Chapter 5, Problem 15P

To determine

The coefficient of static friction between the coin and the turntable.

Answer to Problem 15P

Solution:

0.21

Explanation of Solution

Given:

Radius = $\text{13}\text{.0}\text{cm}$

Tangentialacceleration $\approx 0{\text{m/s}}^2$ (speed of turntable is slowly increased)

Angular speed after which coin starts sliding $=38.0\text{ rpm}$

Formula Used:

Frictional force, ${\text{F}}_{\text{s}}\text{=}{\text{μ}}_s\text{N}$

Where, coefficient of static friction

Static friction force provides centripetal force.

Centripetal force on coin= $m{\omega }^{2}r$

m : a mass of coin

  $\omega :$ The angular speed of the coin

r: radius of

  

  $\begin{array}{l}\omega =38.0\text{ rpm}\\ \frac{38.0\times 2\pi }{60}=3.98\text{rad/s}\end{array}$

Centripetal force on coin, $=m{\omega }^{2}r$

  $F_s={\mu }_{s}N$

Where $F_s$ : static friction force

  ${\mu }_s$ : Static friction coefficient

N : normal reaction

Apply Newton’s law in the vertical direction

  $\begin{array}{l}N-mg=0\\ N=mg\\ F_s={\mu }_s.mg\end{array}$

Centripetal force on the coin $=F_s$

Calculations:

  $\begin{array}{l}m{\omega }^{2}r={\mu }_s.mg\\ {\mu }_s=\frac{{\omega }^{2}r}{g}\\ =\frac{{\left(\frac{38.0\times 2\pi }{60}\right)}^2\times \left(0.13\right)}{\left(9.8\right)}\\ =0.21\end{array}$

Conclusion:

Hence, coefficient of static friction between turntable and coin $=0.21$ .