A disc of a mass `M` and radius `R` can rotate freely in vertical plane about a horizontal axis at `O`. Distant `r` from the centre of disc as shown in the figure. The disc is released from rest in the shown position.

The angular acceleration of disc when `OC` rotates by an angles of `37^(@)` is

Options

A. `(8rg)/(5[R^(2)+2r^(2)])`
B. `(5rg)/(4[R^(2)+2r^(2)])`
C. `(10rg)/(3[R^(2)+2r^(2)])`
D. `(8rg)/(5R^(2))`

Answer

A

Explanation

From `tau=Ialpha`
`rArrMg xxr cos37^(@)=[(MR^(2))/2+Mr^(2)]alpha`
`rArralpha=(8 rg)/(5[R^(2)+2r^(2)])`
From energy -conservation.
`(I omega^(2))/2=Mgxxr sin 37^(@)`
`rArr[(MR^(2))/2+Mr^(2)](omega^(2))/2=Mgrxx3/5`
`rArr omega=sqrt((12 gr)/(5[R^(2)+2r^(2)]))`
From `FBD` of disc,
`R_(x)-Mg sin 37^(@)=Ma_(r)=Momega^(2)r`

`Mg cos37^(@)-R_(y)=Ma_(t)=Mralpha`
`rArrR_(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)])`
`[sqrt(g(R^(2)+6r^(2))^(2)+(4R^(2))^(2))]`