A figure skater stands on one spot on the ice(assumed frictionless) and spins around with her arms extended. When she pulls in her arms, her moment of inertia decreases to 3/4 of the initial value. Assume that her spin angular momentum is conserved. What is the ratio of her final rotational kinetic energy to her initial rotational kinetic energy?
4/3
I omega, angular momentum, does not change so:
I1 omega1 = I2 omega2
I1 omega1 = (3/4) I1 omega2
so
omega2 = omega 1 * 4/3
then
original ke = (1/2) I1 omega1&2
final Ke = (1/2) I2 omega2^2
so
final Ke=(1/2)(3/4)I1 *(4/3)^2 omega1^2
= 4/3 of original
(she had to do work to pull her arms in)
Thank you Damon
To solve this problem, we can use the conservation of angular momentum and the equation for rotational kinetic energy.
The conservation of angular momentum states that the initial angular momentum of an object is equal to its final angular momentum, as long as there are no external torques acting on it. Mathematically, we can express this as:
I₁ω₁ = I₂ω₂
Where I₁ and I₂ are the initial and final moments of inertia, and ω₁ and ω₂ are the initial and final angular velocities, respectively.
In this case, the figure skater is standing on one spot and spinning around with her arms extended. When she pulls her arms in, her moment of inertia decreases to 3/4 of its initial value. Let's call the initial moment of inertia I and the final moment of inertia 3/4I.
The ratio of the final moment of inertia to the initial moment of inertia is:
(3/4I) / I = 3/4
Now, let's apply conservation of angular momentum. Since the skater is spinning without any external torques acting on her, her initial angular momentum is equal to her final angular momentum:
I₁ω₁ = I₂ω₂
Substituting the values, we have:
I * ω₁ = (3/4I) * ω₂
The moment of inertia I cancels out, giving us:
ω₁ = (3/4)ω₂
Next, let's consider the equation for rotational kinetic energy:
K₁ = (1/2)I₁ω₁²
K₂ = (1/2)I₂ω₂²
Where K₁ and K₂ are the initial and final rotational kinetic energies, respectively.
Now, we need to find the ratio of the final rotational kinetic energy (K₂) to the initial rotational kinetic energy (K₁). We can plug in the value of ω₂ from the angular momentum conservation equation:
K₂ = (1/2)(3/4I)ω₂²
Simplifying:
K₂ = (3/8)Iω₂²
To find the ratio, we divide K₂ by K₁:
(K₂ / K₁) = [(3/8)Iω₂²] / [(1/2)I₁ω₁²]
Since we know that ω₁ = (3/4)ω₂, we can substitute this into the equation:
(K₂ / K₁) = [(3/8)Iω₂²] / [(1/2)I(3/4)²ω₂²]
Simplifying further:
(K₂ / K₁) = (3/8)ω₂² / (1/2)(3/4)²ω₂²
The ω₂² terms cancel out, giving us:
(K₂ / K₁) = (3/8) / (1/2)(3/4)²
(K₂ / K₁) = (3/8) / (1/2)(9/16)
(K₂ / K₁) = (3/8) / (9/32)
(K₂ / K₁) = (3/8) * (32/9)
(K₂ / K₁) = 32/24
Simplifying the fraction, we get:
(K₂ / K₁) = 4/3
Therefore, the ratio of the final rotational kinetic energy to the initial rotational kinetic energy is 4/3.