Question

The figure 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 0.439R; disk B has radius 0.240R; and disk C has radius 1.69R. 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?

Answer 1

The formula for angular momentum L is:

(1) L = I * ω,

where:

(2) I = moment of inertia

and

(3) ω = angular speed

Therefore,

(4) LC / LB = (IC / IB) * (ωC / ωB)

For a disk,

(5) I = m * r^2 / 2 [see Source 1]

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:

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

Ratioing (5) and then substituting (6):

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

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

= (rC / rB)^4

Now we must find ωC / ωB. Using the pulley system geometry:

(8) ωC = (rB / rAh) * (rA / rC) * ωB, so

(9) ωC / ωB = [(rB / rAh) * (rA / rC) * ωB] / ωB

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

Now we can substitute (7) and (9) into (4):

(10) LC / LB = (IC / IB) * (ωC / ωB)

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

Since all the terms are now expressed in ratios of radii, we can substitute the given values:

= ((1.69 / 0.240)^4) * ((0.240 / 0.439) * (1.00 / 1.69))

= 795.4  <<<=== ratio of LC to LB................Answer.

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