In the above question disc B is brought in contact with disc A, they acquire a common angular velocity in time t. The average frictional torque on one disc by the other during this period is :

In the above question disc B is brought in contact with disc A, they a

1.	In the above question disc B is brought in contact with disc A, they acquire a common angular velocity in time t. The average frictional torque on one disc by the other during this period is :
Answer» `(2Iomega)/(3t)` `(9Iomega)/(2t)` `(9Iomega)/(4T)` `(3Iomega)/(2t)` Solution :LET `omega`. be the angular VELOCITY when the two discs GET combined. Then by conservation of angular momentum, `(I+2I)omega.=Ixx2omega+2Iomega` or `omega.=(4omega)/(3)""...(i)` Now `omega.=omega+at=omega+(tau)/(2I)xxt` (where `tau` is the average FRICTIONAL torque) or `(4omega)/(3)-omega=(tau)/(2I).t` `tau=(2Iomega)/(3t)`

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In the above question disc B is brought in contact with disc A, they acquire a common angular velocity in time t. The average frictional torque on one disc by the other during this period is :

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