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