CBSE Class 12-science Answered
Sir plz solve the question.
Asked by Soma | 21 Feb, 2019, 12:03: PM
Expert Answer
Figure shows various forces acting on a rolling circular object.
Weight which is the force die to acceleration due to gravity is resolved into two components,
one component (mg sinθ) parallele to inclined plane and other one ( mg cosθ) normal to inclined plane.
Normal Reaction force is N and friction force is μN = μ mg cosθ
linear acceleration a is obtained from m×a = (mg sinθ - μ mg cosθ)
Hence linear acceleration a = (g sinθ - μ g cosθ)
for rotational motion, Torque τ due to friction force is giving angular acceleration α
Hence τ = I×α
It is assumed the circular object is having moment of inertia = (1/2)mR2 .
Torque is product of friction force and perpendicular distance which is radius of circular object.
Linear acceleration is R times angular acceleration
hence we have, μ mg cosθ × R = (1/2)mR2 × (g sinθ - μ g cosθ)/R ........................(1)
from eqn.(1) we get minimum friction coefficient μ = (1/3)tanθ
Answered by Thiyagarajan K | 21 Feb, 2019, 02:35: PM
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