2.6: Tangential and Normal Components of Acceleration
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This section breaks down acceleration into two components called the tangential and normal components. Similar to how we break down all vectors into ˆi, ˆj, and ˆk components, we can do the same with acceleration. The addition of these two components will give us the overall acceleration.
Introduction
We're use to thinking about acceleration as the second derivative of position, and while that is one way to look at the overall acceleration, we can further break down acceleration into two components: tangential and normal acceleration. The tangential acceleration, denoted aTallows us to know how much of the acceleration acts in the direction of motion. The normal acceleration aN is how much of the acceleration is orthogonal to the tangential acceleration.
Remember that vectors have magnitude AND direction. The tangential acceleration is a measure of the rate of change in the magnitude of the velocity vector, i.e. speed, and the normal acceleration are a measure of the rate of change of the direction of the velocity vector.
This approach to acceleration is particularly useful in physics applications, because we need to know how much of the total acceleration acts in any given direction. Think for example of designing brakes for a car or the engine of a rocket. Why might it be useful to separate acceleration into components?
Theoretical discussion with descriptive elaboration
We can find the tangential accelration by using Chain Rule to rewrite the velocity vector as follows:
v=drdt=drdsdsdt=Tdsdt
Now, since acceleration is simply the derivative of velocity, we find that:
a=dvdt=ddt(Tdsdt)=d2sdt2T+dsdtdTdt=d2sdt2T+dsdt(dTdsdsdt)=d2sdt2T+dsdt(κNdsdt)=d2sdt2T+κ(dsdt)2N
dTds=κN
This, in turn, gives us the definition for acceleration by components.
If the acceleration vector is written as
a=aTT+aNN,
then
aT=d2sdt2=ddt|v| and aN=κ(dsdt)2=κ|v|2
To calculate the normal component of the accleration, use the following formula:
aN=√|a|2−a2T
We can relate this back to a common physics principal-uniform circular motion. In uniform circulation motion, when the speed is not changing, there is no tangential acceleration, only normal accleration pointing towards the center of circle. Why do you think this is? Hint: look in the introduction section for the difference between the two components of acceleration.
Without finding T and N, write the accelration of the motion
r(t)=(cost+tsint)ˆi+(sint−tcost)ˆj for t>0.
To solve this problem, we must first find the particle's velocity.
v=drdt=(−sint+sint+tcost)ˆi+(cost−cost+tsint)ˆj=(tcost)ˆi+(tsint)ˆj
Next find the speed.
|v|=√t2cos2t+t2sin2t=√t2=|t|
When t>0, |t| simply becomes t.
We know that aT=ddt|v|, which we can use to find that ddt(t)=1.
Now that we know aT, we can use it to find aN using Equation ???.
a=(cost−tsint)ˆi+(sint+tcost)ˆj
|a|2=t2+1
aN=√(t2+1)−(1)=t
By substituting this back into the original definition, we find that |a|=(1)T+(t)N=T+tN
Write ain the form a=aTT+aNN without finding T or N.
r(t)=(t+1)ˆi+2tˆj+t2ˆk
v=tˆi+2ˆj+2tˆk
|v|=√5+4t2
aT=12(5+4t2)−12(8t)
=4t(5+4t2)−12
aT(1)=4√9=43
a=2ˆk
a(1)=2ˆk
a(t)=2ˆk
Now we use Equation ???:
aN=√|a|2−a2T=√22−(43)2=√209=2√53
a(1)=43T+2√53N
References
- Weir, Maurice D., Joel Hass, and George B. Thomas. Thomas' Calculus: Early Transcendentals. Boston: Addison-Wesley, 2010. Print.