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

Answer

9

Explanation

As the belt does not slip `v_(P)=v_(Q)`
i.e., `r_(A) omega_(A)=r_(C)omega_(C)` [as `v=romega`]…………i
According to the given problem if `r_(A)=r,r_(C)=3r`, so eqn i becomes
`omega_(A)=3omega_(C)`………..ii
If both the wheels have the same rotational kinetic energy,
`1/2I_(A)omega_(A)^(2)=1/2I_(C)omega_C^(2)`
`(I_(A))/(I_(C))=[(omega_(C))/(omega_(A))]^(2)-[1/3]^(2)=1/9`