what is the magnitude & tangent of the direction of velocity of the particle at t = 2.0 s?
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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
NOTE
dTds=κN
This, in turn, gives us the definition for acceleration by components.
DEFINITION: ACCELERATION VECTOR
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.
EXAMPLE 2.6.1
Without finding T and N, write the accelration of the motion
r(t)=(cost+tsint)ˆi+(sint−tcost)ˆjfor 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 aNusing Equation 2.6.11.
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
EXAMPLE 2.6.2
Write ain the form \mathbf{a}=a_T\mathbf{T}+a_N\mathbf{N}without finding T or N.
\mathbf{r}(t)=(t+1) \hat{ \textbf{i} } +2t \hat{ \textbf{j} } +t^2 \hat{ \textbf{k} }
\mathbf{v}=t \hat{ \textbf{i} } +2 \hat{\textbf{j}} +2t \hat{ \textbf{k} }
|\mathbf{v}|=\sqrt{5+4t^2}
a_T=\dfrac{1}{2}\left ( 5+4t^2 \right )^{-\dfrac{1}{2}}(8t)
=4t(5+4t^2)^{-\dfrac{1}{2}}
a_T(1)=\dfrac{4}{\sqrt{9}}=\dfrac{4}{3}
\mathbf{a}=2 \hat{ \textbf{k} }
\mathbf{a}(1)=2 \hat{ \textbf{k} }
\mathbf{a}(t)=2 \hat{ \textbf{k} }
Now we use Equation \ref{Normal}:
\begin{align} a_N=\sqrt{|\mathbf{a}|^2-a_T^2}&=\sqrt{2^2-\left ( \dfrac{4}{3} \right )^2} \\ & =\sqrt{\dfrac{20}{9}} \\ &=\dfrac{2\sqrt{5}}{3} \end{align}
\mathbf{a}(1)=\dfrac{4}{3}\mathbf{T}+\dfrac{2\sqrt{5}}{3}\mathbf{N}
References
Weir, Maurice D., Joel Hass, and George B. Thomas. Thomas' Calculus: Early Transcendentals. Boston: Addison-Wesley, 2010. Print.
Contributors
Alagu Chidambaram (UCD)
Integrated by Justin Marshall.
Back to top
2.5: Velocity and Acceleration
2.7: Parametric Surfaces
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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
NOTE
dTds=κN
This, in turn, gives us the definition for acceleration by components.
DEFINITION: ACCELERATION VECTOR
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.
EXAMPLE 2.6.1
Without finding T and N, write the accelration of the motion
r(t)=(cost+tsint)ˆi+(sint−tcost)ˆjfor 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 aNusing Equation 2.6.11.
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
EXAMPLE 2.6.2
Write ain the form \mathbf{a}=a_T\mathbf{T}+a_N\mathbf{N}without finding T or N.
\mathbf{r}(t)=(t+1) \hat{ \textbf{i} } +2t \hat{ \textbf{j} } +t^2 \hat{ \textbf{k} }
\mathbf{v}=t \hat{ \textbf{i} } +2 \hat{\textbf{j}} +2t \hat{ \textbf{k} }
|\mathbf{v}|=\sqrt{5+4t^2}
a_T=\dfrac{1}{2}\left ( 5+4t^2 \right )^{-\dfrac{1}{2}}(8t)
=4t(5+4t^2)^{-\dfrac{1}{2}}
a_T(1)=\dfrac{4}{\sqrt{9}}=\dfrac{4}{3}
\mathbf{a}=2 \hat{ \textbf{k} }
\mathbf{a}(1)=2 \hat{ \textbf{k} }
\mathbf{a}(t)=2 \hat{ \textbf{k} }
Now we use Equation \ref{Normal}:
\begin{align} a_N=\sqrt{|\mathbf{a}|^2-a_T^2}&=\sqrt{2^2-\left ( \dfrac{4}{3} \right )^2} \\ & =\sqrt{\dfrac{20}{9}} \\ &=\dfrac{2\sqrt{5}}{3} \end{align}
\mathbf{a}(1)=\dfrac{4}{3}\mathbf{T}+\dfrac{2\sqrt{5}}{3}\mathbf{N}
References
Weir, Maurice D., Joel Hass, and George B. Thomas. Thomas' Calculus: Early Transcendentals. Boston: Addison-Wesley, 2010. Print.
Contributors
Alagu Chidambaram (UCD)
Integrated by Justin Marshall.
Back to top
2.5: Velocity and Acceleration
2.7: Parametric Surfaces
Recommended articles
13.4: Motion in SpaceWe have now seen how to describe curves in the plane and in space, and how to determine their properties, such as arc length and curvature. All of thi...
13.4: Motion Along a Curve
2.2: Arc Length in SpaceFor this topic, we will be learning how to calculate the length of a curve in space. The ideas behind this topic are very similar to calculating arc l...
2.3: Curvature and Normal Vectors of a CurveFor a parametrically defined curve we had the definition of arc length. Since vector valued functions are parametrically defined curves in disguise, w...
2.7: Parametric SurfacesWe have now seen many kinds of functions. When we talked about parametric curves, we defined them as functions from \mathbb{R} to \mathbb{R
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A tangential unit vector can be defined at each point of the trajectory, in the direction tangent to it and in which increases. Figure 3.1 shows the tangential unit vector in three points A, B and P of a trajectory.
Tangential unit vector in points of a trajectory
Figure 3.1: Tangential unit vector in three points of the trajectory.
Note that there are two different tangential unit vectors at point P. One of them is tangent to the curve between B and P and the other is tangent to the curve between P and Q. The velocity vector of an object that follows that trajectory will always be in the direction of the tangential unit vector, or in the opposite direction. In points such as P, where there are two different tangential unit vectors, the velocity will necessarily be zero; the object will be momentarily at rest at that point, starting to move again in a different direction from the one it had before it stopped.
In points where the velocity is different from zero, there is always only one tangential unit vector , which determines the direction of the velocity vector. Namely, the velocity vector can be written,
(3.1)
As mentioned in Chapter 2, the velocity vector is given by the derivative of the position vector
(3.2)
Since the position vector depends on the choice of the origin of the reference frame, it does not have any relation to the tangential unit vector (see Figure 3.2). However, the displacement vector is independent of the choice of origin and therefore, equation 3.2 ensures that the velocity vector is also independent of the choice of the origin.
Tangential unit vector in points of a trajectory
Figure 3.1: Tangential unit vector in three points of the trajectory.
Note that there are two different tangential unit vectors at point P. One of them is tangent to the curve between B and P and the other is tangent to the curve between P and Q. The velocity vector of an object that follows that trajectory will always be in the direction of the tangential unit vector, or in the opposite direction. In points such as P, where there are two different tangential unit vectors, the velocity will necessarily be zero; the object will be momentarily at rest at that point, starting to move again in a different direction from the one it had before it stopped.
In points where the velocity is different from zero, there is always only one tangential unit vector , which determines the direction of the velocity vector. Namely, the velocity vector can be written,
(3.1)
As mentioned in Chapter 2, the velocity vector is given by the derivative of the position vector
(3.2)
Since the position vector depends on the choice of the origin of the reference frame, it does not have any relation to the tangential unit vector (see Figure 3.2). However, the displacement vector is independent of the choice of origin and therefore, equation 3.2 ensures that the velocity vector is also independent of the choice of the origin.
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