Engineering Example 1

A road has a bend with radius of curvature 100 m. The road is banked at an angle of $1 0^{\circ}$ . At what speed should a car take the bend in order not to experience any (net) side thrust on the tyres?

Mathematical statement of the problem

Figure 28 below shows the forces on the car.

Figure 28 :

{A vehicle rounding a banked bend in the road.}

In the figure $R$ is the reactive force of the ground acting on the vehicle. The vehicle provides a force of $m g$ , the weight of the vehicle, operating vertically downwards. The vehicle needs a sideways force of $\frac{m v^{2}}{r}$ in order to maintain the locally circular motion.

We have used the following assumptions:

The sideways force needed on the vehicle in order to maintain it in circular motion (called the centripetal force) is $\frac{m v^{2}}{r}$ where $r$ is the radius of curvature of the bend, $v$ is the velocity and $m$ the mass of the vehicle.
The only force with component acting sideways on the vehicle is the reactive force of the ground. This acts in a direction normal to the ground. (That is, we assume no frictional force in a sideways direction.)
The force due to gravity of the vehicle is $m g$ , where $m$ is the mass of the vehicle and $g$ is the acceleration due to gravity ( $\approx 9.8 m s^{- 2}$ ). This acts vertically downwards.
The problem we need to solve is ‘What value of $v$ would be such that the component of the reactive force of the ground exactly balances the sideways force of $\frac{m v^{2}}{r}$ ?’ This will give us the maximum velocity at which the vehicle can take the bend.

Mathematical analysis

We can split the reactive force of the ground into two components. One component is in the horizontal direction and the other in a vertical direction as in the following figure:

Figure 29 :

{Reaction forces on the car}

The force of $\frac{m v^{2}}{r}$ must be provided by a component of the reactive force in the horizontal direction i.e.

$R sin (1 0^{\circ}) = \frac{m v^{2}}{r} (1)$

However the reactive force must balance the force due to gravity in the vertical direction therefore

$R cos (1 0^{\circ}) = m g (2)$

We need to find $v$ from the above equations. Dividing Equation (1) by Equation (2) gives

$tan (1 0^{\circ}) = \frac{v^{2}}{r g} \Rightarrow v^{2} = r g tan (1 0^{\circ})$

We are given that the radius of curvature is 100 m and that $g \approx 9.8 m s^{- 2}$ . This gives

$v^{2} \approx 100 \times 9.8 \times 0.17633$

$\Rightarrow v^{2} \approx 172.8$

$\Rightarrow v \approx 13.15 m s^{- 1}$ (assuming $v$ is positive)

Interpretation

We have found that the maximum speed that the car can take the bend in order not to experience any side thrust on the tyres is $13.15 m s^{- 1}$ . This is $13.15 \times 60 \times 60 ∕ 1000$ kph $= 47.34$ kph. In practice, the need for a margin of safety would suggest that the maximum speed round the bend should be $13 m s^{- 1}$ .

Exercises

A bend on a stretch of railway track has a radius of 200 m. The maximum sideways force on the train on this bend must not exceed 0.1 of its weight.
1. What is the maximum possible speed of the train on this bend?
2. How far before this bend should a train travelling at $30 m s^{- 1}$ begin to decelarate given that the maximum braking force of the train is 0.2 of its weight?
3. What modelling assumptions have you made? Comment on their validity.
The diagram shows a portion of track of a one-way fairground ride on which several trains are to run. $A B$ and $C D$ are straight. $B C$ is a circular arc with the dimension shown. Because $B C$ is also on a bridge, safety regulations require that the rear of one train must have passed point $C$ before the front of the next train passes point $B$ . Trains are 30 m long.

If the maximum sideways force on a train can be no more than 0.1 of its weight, find the shortest time it can take for a train to travel from $B$ to $C$ . Hence find the minimum time between the front of one train passing point $B$ and its rear end passing point $C$ . Recommend a minimum distance between trains.

1. According to Equation (2.2) on page 35, during travel round the bend the sideways force on the train is given by
  $M r ω^{2} = \frac{M v^{2}}{r}$
  
  The weight of the train is $M g$ . Given that $\frac{M v^{2}}{r} \leq 0.1 M g$ , the maximum possible speed, $v_{max}$ , is given by $v_{max} = \sqrt{0.1 r g}$ .
  
  Using $r = 200$ m and $g = 9.81 m s^{- 2}$ , this implies that $v_{max} = 14.0 m s^{- 1}$ to 3 s.f.
2. Given initial speed is $30 m s^{- 1}$ , and final speed is $14 m s^{- 1}$ and maximum braking force is 0.2 $M g$ , implying acceleration is $- 0.2 g$ . Then, using the formula ‘ $v^{2} = u^{2} + 2 a s$ ’, where $u$ is initial speed, $v$ its final speed, $a$ is acceleration and $s$ is distance travelled, gives
  $20 g = 900 - 0.4 g s$ or $s = 179.358$ m.
  
  This suggests that braking should begin about 180 m before the start of the bend.
3. Assumptions include constant maximum braking, negligible thinking time and no skidding.
The shortest time on the circular bend will be taken when the train is moving at the maximum possible speed.
This will occur when $\frac{M v^{2}}{r} = 0.1 M g$ . If $r = 144$ m and $g = 9.81 m s^{- 2}$ , this implies $v_{max} = 11.885 m s^{- 1}$ .

The length of $B C$ is $144 \frac{π}{3} = 150.796$ m.

The time taken for any point on the train to move from $B$ to $C$ is $\frac{150.796}{11.885} = 12.687$ s.

So, given that the length of each train is 30 m, to make sure that the rear of one train has passed $C$ before the front of the next train arrives at $B$ , a minimum time between the trains of $(12.687 + \frac{30}{11.885})$ s $= 15.212$ s is required. After including a small safety margin, each train should be 16 s apart. Assuming that the trains are moving at a constant speed of $11.885 m s^{- 1}$ , this implies that they should be at least 190 m apart.

3 Engineering Example 1

3.1 Car velocity on a bend

Exercises

3 Engineering Example 1

3.1 Car velocity on a bend

Exercises

Answer