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Question

A coin is pushed down tangentially from a position θ on a cylindrical surface, with a velocity v as shown in Fig. If the coefficient of friction between the coin and surface is μ, find the tangential acceleration of the coin.
984605_6e8b4b66bd8745b58c2fd9c6a53b91b7.png

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Solution

As the coin slides down, friction becomes kinetic and acts up along the plane. Aaprt from this, this coin experiences mg and N as shown in Fig.
Force equation in radial direction:
Fr=Nmgcosθ=mar (i)
Ft=mgsinθfk=ma1 (ii)
Law of kinetic friction: fk=μN (iii)
We know radial acceleration of the particle ar=mv2R (iv)
Substititing fk from Eq. (iii), and ar from Eq. (iv) in (ii), we have
at=gsinθμNm
Now, substituting N from Eq. (i) in Eq. (iv), we have
qt=g(sinθμcosθ)μmv2R
1027112_984605_ans_dac394ef9895466e962939cd07f4055c.png

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