A coin is pushed down tangentially from position `theta` on a cylindrical surface with a velocity `v` as shown in Fig. If the coefficient of friction between the coin and surface is `mu` find the tangential acceleration of the coin

Explanation

As the coin slides down , friction brcause kinetic and acts up along the plane . Apart from this coin experiences `mg darr` and `N uarr` as shown in Fig
Force equation in radial direction :
`sum F_(r) = N - mg cos theta = ma_(r)`…..(i)
`F_(t) = mg sin theta - f_(k)= ma_(r)`…..(ii)
Law of kinetic friction `f_(k) = mu N` .....(iii)
We know radius acceleration of the particle `a_(r) = (mv^(2))/(R )`.....(iv)

Substititing `f_(k)` Eq (iii) and `a_(r)` from Eq (iv) in (ii) , we have
`a_(r) = g sin theta - (mu N)/(m)`
Now , Substititing `N` from wq (i) in in Eq (v) we have
`a_(t) = g (sin theta - mu cos theta)- (mu m v^(2))/(R )`