The coefficient of static friction for the tires of a race car is 0.950 and the coefficient of kinetic friction is 0.800. The car is on a level circular track of 50.0 m radius on a planet where compared to Earth's . If the car is to be able to travel at the same speed on the planet as on Earth, the radius of the track on the planet must be ____ times as large as the radius of the track on Earth.
A) 0.250
B) 0.500
C) 1.00
D) 2.00
E) 4.00

Question

The coefficient of static friction for the tires of a race car is 0.950 and the coefficient of kinetic friction is 0.800. The car is on a level circular track of 50.0 m radius on a planet where  compared to Earth's  . If the car is to be able to travel at the same speed on the planet as on Earth, the radius of the track on the planet must be ____ times as large as the radius of the track on Earth.
A) 0.250 
B) 0.500 
C) 1.00 
D) 2.00 
E) 4.00

Quizplus · Accepted Answer

The maximum speed of the car on the track on Earth is determined by the maximum centripetal force it can sustain without slipping:

$F_{c,max} = \mu_smg$

where $\mu_s$ is the coefficient of static friction, $m$ is the mass of the car, $g$ is the acceleration due to gravity. At maximum speed $v$, the centripetal force acting on the car is provided by the frictional force:

$F_{c} = \mu_kmg$

where $\mu_k$ is the coefficient of kinetic friction. Equating the two forces and solving for $v$, we get:

$v = \sqrt{\mu_sgr}$

where $r$ is the radius of the track.

For the same value of $v$, the centripetal acceleration on the planet is 11eb6adc_ed59_6518_b0cf_23760c811831_TB8245_11 times smaller than on Earth, because $g_{	ext{planet}} = 11eb6adc_ed59_6518_b0cf_23760c811831_TB8245_11 g_{	ext{Earth}}$. Therefore, the radius of the track on the planet must also be 11eb6adc_ed59_6518_b0cf_23760c811831_TB8245_11 times smaller in order to keep the centripetal force the same:

$r_{	ext{planet}} = \frac{1}{11eb6adc_ed59_6518_b0cf_23760c811831_TB8245_11} r_{	ext{Earth}}$

Substituting this expression into the previous equation, we get:

$v_{	ext{planet}} = \sqrt{\mu_s g_{	ext{planet}} r_{	ext{planet}}} = \sqrt{\mu_s (g_{	ext{Earth}}/11eb6adc_ed59_6518_b0cf_23760c811831_TB8245_11) (r_{	ext{Earth}}/11eb6adc_ed59_6518_b0cf_23760c811831_TB8245_11)} = \sqrt{\mu_s g_{	ext{Earth}} r_{	ext{Earth}}}/11eb6adc_ed59_6518_b0cf_23760c811831_TB8245_11$

To keep $v_{	ext{planet}} = v_{	ext{Earth}}$, we need:

$\sqrt{\mu_s g_{	ext{Earth}} r_{	ext{Earth}}} = \sqrt{\mu_s g_{	ext{Earth}} r_{	ext{planet}}}/11eb6adc_ed59_6518_b0cf_23760c811831_TB8245_11$

Squaring both sides and solving for $r_{	ext{planet}}$, we get:

$r_{	ext{planet}} = \frac{1}{11eb6adc_ed59_6518_b0cf_23760c811831_TB8245_11^2} r_{	ext{Earth}}$

Therefore, the radius of the track on the planet must be 11eb6adc_ed59_6518_b0cf_23760c811831_TB8245_11^2 times smaller than on Earth, or 4 times smaller approximately. The answer is (E) 4.00.

The Coefficient of Static Friction for the Tires of a Race