A solid cylinder of mass m is wrapped with an inextensible light string and is placed on a rough inclined plane as shown in the figure. The frictional force acting between the cylinder and the inclined plane is:
[ the coefficient of static friction, μs is 0.4]

Option 1 -

0

Option 2 -

7mg/2

Option 3 -

5mg

Option 4 -

mg/5

0 4 Views | Posted a month ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    a month ago
    Correct Option - 4


    Detailed Solution:

    μ_min = (I tanθ)/ (I + mR²)
    = [ (mR²/2)tanθ] / [mR²/2 + mR²] = (tanθ)/3 = (tan 60°)/3 = √3/3 = 1.732/3 = 0.5773

    Since given coefficient of static friction is less than μ_min, so body will perform rolling with slipping and kinetic friction will act
    F? = μN = μmg cosθ = (0.4) × mg cos 60° = mg/5

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