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In a car's suspension system, each wheel is connected to a vertical spring; these springs absorb shocks when the car travels on bumpy roads. In one car, each spring has a spring constant of 4.6 × 10⁴ N/m.

If this 1300 kg car is driven at 23 m/s through the bottom of a circular dip in the road that has a radius of 600 m, by how much do these springs compress compared to when the car is driven on a flat road?

User Dstftw
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2 Answers

3 votes

Answer:

Step-by-step explanation:

Given this 1300 kg car is driven at 23 m/s through a circular dip in the road of radius of 600m,

its centrifugal acceleration = speed^2 / radius

= 23^2 / 600

= 0.8817 m/s^2

The centrifugal force in turn = mass * acceleration

= 1300 * 0.8817

= 1146.17 N

This force is absorbed by suspension springs of the four wheels.

Given spring constant of 4.6 x 10^4 N/m and 4 springs total,

the extra compression = Force / number of springs / spring constant

= 1146.17 / 4 / 4.6 x 10^4

= 0.006229 m

= 6.23 mm

Note that this is the extra compression on springs during drive-through of the dip compared to when the car is driven on a flat road. There is a constant compression on springs due to weight of the car which is the same whether the car is driving through a dip or on a flat road.

User John Towers
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3 votes

Answer:

6.23 mm

Step-by-step explanation:

The vertical acceleration at the bottom of the dip (in addition to that required to counter gravity) is ...

a = v²/r = (23 m/s)²/(600 m) = 529/600 m/s²

Then the force on the car (in addition to that required to counter gravity) is ...

F = Ma = (1300 kg)(529/600 m/s²) = 1146 1/6 N

The additional force supplied by each of the 4 springs is 1/4 of this, so the additional spring compression is ...

((1146 1/6 N)/4)/(4.6×10^4 N/m) ≈ 0.00623 m = 6.23 mm

The springs are compressed an additional 6.23 mm to provide the required vertical acceleration.

_____

We have to assume that the path followed by the car's center of mass has a radius of 600 m. We're also assuming that the rotational motion of the car about its center of mass does not require any differential force between front and back springs, as it will be at a constant rate for a constant speed through the dip. And, we assume that the mass of the car is evenly among the four springs (often not the case).

User Runejuhl
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