Question

A cylinder of mass `10 kg` and radius `15 cm` is rolling perfectly on a plane of inclination `30^(@)`. The coefficient of static friction is `mu_(s) = 0.25`.
(a) How much is the force of friction acting on the cylinder ?
(b) What is the work done against friction during rolling ?
(c ) If the inclination `theta` of plane is increased, at what value of `theta` does the cylinder begin to skid and not roll perfectly ?

Answer 1

Here, `m = 10 kg, r = 15 cm = 0.15 m`
`theta = 30^(@), mu_(s) = 0.25`
Acceleration of the cylinder down the incline, `a = (2)/(3) g sin theta = (2)/(3) xx 9.8 sin 30^(@) = (9.8)/(3) m//s^(2)`
(a) Force of friction, `F = mg sin theta - ma = m(g sin theta - a) = 10 (9.8 sin 30^(@) - (9.8)/(3)) = 16.4 N`
(b) During rolling, the point of contact is at rest. Therefore, work done against friction is zero.
(c ) For rolling without slipping/skidding, `mu = (1)/(3) tan theta`
`tan theta = 3 mu = 3 xx 0.25 = 0.75`
`theta = 37^(@)`