Question

A coin is pushed down tangentially from a 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.

Answer 1

As the coin slides down, friction becomes kinetic and acts up along the plane. Aaprt from this, this coin experiences $mg\downarrow$ and $N\uparrow$ as shown in Fig.
Force equation in radial direction:
$\sum F_r=N-mgcos\theta =m{a}_r$ (i)
$\sum F_t=mgsin\theta -f_k=m{a}_1$ (ii)
Law of kinetic friction: $f_k=\mu N$ (iii)
We know radial acceleration of the particle $a_r=\frac{m{v}^2}{R}$ (iv)
Substititing $f_k$ from Eq. (iii), and $a_r$ from Eq. (iv) in (ii), we have
$a_t=gsin\theta -\frac{\mu N}{m}$
Now, substituting N from Eq. (i) in Eq. (iv), we have
$qt=g(sin\theta -\mu cos\theta )-\frac{\mu m{v}^2}{R}$