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Question: The upper half of an inclined plane of inclination θ is perfectly smooth while the lower half is rou...

The upper half of an inclined plane of inclination θ is perfectly smooth while the lower half is rough. A body starting from the rest at top comes back to rest at the bottom if the coefficient of friction for the lower half is given by

A

μ = sin θ

B

μ = cot θ

C

μ = 2 cos θ

D

μ = 2 tan θ

Answer

μ = 2 tan θ

Explanation

Solution

For upper half

v2=u2+2al/2=2(gsin⁡θ)l/2=glsin⁡θv ^ { 2 } = u ^ { 2 } + 2 a l / 2 = 2 ( g \sin \theta ) l / 2 = g l \sin \theta

For lower half

⇒ 0=u2+2g(sin⁡θ−μcos⁡θ)l20 = u ^ { 2 } + 2 g ( \sin \theta - \mu \cos \theta ) \frac { l } { 2 }

⇒ −glsin⁡θ=gl(sin⁡θ−μcos⁡θ)- g l \sin \theta = g l ( \sin \theta - \mu \cos \theta )

⇒ μcos⁡θ=2sin⁡θ⇒μ=2tan⁡θ\mu \cos \theta = 2 \sin \theta \Rightarrow \mu = 2 \tan \theta