The figure below shows three rotating, uniform disks that are coupled by belts.

Question

The figure below shows three rotating, uniform disks that are coupled by belts. One belt runs around the rims of disks A and C. Another belt runs around a central hub on disk A and the rim of disk B. The belts move smoothly without slippage on the rims and hub. Disk A has radius R; its hub has radius 0.4400R; disk B has radius 0.2600R; and disk C has radius 1.900R. Disks B and C have the same density (mass per unit volume) and thickness. What is the ratio of the magnitude of the angular momentum of disk C to that of disk B? L_C/L_B =

Explanation / Answer

The formula for angular momentum L is:

L = I * ,

where:
I = moment of inertia
and
= angular speed

Therefore,

LC / LB = (IC / IB) * (C / B) --- (1)

For a disk,

I = m * r^2 / 2

Therefore we need to find mC / mB. Since the densities of disks B and C are the same, their masses are proportional to their volumes V; the volumes are in turn proportional to the squares of their radii, since they have the same thickness:

mC / mB = rC^2 / rB^2 = (rC / rB)^2


IC / IB = (mC / mB) * rC^2 / rB^2

= (rC / rB)^2 * rC^2 / rB^2

= (rC / rB)^4


Now we must find C / B. Using the pulley system geometry:

C = (rB / rAh) * (rA / rC) * B, so

C / B = [(rB / rAh) * (rA / rC) * B] / B

= (rB / rAh) * (rA / rC)


Now we can substitute these into (1):

LC / LB = (IC / IB) * (C / B)

= ((rC / rB)^4) * ((rB / rAh) * (rA / rC))

(1.9/0.26)^4*(0.26/0.44)*(1/1.9) = 886.92

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