Question

An inclined plane makes an angle of 30 with the horizontal. A solid sphere rolling down this inclined plane from rest without slipping has a linear acceleration equal to

Answer 1

We know that :

a = g sin@ - u g cos@

So,

a = 9.8 X 1/2 - 0 X √(3)/2

a = 9.8/2

a = 4.9 m/s2

Answer 2

Answer:

5g /14  

Explanation:

Net force on c.o.m (center of mass) = mg.sinФ

In this case friction force is acting upwards.

For linear motion,     mg.sinФ - f = ma .......1

for angular motion

            fR = Iλ     ........2        

from 1 & 2

mg.sinФ = ma + Iλ/R         ...........3

the body is rolling without slipping then we apply a = λR

mgsinΦ= ma + 2ma/5                              (Isphere = 2mR²/5)

        a = 5gsinФ/7

         a  =5g/14                                        (sin30 = 1/2)