A coin is pushed down tangentially from an angular position θ on a circular surface, with velocity v as shown. If the cofficient of friction between the coin and surface is μ, find the tangential acceleration of the coin.
A
gsinθ
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B
gsinθ−μgcosθ
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C
gsinθ−μgcosθ−μv2/R
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D
gsinθ−μv2/R
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Solution
The correct option is Cgsinθ−μgcosθ−μv2/R
FBD of coin is shown above.
As coin slides down, friction acts along the plane.
Therefore, in radial direction: N−mgcosθ=mv2/R N=[mv2/R+mgcosθ]…(i)
In tangential direction: (mgsinθ)−f=mat mat=mgsinθ−μN (kinetic friction f=μN)
Substituting (i) mat=mgsinθ−μ[mv2R+mgcosθ] ⇒at=gsinθ−μv2R−μgcosθ