Question

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?

Answer 1

Answer:

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.