Question

You're driving a vehicle of mass 1350kgand you need to make a turn on a flat road. The radius of curvature of the turn is. The coefficient of static friction and the coefficient of kinetic friction are both 0.25.

(a) What is the fastest speed you can drive and still make it around the turn? Invent symbols for the various quantities and solve algebraically before plugging in numbers.

(b) Which of the following statements are true about this situation?

(1) The net force is nonzero and points away from the centre of the kissing circle. (2) The rate of change of the momentum is nonzero and points away from the centre of the kissing circle.

(3) The rate of change of the momentum is nonzero and points toward the centre of the kissing circle.

(4) The momentum points toward the centre of the kissing circle.

(5) The centrifugal force balances the force of the road, so the net force is zero. (6) The net force is nonzero and points two and the centre of the kissing circle.

(c) Look at your algebraic analysis and answer the following question. Suppose that your vehicle had a mass five times as big$\left(6750kg\right)$. Now what is the fastest speed you can drive and still make it around the turn?

(d) Look at your algebraic analysis and answer the following question. Suppose that you have the originalvehicle but the turn has a radius twice as large (152 m). What is the fastest speed you can drive and still make it around the turn? This problem shows why high-speed curves on freeways have very large radii of curvature, but low-speed entrance and exit ramps can have smaller radii of curvature.

Short answer

a) The fastest speed is${\mathrm{v}}_{\max }=13.64\mathrm{m}/\mathrm{s}$ .

b) The true choices are 4 and 5.

c) The fastest speed you can drive and still make it around the turn is$\mathrm{v}=13.64\mathrm{m}/\mathrm{s}$.

d) The fastest speed you can drive and still make it around the turn is${\mathrm{v}}_{\max }=19.30\mathrm{m}/\mathrm{s}$

Step-by-step solution

Given data

A vehicle of mass$1350\mathrm{kg}$ is made a turn on a flat road.The radius of curvature of the turn is $76\mathrm{m}$.The coefficient of static friction and the coefficient of kinetic friction are both

Definition of static and kinetic friction.

The friction that occurs when an object is put on a surface is known as static friction.

Kinetic friction is caused by an item moving over a surface.

(a) Finding the fastest speed you can drive and still make it around the turn

The vehicle experiences an applied force by the floor; this force is the friction force f which is given by

$f={\mu }_{k}a$

Where a is the acceleration of the vehicle

The friction coefficient is the ratio of friction to the normal force${\mu }_k=f/F$where the normal force on the vehicle is due to the gravitational force $F=mg$.

Therefore, the friction force applied to the vehicle

role="math" localid="1656677686271" $$\begin{gathered} f=F{\mu }_k \\ =mg{\mu }_k \end{gathered}$$

Where F the normal is force andis the mass of the vehicle.

The centrifugal force$F_c$equals the rate change, which is the change in direction owing to the perpendicular rate of change.

The magnitude of the perpendicular rate change at speeds much less than the speed of light is given by

$$\begin{gathered} F_c=\frac{d\stackrel{\rightharpoonup }{p}}{dt} \\ =\frac{m{v}^2}{R} \end{gathered}$$

At the maximum speed, the net force acting on the vehicle is zero,$F_{net}=0$.

This net force equals the sum of the centrifugal forceand the friction forcebut both are in the opposite direction of each other where the friction force trying to stop the vehicle while the centrifugal force increases its speed, so the net force is given by

$$\begin{gathered} F_{net}=F_c-f \\ 0=F_c-f \\ {F}_c=f \\ {\frac{\cancel{h}v}{R}}^2=\cancel{h}g{\mu }_k \\ v=\sqrt{{\mu }_{k}Rg} \end{gathered}$$

Now put the values for, and to get the maximum speed

role="math" localid="1656677490391" $$\begin{gathered} v_{max}=\sqrt{{\mu }_{k}Rg} \\ =\sqrt{\left(0.25\right)\left(76m\right)\left(9.8m/s^2\right)} \\ =13.64m/s \end{gathered}$$

(b) Find the true statement

First statement is False, because the net force is zero and the vehicle's momentum is moving toward the kissing circle's centre.

Second statement is False, as indicated by option (1)

Third statement is False; even if the route is near the centre, the net force is zero.

Fourth statement is True, since the momentum's rate of change is moving toward the path's centre.

Fifth statement is true, because the centrifugal force opposes the friction force and the net force is zero, as stated in section (a).

Sixth statement is False, because there is no net force.

(c) Finding the fastest speed you can drive and still make it around the turn when mass will be five times

From part (a), the speed of the vehicle is given by

$v=\sqrt{{\mu }_{k}Rg}$

Because the speed does not rely on the mass of the vehicle, as indicated by equation (1), the speed will remain constant even if the mass grows fivefold.

$v=13.64m/s$

(d) Finding the fastest speed you can drive and still make it around the turn when radius will be double

The speed of the vehicle,

$v=\sqrt{{\mu }_{k}Rg}$

The speed depends on the radius of the path.

Now put the values for${\mu }_k$, R and g to get the maximum speed where$\mathrm{R}=152\mathrm{m}$

$$\begin{gathered} v_{max}=\sqrt{{\mu }_{k}Rg} \\ =\sqrt{\left(0.25\right)\left(152m\right)\left(9.8m/s^2\right)} \\ =19.30m/s \end{gathered}$$

Thus, the fastest speed you can drive and still make it around the turn is ${\mathrm{v}}_{\max }=19.30\mathrm{m}/\mathrm{s}$