Answer :
Answer:
Explanation:
Given
both ring and disc have mass m and radius r
Let [tex]\theta [/tex]be the inclination of plane with horizontal
[tex]f_r[/tex] be the friction force which provides torque to object
[tex]mgsin\theta -f_r=ma[/tex]
and [tex]I\alpha =f_r\times r[/tex] ---1
where I=moment of inertia
[tex]mgsin\theta -\frac{I\aplha }{r}=ma[/tex]
and we know for pure rolling a=\alpha \times r[/tex]
thus
[tex]a=\frac{gsin\theta }{1+\frac{I}{mr^2}}[/tex]
Thus Linear acceleration of ring with [tex]I=mr^2[/tex]
[tex]a=\frac{gsin\theta }{2}[/tex]
For disc [tex]I=\frac{mr^2}{2}[/tex]
[tex]a=\frac{2gsin\theta }{3}[/tex]
Angular acceleration for ring
[tex]\alpha =\frac{gsin\theta }{2r}[/tex]
For disc
[tex]\alpha =\frac{2gsin\theta }{3r}[/tex]