Consider a thin uniform circular ring rolling down in an inclined plane without slipping. Compute the linear acceleration along the inclined plane if the angle of inclination is 45'

Explanation

The linear acceleration along the inclined plane can be computed by `a=(g sin theta)/(1+(K^(2))/(R^(2)))`
For a thin uniform circular ring, axis passing through its center is `I=MR^(2)`
`:.K^(2)=R^(2) rArr (K^(2))/(R^(2))=1`
And the angle of inclination `theta=45^(@) rArr sin 45^(@)=(1)/(sqrt(2))`
Hence, `a=((g)/(sqrt(2)))/(1+1), a=(g)/(2 sqrt(2)) ms^(-2)`