CBSE Class 11-science Answered
The minimum coefficient of friction for a solid sphere to roll without slipping on an inclined plane of inclination 45° , is?
Asked by sureshsonusharma511980 | 16 Oct, 2019, 06:28: PM
Expert Answer
Figure shows a sphere rolling on an inclined plane without slipping.
Down word force, which is weight mg, is resolved into two components.
One component mgcosθ is pressing against the inclined plane to get equal reaction force R.
Other component mgsinθ is parallel to inclined plane that is driving the sphere downwards
against friction force μR , where μ is the friction coefficient.
Net force driving the sphere downwards, F = mgsinθ - μR = ( mgsinθ - μ mg cosθ )
hence acceleration of sphere, a = F/m = g( sinθ - μ cosθ ) ..........................(1)
Sphere is rolling due to the torque created by friction force μR
For rotational motion, we have, τ = Iα ...............(2)
where τ is the torque, τ = μR × r ......................(3)
where r is radius of sphere.
I is moment of inertia of sphere, I = (2/5) mr2
α is angular acceleration of sphere, α = a/r.
eqn.(2) is rewritten as, μR × r = μ mg cosθ × r = (2/5) mr2 × a/r ...................(4)
eqn.(4) is simplified as, μ g cosθ = (2/5)a or a = (5/2) μ g cosθ ..................(5)
By equating (1) and (5), we get, (5/2) μ g cosθ = g( sinθ - μ cosθ ) or μ = (2/7)tanθ
we are given that θ = 45o , hence μ = (2/7)
Answered by Thiyagarajan K | 16 Oct, 2019, 11:02: PM
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