Question

Your answer is partially correct. Try again. 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 0.451 R; disk B has radius 0.201R; and disk C has radius 1.90R. 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? Number Units. the tolerance is +/-2%

Answer 1

Here Hub of radius A = 0.541R

B = 0.201R

C = 1.90R

L = I * ω

where:
I = moment of inertia
and
ω = angular speed

Therefore,

LC / LB = (IC / IB) * (ωC / ωB)

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 ωC / ωB. Using the pulley system geometry:
ωC = (rB / rAh) * (rA / rC) * ωB, so

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

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

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:

= ((1.90/ 0.201)^4) * ((0.201 / 0.541) * (1.00 / 1.90))

= 1561.05 => ratio of LC to LB

Hopefully this is correct answer.