A disc of mass `M` and radius `R` can rotate freely in a vertical plane about a horizontal axis at `O` distance `r` from the centre of the disc as shown in Fig. The disc is released from rest in the shown position. Answer the following questions based on the above information

Reaction force exerted by the hinge on the disc at this instant is

Options

A. `(Mg)/(5(R^(2)+2r^(2)))xxsqrt(9(R^(2)+6r^(2))^(2)+(4R^(2))^(2))`
B. `(Mg)/(5(R^(2)+2r^(2)))xx3(R^(2)+6r^(2))`
C. `(4Mg)/(5(R^(2)+2r^(2)))xxR^(2)`
D. `(Mg)/(5(R^(2)+2r^(2)))xx4R^(2)`

Answer

A

Explanation

From `tau=Ialpha`

`Mgxxrcos37^@=[(MR^(2))/2+Mr^(2)]alpha`
`alpha=(8rg)/(5[R^(2)+2^(2)])`
From energy conservation
`(Iomega^(2))/2=Mgxxrsin37^@`
`[(MR^(2))/2+Mr^(2)](omega^(2))/R=Mgrxx3/5`
`omega=sqrt((12gr)/(5[R^(2)+2r^(2)]))`
For `FBD` of the disc
`R_(x)-mgsin37^@=Ma_(r)=momega^(2)r`

`Mgcos37^@-R_(y)=Ma_(t)=Mralpha`
`R_(x)=(3Mg)/5[(R^(2)+6r^(2))/(R^(2)+2r^(2))], R_(y)=(Mg)/5[(4R^(2))/(R^(2)+2r^(2))]`
`R=sqrt(R_(x)^(2)+R_(y)^(2))=(Mg)/(5[R^(2)+2r^(2)])xxsqrt(9(R^(2)+6r^(2))^(2)+(4R)^(2))`