Question

Two bicycle tires are set rolling with the same initial speed of 3.90 m/s along a long, straight road, and the distance each travels before its speed is reduced by half is measured. One tire is inflated to a pressure of 40 psi and goes a distance of 18.6 m ; the other is at 105 psi and goes a distance of 93.1 m . Assume that the net horizontal force is due to rolling friction only and take the free-fall acceleration to be g = 9.80 m/s2 . What is the coefficient of rolling friction (Uk) for the tire under low pressure?

Answer 1

Answer:

0.03129

Explanation:

$\mu$ = Coefficient of rolling friction

g = Acceleration due to gravity = 9.8 m/s²

Initial speed of both the bicycles is

u = 3.9 m/s

Final velocity

$v=\frac{u}{2}=\frac{3.9}{2}=1.95\ m/s$

a = Acceleration =$\mu g$

Equation of motion

$v^2-u^2=2as\\\Rightarrow v^2-u^2=2\mu gs\\\Rightarrow \mu=\frac{v^2-u^2}{2gs}\\\Rightarrow \mu=\frac{1.95^2-3.9^2}{2\times 9.8\times 18.6}\\\Rightarrow \mu=0.03129$

The coefficient of rolling friction for the tire under low pressure is 0.03129

Answer 2

Answer:

u_k = 0.0313  

Explanation:

Given:

- Distance traveled by low pressure tire s = 18.6 m

- Initial velocity v_i = 3.90 m/s

- Final velocity v_f = 0.5*v_i

- Coefficient of friction: u_k

Find:

- What is the coefficient of rolling friction (Uk) for the tire under low pressure?

Solution:

- Make a free body diagram of the tire. Three forces act on the tire, Weight of the tire W, Friction force at the circumference between the road and tire parallel to road in direction of motion F_f. And the normal contact force exerted by the road on the tire F_n. Using Newton's Second law of motion we have:

- In direction of motion          F_f = m*a

- Apply equilibrium conditions on the tire in vertical direction:

                                                F_n - W = 0

                                                F_n = W

                                                F_n = m*g

- The friction force  F_f is a fraction of normal contact force F_n with a constant of coefficient of friction as follows:

                                                F_f = u_k*F_n

                                                F_f = u_k*m*g

- Hence our equation of motion becomes:

                                                u_k*m*g = m*a

                                                u_k = a / g

- The deceleration of tire as it rolls on the road can be computed by using the third equation of motion:

                                           v_f^2 - v_i^2 = 2*a*s

Where v_f = 0.5*v_i,

                                           0.25*v_i^2 -*v_i^2 = 2*a*s

                                            a = -0.75*v_i^2 / 2*s

- The coefficient of rolling friction can now be calculated:

                                           u_k = |-0.75*v_i^2 / 2*s*g|

- Input values:

                                           u_k = 0.75*3.9^2 / 2*18.6*9.81

                                           u_k = 0.0313