1
$\begingroup$

We know there are two types of accelaration in circular motion, one is centripetal acceleration and the other one is tangential acceleration. The resultant of these two is the net acceleration $a$. But what is the use of this net acceleration? I mean in case of tangential acceleration, it increases linear speed and centripetal acceleration changes direction, but what good does the net acceleration do?

$\endgroup$
1
  • $\begingroup$ It changes the velocity vector. $\endgroup$
    – nasu
    Sep 11, 2021 at 23:53

2 Answers 2

1
$\begingroup$

But what is the use of this net acceleration?

The net force (centripetal force plus tangential) associated with the two accelerations can be use to determine whether or not a vehicle will skid while both accelerating and cornering, compared to cornering alone or accelerating alone.

If the vehicle is both accelerating and cornering, the total friction force will be greater than either the lateral friction associated with cornering alone or the longitudinal friction associated with accelerating alone. Since the total friction force is shared between the two, the vehicle will slip sooner if both accelerating and cornering at the same time, than if only accelerating or only cornering.

This can be illustrated by using the so called Kamm circle of friction. Refer to the figures below of a vehicle accelerating forward (up in the figure) and cornering to the right. $F_{Lat}$ is the centripetal force and $F_{Lon}$ is the tangential force. The centripetal acceleration is then $F_{Lat}/M$ and the tangential acceleration is $F_{Lon}/M$.

The Kamm circle assumes that the coefficient of static friction is the same both longitudinal and laterally, as well as assumes the normal load supported by the tire is the same for both accelerating and cornering.

The figures are an overhead view of one of the vehicle tires. The vehicle direction of motion is upward. The circle radius represents the maximum possible static friction force, or $u_{s}N$ where $N$ is the load (weight) of the vehicle supported by the tire.

Fig 1 shows the vehicle accelerating only where the longitudinal friction force enabling acceleration, $F_{Lon}$, equals the maximum possible static friction force, i.e., where loss of traction is impending.

Fig 2 shows the vehicle only cornering (to the right) where the lateral friction force, $F_{Lat}$ which equals the centripetal force, equals the maximum possible static friction force, i.e., where skidding is impending.

Fig 3 shows the vehicle both accelerating and cornering. As can be seen, the total friction force reaches the maximum possible static friction force sooner that when only accelerating or cornering.

Hope this helps.

enter image description here

$\endgroup$
1
$\begingroup$

Acceleration if a vector quantity. The time derivative of velocity, also a vector:

$$\vec a = \frac{d\vec v}{dt}$$

There are several uses of it. One is to extrapolate the velocity, say in a predictive filter:

$$ \vec v(t+\delta t)\approx \vec v(t) + \delta t\vec a$$

another is to compute the so-called $g$-load:

$$ \frac 1 g ||\vec a|| $$

If $\vec s(t)$, where $\vec v(t) = d\vec s(t)/dt$, is circular, or tracing arc, one can project the acceleration onto the velocity:

$$ a_{\parallel} = \vec a \cdot \frac{\vec v}{||\vec v||}$$

and also the rejection:

$$ a_{\perp} = \vec{a}-a_{\parallel}$$

If you are a Formula One engineer looking at telemetry from the car, you need both to understand your car's performance. $ a_{\perp}$ tells you about you cornering, while $ a_{\parallel}$ is important for understanding braking into and acceleration out of the corner.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.