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Document Title
Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]

Substituting equation (2.57) in equation (2.56), we get:

dr dr dr dθ
= r + r
dt dt dθ dt

From the discussion on circular motion, remember that the radius unit vector r when

differentiated with respect to θ , we get unit tangent vector T . Also, we know that (dθ dt) is the

instantaneous angular velocity (ω). Substituting these facts in the above expression, we get:
dr dr

= r + Tω r - - - - (2.58)
dt dt
Since we seek accelerations, we differentiate equation (2.58) again:

dr
Let us consider the first term in equation . , which is : r
dt

Differentiating the above term by chain rule, we get:

2
d dr d r dr dr
r = r +
dt dt dt 2 dt dt

Recalling the derivative of unit radius vector, which is dr dt = Tω and substituting this, in the

above expression, we get:
2
d dr d r dr
Tω
r = r + - - - - (2.59)
dt dt dt 2 dt
Let us now consider the second term in equation . , that is: Tω r

We shall differentiate the above term using chain rule of differentiation. However, the chain rule
must applied once again, to (ωr). Mathematically, we have,

d dT d ωr dT

Tω r = ωr + T , from circular motion, we have: = −r ω
dt dt dt dt
d dω dr dω dr

2
= Tω r = −r ω ωr + r + ω T = −ω rr + r T + ω T - - - - (2.60)
dt dt dt dt dt
Adding equations (2.59) and (2.60), we get the rate of, rate of change of the radius vector, which is:
2
2
d r d r dr dω dr

2
= r + ωT − ω rr + r T + ω T - - - - (2.61)
dt 2 dt 2 dt dt dt

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