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

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.