Question

Figure 11-51 shows an overhead view of a ring that can rotate about its center like a merry-go-round. Its outer radius R2 is 0.8 m, its inner radius R1 is R2/2,its mass M is 8.8 kg, and the mass of the crossbars at its center is negligible. It initially rotates at an angular speed of 6.5 rad/s with a cat of mass m = M/4on its outer edge, at radius R2. By how much does the cat increase the kinetic energy of the cat-ring system if the cat crawls to the inner edge, at radius R1?

Answer 1

Moment of inertia of the ring=0.5(m)((R2^2)-(R1^2)) (about z axis, ie its rotating axis)
=0.5*8.8*((0.8^2)-(0.4^2))
=2.112 kg(m^2)
Inertia of cat =m(r^2) (r being the distance of the cat from center)
mass of cat=8.8/4=2.2kg
With the cat sitting on outer edge
Kinetic energy of the system= 0.5*2.112*(6.5^2) +0.5*2.2*(0.8^2)(6.5^2) J
=74.36J
In absence of external for the angular momentum of the system remains conserved
Initial angular momentum=(2.112*6.5)+(2.2*0.64*6.5)=22.88 kg(m^2)/s
Let final angular velocity be ω

so final momentum= (2.112*ω)+(2.2*0.16*ω)=2.464ω

So,2.464ω=22.88

or,ω=9.28rad/s
With cat sitting on inner edge

Kinetic energy of the system=0.5*2.112*(9.28^2) +0.5*2.2*(0.4^2)(9.28^2) J

=106.09J

So gain in kinetic energy=106.09-74.36 J=31.73J

Answer 2

where did you get the 0.64 from when calculating initial angular momentum?