Chapter 12, Problem 77PQ

To determine

The coefficient of static friction between the car’s tires and the road.

Answer to Problem 77PQ

The coefficient of static friction between the car’s tires and the road is $0.98$ .

Explanation of Solution

The force of static friction on the automobile must act forward and then more and more inward on the tires, to produce both tangential acceleration and to maintain the centripetal acceleration.

Write the equation for the tangential force.

  $F_t=ma$                                                                                                                  (I)

Here, $F_t$ is the magnitude of the tangential force, $m$ is the mass of the automobile and $a$ is the magnitude of the tangential acceleration.

The centripetal force is provided by the radially inward component of force.

Write the equation for the centripetal force.

  $F_r=m{a}_c$                                                                                                                (II)

Here, $F_r$ is the centripetal force and $a_c$ is the centripetal acceleration of the car’s tires.

Write the equation for $a_c$ .

  $a_c=\frac{v^2}{r}$                                                                                                                 (III)

Here, $v$ is the translational speed and $r$ is the radius of the semicircular curve.

Write the equation for the angular speed.

  $\omega =\frac{v}{r}$

Here, $\omega$ is the angular speed.

Multiply both numerator and denominator of right hand side of equation (III) and put the above equation in it.

  $\begin{array}{c}{a}_c=\frac{v^2}{r^2}r\\ ={\omega }^{2}r\end{array}$

Put the above equation in equation (II).

  $F_r=m{\omega }^{2}r$                                                                                                            (IV)

The centripetal force increases with time as the speed increases and it reaches its maximum value when the automobile has travelled $3/4$ into a semicircular curve or in other words, when its angle has increased by $3\pi /4$ .

Write the equation for $\omega$ in terms of initial angular velocity.

  ${\omega }^2={\omega }_0^2+2\alpha \Delta \theta$

Here, ${\omega }_0$ is the initial angular velocity, $\alpha$ is the angular acceleration and $\Delta \theta$ is the angular displacement.

Put the above equation in equation (IV).

  $F_r=m\left({\omega }_0^2+2\alpha \Delta \theta \right)r$

The automobile starts from rest so that its initial angular velocity must be zero.

Substitute $0$ for ${\omega }_0$ and $3\pi /4$ for $\Delta \theta$ in the above equation.

  $\begin{array}{c}{F}_r=m\left(0+2\alpha \left(\frac{3\pi }{4}\right)\right)r\\ =m\left(\frac{3\pi }{2}\right)r\alpha \end{array}$                                                                                         (V)

Write the equation for the linear acceleration in terms of angular acceleration.

  $a=r\alpha$

Put the above equation in equation (V).

  $F_r=m\left(\frac{3\pi }{2}\right)a$                                                                                                      (VI)

The force of friction when the car slips is equal to the magnitude of the total force found using the Pythagorean theorem.

$f_s=\sqrt{F_t^2+F_r^2}$ (VII)

Here, $f_s$ is the force of friction.

Write the equation for $f_s$ .

  $f_s={\mu }_s{F}_N$                                                                                                          (VIII)

Here, ${\mu }_s$ is the coefficient of static friction and $F_N$ is the normal force

The normal force is equal to the weight of the car.

Write the equation for $F_N$ .

  $F_N=mg$

Here, $g$ is the acceleration due to gravity.

Put the above equation in equation (VIII).

  $f_s={\mu }_{s}mg$                                                                                                            (IX)

Put equations (I), (VI) and (IX) in equation (VII) and rewrite the equation for ${\mu }_s$ .

  $\begin{array}{c}{\mu }_{s}mg=\sqrt{{\left(ma\right)}^2+{\left(m\left(\frac{3\pi }{2}\right)a\right)}^2}\\ =\sqrt{m^2{a}^2+m^2{a}^2{\left(\frac{3\pi }{2}\right)}^2}\\ =ma\sqrt{1+{\left(\frac{3\pi }{2}\right)}^2}\\ {\mu }_s=\frac{a}{g}\sqrt{1+{\left(\frac{3\pi }{2}\right)}^2}\end{array}$                                                                            (X)

Conclusion:

Given that the tangential acceleration is $2.00{\text{ m/s}}^2$ . The value of acceleration due to gravity is $9.81{\text{ m/s}}^2$ .

Substitute $2.00{\text{ m/s}}^2$ for $a$ and $9.81{\text{ m/s}}^2$ for $g$ in equation (X) to find ${\mu }_s$ .

  $\begin{array}{c}{\mu }_s=\frac{2.00{\text{ m/s}}^2}{9.81{\text{ m/s}}^2}\sqrt{1+{\left(\frac{3\pi }{2}\right)}^2}\\ =0.98\end{array}$

Therefore, the coefficient of static friction between the car’s tires and the road is $0.98$ .