- #1

- 330

- 0

If b=radius of the small hoop and 3b is the radius of the large hoop, then the center of mass of the system is -b/2. The moment of inertia (I) of the small hoop I2= m*b^2, and I1=27*m*b^2 for the large one. The moment of inertia of the large hoop about the center of mass of the system is Il=27*m*b^2 + 3*m*(-b/2)^2 ...from the parallel axis theorem. Therefore Il=111*m*b^2/4. The moment of inertia of the small hoop about the center of mass of the system is Is=13*m*b^2/4...parallel axis theorem. The total moment about the center of mass is It = Is + Il = 124*m*b^2/4 = 31*m*b^2

m*g*2*b = .5*4*m*v^2 + .5*It*w^2

=> 2*m*g*b = 2*m*v^2 + 2*It*v^2/b^2. Therefore

m*g*b = m*v^2 + 31*m*v^2. Therefore

g*b = 32*v^2 => v = (g*b/32)^.5. And v for the large hoop is:

vl=6*(g*b/32)^.5

My approach to this question may be wrong as the correct answer

is vl= 6*(g*b/199)^.5. Any help would be welcome.