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 ondisk 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 0.403R; disk B has radius 0.216R;and disk C has radius 1.56R. 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 ofdisk C to that of disk B?



Fig. 11-43
Problem 36.

Answer 1

Given that RA = R, RA' = 0.403R, RB = 0.216R, Rc = 1.56R. the density of B or C = ?,
thickness = t
The ratio of the magnitude of
the angular momentum L of diskC to that of disk B
mass MB = pRB2t?, Mc =pRc^2t?,
moment of inertia IB =MBRB^2/2 =pRB^4t?/2,
Ic =pRc^4t?/2
angular momentum LB = IB ?B =pRB^4t??B/2,
Lc= pRc^4t??c/2
ratio Lc/LB =(Rc?c/RB?B)*(Rc/RB)^3
note ?ARA =?cRc, ?ARA'= ?BRB,
so Rc?c/RB?B =RA/RA'
then
Lc/LB =(RA/RA')*(Rc/RB)^3
= (R/0.403R) (1.56R/0.216R)^3
= 934.77 or 935

Answer 2

the tangential velocity or the velocity of the belt is constant
so

?a*0.471R=?b0.221R

?aR=?c1.51R

?c/?b=0.221/0.471*1.51

ratio=0.310

Answer 3

the tangential velocities will be equal.
hence,
Vt(A) = Vt(C)
Wa * R = Wb * 2.36R

also
Vt(B) = Vt(C)
Wa*.526R = Wb * 2.36R

hence Lc/Lb = 2.36/(.212*.562)