A student sits on a rotating stool holding two 3.0-kg objects. When his arms are extended horizontally, the objects are 1.0 m from the axis of rotation and he rotates with an angular speed of 0.75 rad/s. The moment of inertia of the student plus stool is 3.0 kg⋅m^2 and is assumed to be constant. The student then pulls in the objects horizontally to 0.30 m from the rotation axis. Find the kinetic energy of the student before and after the objects are pulled in.

2.86J

Explanation:

M = 3.0kg

R₁ = 1.0m

R₂ = 0.3m

I₁ = I(mass) + I(student + stool)

I₁ = 2mr₁² + I(student + stool)

I₁ = 2*(3*1²) + 3.0

I₁ = 9.0kgm²

the initial moment of inertia of the system = 9.0kgm²

I₂ = 2mr₂² + I(student + stool)

I₂ = 2*(3 * 0.3²) + 3.0

I₂ = 0.54 + 3.0

I₂ = 3.54kgm²

the final moment of inertia of the system is 3.54kgm²

From conservation of angular momentum

I₁ω₁ = I₂ω₂

ω₂ = (I₁ * ω₁) / I₂

ω₂ = (0.75 * 9) / 3.54

ω₂ = 1.09rad/s

kinetic energy of rotation (k.e) =½ Iω²

K.E = (K.E)₂ - (K.E)₁

k.e = [½ * 3.54 * (1.90)²] - [½ * 9.0 * 0.75²]

K.E = 6.3897 - 2.53125

K.E = 2.85845

K.E = 2.86J