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 μ_s = 0.25. 

1. How much is the force of friction acting on the cylinder? 

2. What is the work done against friction during rolling? 

3. If the inclination θ of the plane is increased, at what value of θ does the cylinder begin to skid, and not roll perfectly?

Answer 1

1. Consider the figure

N= mg cosθ …………….. (1)

f ≤ μ_s N (if not slipping)

= μ_s mg cosθ

also mg sinθ – f = ma ………(2)

f.R = 1 α = \(\frac{mR^2}{2}\) α M.1. of cylinder

⇒ f = \(\frac{ma}{2}\)

∴ a = \(\frac{2gsin \,\theta}{3}\)

∴ f =  \(\frac{mg\,sin \,\theta}{3}\) \(\frac {10 \times 10 \times sin\, 30}{3}\)≈ 16.6 N

2. Since, the cylinder, is rolling, the point of its contact with plane is stationery hence the power delivered by friction is zero, hence the work done is zero.

3. Slipping occurs when

f ≥ μ_s mg cosθ

⇒ \(\frac{mg\,sin \theta}{3}\)≥ μ_s mg cosθ

⇒ tan θ ≥ μ_s × 3 = 0.75

⇒ θ ≥ 37°.