Question

A cylinder of radius r and mass m rolls without slipping down a plane inclined at an angle theta

Answer 1

According to question 

mg sinA - f = ma --------------(1) 

in pure rolling alpha = a/r where alpha = angular acceleration 

Now 

net torque = I alpha where I = moment of inertia 

f r = I alpha 

fr = I a/r 

therefore 

f = I a/r^2. -------------------(2) 

putting value of f in (1) 

mg sin A - I a/r^2 = ma 

here I of solid cylinder = mr^2/2 

mg sin A - ma/2 = ma 

a = 2g sin A / 3 

Now 

f = Ia /r^2 from (2) 

f = ma / 2 which should be less than or equal to u N. otherwise body will slip (where u is coefficient of friction and N is normal acting on cylinder which is equal to mg cos A) 

ma / 2 less than or equal to u mg cos A 

m 2gsin A / 3 2 less than or equal to u mg cos A 

tan A should be less than or equal to 3u

follow me ,!

Answer 2

Explanation:

According to question

mg sinA - f = ma --------------(1)

in pure rolling alpha = a/r where alpha = angular acceleration

Now

net torque = I alpha where I = moment of inertia

f r = I alpha

fr = I a/r

therefore

f = I a/r^2. -------------------(2)

putting value of f in (1)

mg sin A - I a/r^2 = ma

here I of solid cylinder = mr^2/2

mg sin A - ma/2 = ma

a = 2g sin A / 3

Now

f = Ia /r^2 from (2)

f = ma / 2 which should be less than or equal to u N. otherwise body will slip (where u is coefficient of friction and N is normal acting on cylinder which is equal to mg cos A)

ma / 2 less than or equal to u mg cos A

m 2gsin A / 3 2 less than or equal to u mg cos A

tan A should be less than or equal to 3u

Therefore for rolling shutout slipping A should be less than or equal to tan inverse A