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 µs = 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 θ of the plane is increased, at what value of θ does the cylinder begin to skid, and not roll perfectly?" Question 7.31System Of Particles And Rotational Motion

Answer 1

Here,
mass of cylinder (m) = 10 Kg
Radius (r) = 15 cm = 0.15 m
Inclination of plane (∅) = 30°
coefficient of friction (u) = 0.25


(a) friction = mgsin∅/(1 + mr²/I)
For solid cylinder (I) = mr²/2
Fr = mgsin∅/3
= 10 × 9.8 × 1/2 × 1/3
= 16.3 N


(b) force of friction acts perpendicular to the direction of the displacement.
So , work done against during rolling .
W = Fscos90° = 0


(c) for rolling without slipping
use formula,
u = tan∅/( 1 + mr²/I)
For solid cylinder
u = tan∅/3
tan∅ = 3u
= 3×0.25 = 0.75 = tan36°54'
∅ = 36°54' ≈ 37°

Answer 2

Answer:

check the attachment...