Figure 11-43 shows three rotating, uniform disks that are coupled by belts. One

Question

Figure 11-43 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 0.562R; disk B has radius 0.212R; and disk C has radius 2.36R. 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?

Explanation / Answer

The formula for angular momentum L is:

(1) L = I * ?,

where:

(2) I = moment of inertia
and
(3) ? = angular speed

Therefore,

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

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 ?C / ?B. Using the pulley system geometry:

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

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

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


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

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

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

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

= ((1.51 / 0.221)^4) * ((0.221 / 0.512) * (1.00 / 1.51))

= 622.99 = 6.23e2 <<<=== ratio of LC to LB

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