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 - <p>0</p>
Option 2 - <p>7mg/2<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 3 - <p>5mg<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 4 - <p>mg/5</p>
9 Views|Posted 5 months ago
Asked by Shiksha User
1 Answer
V
5 months 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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