1.

A force F acts tangentially on a part of a uniform disc of mass 'm' and radius 'R' hinged at point O. The force exerted by hinge on the portion of disc initially.

Answer»

`((4-PI)/(pi))F`
`(4F)/(pi)`
`F((pi-2)/(pi))`
`(F)/(2)`

Solution :`tau= IALPHA rArr alpha = (FR)/(mR^(2))2= (2F)/(mR)`

POSITION of C.O.M `= (2R)/(3)(sin((pi)/(6)))/(((pi)/(6))) = (2R)/(pi)=R`
Acceleration of C.O.M `= (4F)/(mpi)`
Hence `F+F' = (4R)/(mpi)m`
`F' = ((4-pi)/(pi))F`


Discussion

No Comment Found

Related InterviewSolutions