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'

Consider a thin uniform circular ring rolling down in an inclined plan

1.	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'
Answer» Solution : 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)`