Let us assume a round object of mass m and radius R is is rolling an inclined plane wihout slipping as shown in figure . There are two forces acting on the object along the inclined plane . One is the component of gravitional of gravition force `( m g cos theta)` is the static frictional force ` (f )` .The other by the plane .As the motion is happening along the incline ,we shall write the equation for motion form the free body diagram (FBP) of the object .
for translational motion . `m g sin theta ` is the supporting force F is the opposing force .
`m g sin theta = ma `
for rotational motion .Let us take the torque with respect to the center of the object then ` m g sin theta ` cannot cause torque with through it but the frictional force f can set torque of R f .
` Rf = I alpha `
by using the relation `= a = r alpha` and moment of inerita `I= mK^2` , we get
` Rf = mK^2 (a) /(R ) , f = ma ((K^2)/(R^2))`
Now equation becomes ,
` mg sin theta - ma ((K^2 )/(R^2))= ma`
` ma sin theta = ma + ma ((K^2)/(R^2))`
` a( 1+(K^2)/(R^2))= g sin theta `
After rewriting it for acceleration , we get ,
` a= ( g sin theta ) /( (1 + (k^2)/(R^2))`
we can also find the expression for final velocity of the rolling object object by using third equation of motion for the inclined plane .
`v^2 = u^2 + 2as ` if the body starts rolling from rest `u=0` when h is the vertical height of the incline , the length of the incline s is ` s= (h )/( sin theta)`
` v^2 = 2 ( g sin theta)/((1+ (K^2)/( R^2)) ((h)/( sin theta )) = ( 2 gh ) /( ( 1+ (K^2)/(R^2)))`
By taking square root ,
`V= sqrt( ( 2gh )/( ( 1 + (K^2)/(R^2)))`
the time taken for rolling down the incline could also be from eqation of motion as ` v= u + at ` for the object which starts rolling from rest , u=0 then
` t= (v)/(a) `
` t= (sqrt((2gh )/( (1+(k^2)/(R^2)))) )((1+ (K^2)/(R^2))/( g sin theta))`
`t= sqrt((2h (1+ (K^2)/(R^2)))/( g sin^2 theta))`
The equation suggests that for a given incline the with the least of radius of gyration K will reach the bottom of the incline first .
