InterviewSolution
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InterviewSolution
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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? |
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Answer» 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_(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` |
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