ (K²₁) × ω₁ = (K²₂) × ω₂

ratio,

⇒ (K²₁) / (K²₂) = ω₂ / ω₁

⇒ K₁ / K₂ = √ (ω₂ / ω₁)

K₁ / K₂ = √ (ω₂ / ω₁)

Ratio of initial radius of gyration to final radius gyration is √ (ω₂ / ω₁).

____________________

Answered by nirman95
7

Answer:

Given:

Angular Velocity changes by changing the Moment of Inertia. But no torque is experienced.

To find:

Ratio of final pf gyration to the initial radius of gyration.

Concept:

Radius of gyration is a Radius of a Circular ring having the same Moment of Inertia (along geometric axis) as that of a particular object along a particular axis.

For any body :

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \boxed{ \huge{ \red{I = m {k}^{2} }}}

k : radius of gyration.

Calculation:

Since no external torque is experienced , we can say that the Angular Momentum of the body has been conserved (constant)

 \therefore \: L = constant

 =  > L1 = L2

 =  > I_{1} \omega_{1} = I_{2} \omega_{2}

 =  > (m {k_{1}}^{2} ) \omega_{1} =(m {k_{2}}^{2} )  \omega_{2}

 =  >  {(\dfrac{k_{1}}{k_{2}})}^{2}   =  \dfrac{ \omega_{2}}{ \omega_{1}}

 =  >  \dfrac{k_{1}}{k_{2}}=   \sqrt{\dfrac{ \omega_{2}}{ \omega_{1}}}

So final answer :

 \boxed{ \red{ \sf{ \large{ \dfrac{k_{2}}{k_{1}}=   \sqrt{\dfrac{ \omega_{2}}{ \omega_{1}}}}}}}

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