1.

A circular disc is rolling down an inclined plane without slipping. If the angle of inclination is 30^(@), the acceleration of the disc down the inclined plane is :

Answer»

g
`(g)/(2)`
`(g)/(3)`
`(SQRT(2))/(3)g`

Solution :`a=(gsintheta)/(1+(I)/(MR^(2)))`
For disc, `I=(1)/(2)mr^(2)impliesa=(gsin30^(@))/(1+(1)/(2)(mr^(2))/(mr^(2)))`
`a=(2)/(3)gxx(1)/(2)=(1)/(3)g`


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