Question

A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of $1.70m/s^2$. The car makes it one-quarter of the way around the circle before it skids off the track. From these data, determine the coefficient of static friction between the car and the track.

Short answer

The coefficient of static friction between the car and the track is ${\mu }_s=0.572$.

Step-by-step solution

Newton’s second law

The second law states that the acceleration of an object depends upon the net force acting on the object and the mass of the object.
$\mathit{F}\mathbf{=}\mathit{m}\mathbf{·}\mathit{a}$

Apply Newton’s second law to find the vertical, tangential and radial direction of a car

Consider, the given values,

$v_i=0$

$a_t=1.7m/s^2$

$\Delta x=\frac{1}{4}\left(2\pi r\right)=\frac{\pi r}{2}$

Thus, apply the Newton’s second law to the car in vertical position,

$$\begin{gathered} \sum F_y=n-mg=0 \\ n=mg ................ \left(1\right) \end{gathered}$$

Since, the static frictional force must compensate for the tangential and centripetal direction, when the car travels on a circular path.

Apply the Newton’s second law to a car in tangential direction,

$$\begin{gathered} \sum F_t=f_t \\ =m{a}_t ........ \left(2\right) \end{gathered}$$

f_t is a tangential component of friction.

Apply the Newton’s second law to a car in radial direction,

$$\begin{gathered} \sum F_c=f_c \\ =m{a}_c \\ =m\frac{v_t^2}{r} ........... \left(3\right) \end{gathered}$$

f_c is the radial component of friction.

Hence, by applying particle under constant acceleration model to the car in tangential direction, final tangential velocity v_t can be found,

$$\begin{gathered} v_t^2=v_i^2+2a_t\Delta x \\ {v}_t^2=0+2a_t\left(\frac{\pi r}{2}\right) \\ =a_t\pi r ........... \left(4\right) \end{gathered}$$

Substitute equation (4) into (3)

$$\begin{gathered} f_c=m\frac{a_t\pi r}{r} \\ =\pi m{a}_t ........... \left(5\right) \end{gathered}$$

Use Pythagorean Theorem to find the coefficient of static friction

Since, the tangential and radial components of friction are perpendicular, the static friction can be found by using Pythagorean Theorem,

f_s=$\sqrt{f_c^2+f_T^2}$

Hence, the frictional force is the product of the coefficient of static friction $\left({\mu }_s\right)$and the normal force (n).

Substitute for f_c and f_r from equation (2) and (5),

${\mu }_{s}n=\sqrt{{\left(m{a}_t\right)}^2+{\left(\pi m{a}_t\right)}^2}$

Substitute for normal force (n) from equation (1) and take (ma_t) as common factor.

${\mu }_{s}mg=m{a}_t\sqrt{1+{\pi }^2}$

Solve,

${\mu }_s=\frac{a_t\sqrt{1+{\pi }^2}}{g}$

Substitute the value of a and g in the above equation,

$$\begin{gathered} a_t=1.7 m/s^2, \\ g=9.8 m/s^2 \end{gathered}$$

Therefore,

${\mu }_s=\frac{(1.7)\sqrt{1+{\pi }^2}}{9.8}=0.572$

Therefore, the coefficient of the static friction between the car and the track is data-custom-editor="chemistry" ${\mu }_s=$0.572.