Tangential and normal components of acceleration

Tangential and normal components of acceleration

Lessons

Notes:

Position, Velocity & Acceleration

Normally we say that r(t)r(t) is a vector function, but we can also apply this to Physics and call this the position function.

Recall from Calculus 1 that taking the derivative of a position function gives the velocity function. In other words,

r(t)=v(t)r'(t) = v(t)

Also recall that taking the derivative of the velocity function gives acceleration. Therefore:

v(t)=a(t)=r(t)v'(t) = a(t) = r''(t)


Tangent & Normal Components of Acceleration
Acceleration is made of two components: tangential and normal. The tangential component is the component that is tangent to the curve, and the normal component is the component orthogonal (or perpendicular) to the curve. Putting this into an equation gives us:

a=atT+anN a = a_tT+ a_nN

Where:
aTa_T \to tangential component
aNa_N \to normal component
TT \to vector function tangent to the curve r(t)
NN \to vector function normal (orthogonal) to the curve r(t)r(t) To compute aTa_T and ana_n, we use the following formulas:

aT=r(t)r(t)r(t)a_T = \frac{r'(t) \cdot r''(t)}{||r'(t)||}
aN=r(t)×r(t)r(t)a_N = \frac{||r'(t) \times r''(t)||}{||r'(t)||}

  • Introduction
    Tangential & Normal Components of Acceleration Overview:
    a)
    Position, Velocity & Acceleration
    • r(t)r(t) \to position vector function
    • r(t)r'(t) \to velocity vector function v(t)v(t)
    • r(t)r''(t) \to acceleration vector function a(t)a(t)
    • An example of finding the acceleration function

    b)
    Tangential & Normal Components of Acceleration
    • Two components: Tangential aTa_T & Normal aNa_N
    • aT=r(t)r(t)r(t)a_T = \frac{r'(t) \cdot r''(t)}{||r'(t)||}
    • aN=r(t)×r(t)r(t)a_N = \frac{||r'(t) \times r''(t)||}{||r'(t)||}
    • Can calculate acceleration using a=aTT+aNNa=a_TT+a_NN
    • An example of finding aTa_T & aNa_N


  • 1.
    Finding the Position Vector Function
    Suppose an object's acceleration is given by a(t)=3ti+t2j+e2tka(t)=3t i+t^2j+e^{2t}k. The objects initial velocity is v(0)=i+kv(0)=i+k and the object's initial position is r(0)=ij+kr(0)=i-j+k. Determine the object's velocity and position functions.

  • 2.
    Suppose an object's acceleration is given by a(t)=cos3ti+sin2tj+4t2ka(t)= \cos 3t i+ \sin 2tj+ 4t^2k. The objects initial velocity is v(0)=i+j+2kv(0)=i+j+2k and the object's initial position is r(0)=i+2j+3kr(0)=-i+2j+3k. Determine the object's velocity and position functions.

  • 3.
    Finding the Tangent & Normal Components
    Determine the tangential and normal components of acceleration for the object whose position is given by r(t)=<t,2+3t,2t32>r(t)= \lt t, 2+3t, 2t^{\frac{3}{2}} \gt

  • 4.
    Determine the tangential and normal components of acceleration for the object whose position is given by r(t)=<cos2t,sin2t,t>r(t)= \lt \cos 2t, \sin 2t, t\gt