The acceleration of the centre of mass of uniform solid disc rolling down an inclined plane of angle `alpha` is `xg sin alpha`. Find x

Answer

0.67

Explanation

The acceleration of the body which is rolling down an inclined plane of angle `alpha` is
`a = (g sin theta)/(1+ K^(2)/R^(2))`
where, K = radius of gyration,
R=radius of body, Now, here the body is a uniform solid disc.
So, `K^(2)/R^(2) =1/2 therefore a=(g sin theta)/(1+1/2)` or `a = (g sin theta)/(3//2)` or `a = (2g sin alpha)/3`