1.

A disc of radius a is ridigly attached at its circumference to a rod of length 3a and the combination suspeded vertically from the other end of the rod. It is swinging in the plane of the disc such that the centre of disc has velocity v. If mass of disc and rod is m

Answer»

Kinetic energy of the DISC is `1/2mv^(2)`
Kinetic energy of the disc `1/2mv^(2)+1/2((ma^(2))/2)(v/a)^(2)`
Kinetic energy of the systemof disc and ROD is `[1/2 mv^(2)+1/2((ma^(2))/2)(v/(4a))^(2)]+1/2[(m(3a)^(2))/3(v/(4a))^(2)]`
Kinetic energy of the system of disc and rod is `1/2[(m(3a)^(2))/3+((ma^(2))/2+m(4a)^(2))](v/(4a))^(2)`

Solution :KE of disc `=1/2MV_(C)^(2)+1/2I_(c) omega^(2)`
`=1/2 mv^(2)+/2xx1/2ma^(2)(v/(4a))^(2)`
KE or rod `=1/2I omega^(2) =1/2 (M(3a)^(2))/3(v/(4a))^(2)`
KE of system `=1/2 I` system `omega^(2)`
`=1/2[(ma^(2))/2+(4a)^(2)+(m(3a)^(2))/2](v/(4a))^(2)`


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