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=(g\sin \theta )/(2)`

Step-by-step 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* r=I* \alpha`

`f_r* r=mr^2* \alpha`

in pure Rolling
`a=\alpha r`

`\alpha =(a)/(r)`

`f_r=ma`

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

`mg\sin \theta =ma+ma`

`2ma=mg\sin \theta`

`a=(g\sin \theta )/(2)`