To solve this problem, we need to understand how the kinetic energy of a rotating system changes when the moment of inertia changes. The kinetic energy (K) of a rotating object is given by the formula:
Kinetic Energy in Rotational Motion
The kinetic energy of a rotating object can be expressed as:
K = (1/2) I ω²
where:
- K is the kinetic energy
- I is the moment of inertia
- ω is the angular velocity
Initial Conditions
Initially, the system has a moment of inertia I₀ and an angular velocity ω₀. Therefore, the initial kinetic energy can be expressed as:
K₀ = (1/2) I₀ ω₀²
Change in Moment of Inertia
When the man stretches out his hands, the moment of inertia of the system doubles. Thus, the new moment of inertia I₁ becomes:
I₁ = 2I₀
Conservation of Angular Momentum
In a closed system where no external torques are acting, angular momentum is conserved. The initial angular momentum L₀ can be expressed as:
L₀ = I₀ ω₀
After the moment of inertia doubles, the new angular momentum L₁ is:
L₁ = I₁ ω₁ = 2I₀ ω₁
Since angular momentum is conserved, we have:
L₀ = L₁
Thus, we can set the two expressions equal to each other:
I₀ ω₀ = 2I₀ ω₁
Dividing both sides by I₀ (assuming I₀ ≠ 0), we find:
ω₁ = (1/2) ω₀
Final Kinetic Energy Calculation
Now we can calculate the final kinetic energy K₁ using the new moment of inertia and the new angular velocity:
K₁ = (1/2) I₁ ω₁²
Substituting the values we have:
K₁ = (1/2) (2I₀) ((1/2) ω₀)²
This simplifies to:
K₁ = (1/2) (2I₀) (1/4) ω₀²
K₁ = (1/2) I₀ (1/2) ω₀²
K₁ = (1/4) I₀ ω₀²
Recalling that K₀ = (1/2) I₀ ω₀², we can express K₁ in terms of K₀:
K₁ = (1/4) (2K₀) = (1/2) K₀
Final Result
The final kinetic energy of the system after the man stretches out his hands is:
K₁ = (1/2) K₀
This means that the kinetic energy of the system is reduced to half of its initial value when the moment of inertia doubles and the angular velocity is halved.
