Two uniform disks are mounted (like merry-go-round) on frictionless bearings on

Question

Two uniform disks are mounted (like merry-go-round) on frictionless bearings on the same axle and can be brought together so that they couple and rotate as one unit. The first disk, with mass M and radius R is set spinning clockwise at angular speed of Wo. The second disk, with mass 3M and Radius R/2 is set spinning counterclockwise with the sample angular speed of Wo. They then coupld together. (the rotaional inertia of a disk is mr^2/2). What is their angular speed and direction of rotation after coupling? How much is the energy loss of the two-disk system caused by such coupling?

Explanation / Answer

so it's conservation of angular momentum. Iw=mr^2/2*w First: M*R^2/2*Wo Second: 3m*(R/2)^2*Wo Adding together we get M*R^2/2*Wo-3M*(R/2)^2*Wo=(M*R^2/2+3M*(R/2)^2/2)*W Solve it out we get: M*R^2/2*Wo(1-3/2)=M*R^2/2(1+3/2)*W (1-3/2)/(1+3/2)*Wo=W -0.2*Wo=W (so going the way of the second one) then for energy: Iw^2+Iw^2 - Icombinedw^2= loss Wo^2*(M*R^2/2*-3M*(R/2)^2)-(M*R^2/2+3M*(R/2)^2/2)*0.2^2*Wo^2 Can simplify it more, but we want loss, not a ratio right?

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