# If the frictional force is the only force that causes the system to lose energy, what is the fractional change in the kinetic energy?

Disk A in (Figure 1) has radius RA and thickness h and is initially rotating clockwise, as viewed from above, at ωi/2. Disk B, made of the same material as A, has radius RB=RA/2 and thickness h and is initially rotating counterclockwise at ωi. The two disks are constrained to rotate about the same axis, which runs through their centers. There is friction between the disks, and so once B slides down and touches A, they spin at the same rotational speed. (Looking for (Kf-Ki)/Ki)

### 1 Answer

- WhomeLv 72 months agoFavorite Answer
Assume disk A is spinning in the positive direction.

Conservation of angular momentum during the collision phase.

m = mass

R = Radius of disc A

ρ = material density

h = material thickness

Moment of inertia of a uniform disk is I = ½mr²

mass of a disk = ρV = ρπr²h

Moment of inertia of disk A, IA = ½(ρπR²h)R² = ρπR⁴h/2

Moment of inertia of disk B, IB= ½ρπ(R/2)²h(R/2)² = ρπR⁴h/32

as the term "ρπR⁴h" is common, let's call it Ψ

IA = Ψ/2

IB = Ψ/32

IA(½ωi) + IB(-ωi) = (IA + IB)ωf

Ψ/2(½ωi) + Ψ/32(-ωi) = (Ψ/2 + Ψ/32)ωf

Ψ/4(ωi) - Ψ/32(ωi) = (Ψ/2 + Ψ/32)ωf

7Ψ/32(ωi)= (17Ψ/32)ωf

ωf = 7ωi/17

so both discs spin at about 41% of disk A's initial angular velocity.

(Kf-Ki)/Ki) = [½(Ψ/2 + Ψ/32)(7ωi/17)² - (½Ψ/2(½ωi)² + ½Ψ/32(-ωi)²)] / (½Ψ/2(½ωi)² + ½Ψ/32(ωi)²)

½Ψ(ωi²) is common in all terms so divides out to 1

(Kf-Ki)/Ki) = [(17/32)(49/289) - (1/8 + 1/32)] / (1/8 + 1/32)

(Kf-Ki)/Ki) = -0.42352941

or a little over 42% loss of kinetic energy to friction

OOps, I erred in a statement "both discs spin at about 41% of disk A's initial angular velocity." As disk A was originally doing ½ωi the percentage is |7/17 - 1/2| / 1/2 = 82% of initial velocity. Does not affect the results.