A man stands on a platform that is rotating (without friction) with an angular speed of 1.2 rev/s; his arms are outstretched and he holds a brick in each had. The rotational inertia of the system consisting of the man, bricks, and platform about the central vertical axis of the platform is 6.0 kgm2. If by moving the bricks the man decreases the rotational inertia of the system to 2.0 kgm2, what are (a) the resulting angular speed of the platform and (b) the ratio of the new kinetic energy of the system to the original kinetic energy?

A. ω₂ = 3.6rev/s

B. K₂ / K₁ = 3

Explanation:

Data;

ω₁ = 1.2rev/s

I₁ = 6.0kgm²

ω₂ = ?

I₂ = 2.0kgm²

for conservation of angular momentum,

Initial angular momentum = final angular momentum

L₁ = L₂

I₁ω₁ = I₂ω₂

solving for ω₂,

ω₂ = (I₁ * ω₁) / I₂

ω₂ = (6.0 * 1.2) / 2.0

ω₂ = 3.6rev/s

the resulting angular speed of the platform is 3.6rev/s.

b. the ratio of the new kinetic energy to the system kinetic energy

let the new kinetic energy = K₂

original kinetic energy = K₁

Kinetic energy of a rotational body = ½*Iω²

K₂ / K₁ = ½ I₁ω₁² = ½I₂ω²

K₂ / K₁ = I₁ω₁² / I₂ω₂²

K₂ / K₁ = (2.0 * 3.6²) / (6.0 * 1.2²)

K₂ / K₁ = 25.92 / 8.64

K₂ / K₁ = 3

The ratio of the new kinetic energy to the old kinetic energy is 3.

b. latent learning (the word "latent" itself means something that exists but isn't obvious, only under certain conditions! )

answer: electrons