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 center 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`

`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))`