Question

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

• A. μ = sin θ
• B. μ = cot θ
• C. μ = 2 cosθ
• D. μ = 2 tan θ

Answer 1

Answer: (D)

μ = 2 tan θ

Answer 2

For upper half by the equation of motion $v ^ { 2 } = u ^ { 2 } + 2 a s$

$v ^ { 2 } = 0 ^ { 2 } + 2 ( g \sin \theta ) l / 2$ $= g l \sin \theta$

[As $u = 0 , s = l / 2 , a = g \sin \theta ]$

For lower half

$0 = u ^ { 2 } + 2 g ( \sin \theta - \mu \cos \theta )$ l /2 [As

$v = 0 , s = l / 2 , a = g ( \sin \theta - \mu \cos \theta )$ ]

⇒ $0 = g l \sin \theta + g l ( \sin \theta - \mu \cos \theta )$ [As final velocity of upper half will be equal to the initial velocity of lower half]

⇒ $2 \sin \theta = \mu \cos \theta$ ⇒ $\mu = 2 \tan \theta$