A thread is passing through a hole at the centre of frictionless table. At the upper end a block of mass `0.5 kg` is tied and a block of mass `8.0 kg` is tied at the lower end which is freely hanging. The smaller mass is rotate, on the table with a constant angular velocity about the axis passing through the hole so as to balance the heavier mass. If the mass of the hanging block is changed from `8.0 kg` to `1.0 kg`, what is the fractional change in the radius and the angular velocity of the smaller mass so that it balances the hanging mass again?

Explanation

For circular motion of a body tied to a string on a horizontal plane
`((mv^(2))/r)=T`
Here as tension is provided by the hanging mass `M`.
i.e., `T=g`, therefore
So,`((mv^(2))/r)=Mg`
According to the given problem,
`((mv_(1)^(2)//r_(1)))/((mv_(2)^(2)//r_(2)))=(M_(1)g)/(M_(2)g)=8/1` ...........i
Now as force `T` is central, so angular momentum is also conserved i.e.,
`mv_(1)r_(1)=mv_(2)r_(2)`..........ii
So substituting the value of `v_(1)//v_(2)` from Eq. i Eq. i
`[(r_(2))/(r_(1))]xx(r_(2))/(r_(1))=8/1`.............iii
i.e., `(r_(2))/(r_(1))=2`
so that `(/_\r)/(r_(1))=(r_(2)-r_(1))/(r_(1))=(r_(2))/(r_(1))-1=2=1`
Furthermore as in circular motion `v=romega`
so, `(omega_(2))/(omega_(1))=(v_(2))/(v_(1))xx(r_(2))/(r_(1))=[(r_(1))/(r_(2))]^(2)`
[as from Eq. ii `(v_(2))/(v_(1))=[(r_(1))/(r_(2))]`
`implies(omega_(1))/(omega_(2))=[1/2]^(2)=1/4`
[as from eq. iii `(r_(2))/(r_(1))=2`]
So, `(/_\omega)/(omega)=(omega_(2)-omega_(1))/(omega_(1))=(omega_(2))/(omega_(1))=1=1/4-l=-3/4`