Question

A bicyclist travels in a circle of radius 25.0 mat a constant speed of 9.00 m/s. The bicycle–rider mass is 85.0 kg. Calculate the magnitudes of

(a) the force of friction on the bicycle from the road and

(b) the netforce on the bicycle from the road.

Short answer

1. The force of friction on the bicycle from the road is 275 N.
2. The net force is 877 N.

Step-by-step solution

Given data

• Radius of the circle: R = 25.0 m.
• Velocity of the bicycle: v = 9.00 m/s.
• Mass of bicycle-rider: m = 85.0 kg.

Understanding the concept

It is the force of friction that provides the centripetal force required for circular motion. The free-body diagram is shown below. The magnitude of the acceleration of the cyclist as it moves along the horizontal circular path is given by$\frac{{\mathbf{v}}^{\mathbf{2}}}{\mathbf{R}}$, where v is the speed of the cyclist and R is the radius of the curve.

Formula:

$f_s=\frac{m{v}^2}{R}$

Draw the free body diagram and write force equations

The horizontal component of Newton’s second law is $f_s=\frac{m{v}^2}{R}$, where $f_s$is the static friction exerted horizontally by the ground on the tires.

Similarly, $F_N$if is the vertical force of the ground on the bicycle and m is the mass of the bicycle and rider, the vertical component of Newton’s second law leads to $F_N=mg=85\times 9.8=833\mathrm{N}$.

(a) Calculate the magnitude of the force of friction on the bicycle from the road

Force of friction on the bicycle from the road:

$f_s=\frac{m{v}^2}{R}$

Substitute the values in the above expression, and we get,

role="math" localid="1660976789922" $$\begin{gathered} f_s=\frac{85.0\mathrm{kg}\times \left(9.00\mathrm{m}/{\mathrm{s}}^2\right)}{25.0\mathrm{m}} \\ {f}_s=275.4\mathrm{N} \\ \approx 275 \end{gathered}$$

Thus, the force of friction on the bicycle from the road is 275 N.

(b) Calculate the magnitude of the net force on the bicycle from the road

Since the frictional force $f_s$and the normal force exerted by the road $F_N$are perpendicular to each other, the magnitude of the force exerted by the ground on the bicycle is,

$F=\sqrt{f_s^2+F_N^2}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} F=\sqrt{{275}^2+{833}^2} \\ F=877\mathrm{N} \end{gathered}$$

The force exerted by the ground on the bicycle F is at an angle:

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{275\mathrm{N}}{833\mathrm{N}}\right) \\ =18.3° \end{gathered}$$

It will make an 18.3-degree angle with respect to the vertical axis.

Thus, the net force is 877 N.