91Ó°ÊÓ

When an object of mass m moves in a circular path of radius r with a constant speed v, then thenet force required to keep moving the object in circular motion acts on the object towards the center of the circle and it is given as:

$F_{\mathrm{R}}=\frac{m{v}^2}{r}$

This type of motion of the object is termed as the uniform circular motion.

Here,coin is moving in a circular motion due to static frictional force acting the coin and the turntable.

The radius of the circular path of the coin is, $r=13.0\mathrm{cm}\left(\frac{{10}^{-2}\mathrm{m}}{1\mathrm{cm}}\right)=13.0×{10}^{-2}\mathrm{m}$.

The maximum rotational speed of the turntable at which the coin slides off is, $N=38.0\mathrm{rpm}$.

The maximum velocity of the coin at which it slides off is equal to the linear velocity of the turntable at that point, it can be expressed as,

$v=N×2\mathrm{π}r$

On putting the given values, you will get:

$\begin{array}{c}v=\left(\frac{38.0\mathrm{rev}}{1\min }\right)×\left(\frac{1\min }{60\mathrm{s}}\right)×\left(\frac{2\mathrm{π}\left(13.0×{10}^{-2}\mathrm{m}\right)}{1\mathrm{rev}}\right)\\ =0.517\mathrm{m}/\mathrm{s}\end{array}$

There is no acceleration of the coin in the vertical direction, therefore normal force of the coin is equal to the weight of the coin,it can be expressed as,

$N=mg$

The only force acting on the coin in the horizontal direction is the frictional force acting between the coin and the turntable. Thus, this frictional force provides the centripetal force $\left(F_{\mathrm{R}}\right)$to the coin. So, at the point of sliding off, the equation can be expressed as,

role="math" localid="1646117006980" $\begin{array}{rcl}{F}_{\mathrm{R}}& =& \mathrm{μ}N\\ \frac{m{v}^2}{r}& =& \mathrm{μ}mg...\left(i\right)\end{array}$

Here, $\mathrm{μ}$is the coefficient of static friction between the coin and the turntable.

Using (i), the expression of coefficient of static friction can be written as:

$\mathrm{μ}=\frac{v^2}{rg}$

On putting the given values, you will get:

$\begin{array}{c}\mathrm{μ}=\frac{{\left(0.517\mathrm{m}/\mathrm{s}\right)}^2}{\left(13.0×{10}^{-2}\mathrm{m}\right)×\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}\\ =0.21\end{array}$

Thus, the value of coefficient of static friction is 0.21.