Question

A coin is pushed down tangentially from an angular position $\theta$ on a circular surface, with velocity $v$ as shown. If the cofficient of friction between the coin and surface is $\mu$, find the tangential acceleration of the coin.

• A. $g\sin \theta$
• B. $g\sin \theta -\mu g\cos \theta$
• C. $g\sin \theta -\mu g\cos \theta -\mu v^2/R$
• D. $g\sin \theta -\mu v^2/R$

Answer 1

The correct option is C $g\sin \theta -\mu g\cos \theta -\mu v^2/R$


$FBD$ of coin is shown above.
As coin slides down, friction acts along the plane.
Therefore, in radial direction:
$N-mg\cos \theta =m{v}^2/R$
$N=[m{v}^2/R+mg\cos \theta ]\dots \left(i\right)$
In tangential direction:
$\left(mg\sin \theta \right)-f=m{a}_t$
$m{a}_t=mg\sin \theta -\mu N$ (kinetic friction $f=\mu N$)
Substituting (i)
$m{a}_t=mg\sin \theta -\mu [\frac{m{v}^2}{R}+mg\cos \theta ]$
$\Rightarrow a_t=g\sin \theta -\frac{\mu v^2}{R}-\mu g\cos \theta$