Two wheels `A` and `C` are connected by a belt `B` as shown in figure. The radius of `C` is three times the radius of `A`. What would be the ratio of the rotational inertia `(I_(A)//I_(C))` if both the wheels have the same rotational kinetic energy?

Explanation

As the belt does not slip, `v_(p) = v_(0)`
i.e., `r_(A) omega_(A) = r_(C) omega_(C)`………….(i)
According to the given problem if `r_(A) = r, r_(C)=3r`, so Eq. (i) becomes
`omega_(A) = 3omega_(C)`………(ii)
If bot the wheels have the same rotational kinetic energy, then
`1/2 I_(A) omega_(A)^(2) = 1/2I_(C) omega_(C)^(2)`

`I_(A)/I_(C) = [omega_(C)/omega_(A)]^(2) = [1/3]^(2) = 1/9`