A uniform disk of mass 10 m and radius 3.0 r can rotate freely about its fixed c

Question

A uniform disk of mass 10m and radius 3.0r can rotate freely about its fixed center like a merry-go-round. A smaller uniform disk of mass m and radius r lies on top of the larger disk, concentric with it. Initially the two disks rotate together with an angular velocity of 19 rad/s. Then a slight disturbance causes the smaller disk to slide outward across the larger disk, until the outer edge of the smaller disk catches on the outer edge of the larger disk. Afterwards, the two disks again rotate together (without further sliding).

(a) What then is their angular velocity about the center of the larger disk?

(b) What is the ratio K/K0 of the new kinetic energy of the two-disk system to the system's initial kinetic energy?

Explanation / Answer

a) Calculation of w, the angular velocity:

I1w1 = I2w2 ----->[1]

where, I1w1 and I2w2 are the angular momenta

Calculating moments of inertia:

I1 = 1/2*10m*[3r]2 +1/2*mr2 = 91/2*mr2

I2 = 1/2*10m*[3r]2 +1/2*mr2 + 1/2*2m*[2r]2 = 99/2*mr2

Putting the values back in the equation,

91w1 = 99w2

w2 = (91/99)*19

w2 = 17.4646 rad/s

angular velocity about the center of the larger disk 17.4646 rad/s

__________________

b)

Calculation of ratio

K/K0 = I1w12/I2w22

K/K0 = [91/2*mr2*192] /  [99/2*mr2*17.46462]

K/K0 = 1.0879

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