Question

A uniform thin circular ring rolls without slipping down an incline making an angle θ with the horizontal. What is its acceleration? (Enter the magnitude. Use any variable or symbol stated above along with the following as necessary: g for the acceleration of gravity.)

Answer 1

Answer:$a=\frac{g\sin \theta }{2}$

Explanation:

Given

inclination is $\theta$

let M be the mass and r be the radius of uniform circular ring

Moment of Inertia of ring $I=mr^2$

Friction will Provide the Torque to ring

$f_r\times r=I\times \alpha$

$f_r\times r=mr^2\times \alpha$

in pure Rolling $a=\alpha r$

$\alpha =\frac{a}{r}$

$f_r=ma$

Form FBD $mg\sin \theta -f_r=ma$

$mg\sin \theta =ma+ma$

$2ma=mg\sin \theta$

$a=\frac{g\sin \theta }{2}$