ANALYSIS OF PARABOLIC MOTION ACCELERATED AND DECELERATED MOTION
In analyzing the Parabolic Motion of a cam follower, we should consider the figure:
Inspection:
In the figure on the left side the AC is the accelerated motion while CB is the constant decelerated
motion. The distance of AC is equal to the distance of CB which is equal to h/2.
Parabolic Equation is s = C θ2 for the Equation 1
𝟏
s = Vo t + 𝟐 a t 2
where the following represents:
θ = the cam angle in radians.
C = is a constant
S = follower displacement at any time (t)
DISPLACEMENT EQUATION
a. For the displacement equation, the follower displacement from A to C and 0 ≤ θ ≤ β/2 where β is
the cam angle turned through while the follower is a given rise of distance h in radians.
In this equation the follower displacement (s) is equals to h/2 and the cam that has turned
through an angle (θ) is equals to β/2, so we can now derive the equation 2 from equation 1 by
substituting:
ℎ
2
=
𝛽
𝐶( )2
2
ℎ ℎ
𝐶 = ( )( )2
2 𝛽
𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1
𝐶=
2ℎ
𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2
𝛽2
The displacement equation would be:
𝑠=(
2ℎ 2
)𝜃
𝛽2
𝛽
2
b. From C to B follower displacement and ≤ θ ≤ β
𝜃 2
𝑠 = ℎ [1 − 2 (1 − ) ]
𝛽
VELOCITY EQUATION
a. For the velocity equation, the follower distance from A to C and 0 ≤ θ ≤ β/2 where, dθ / dt = ῳ =
angular velocity of the cam in radians/seconds.
𝑉=
𝑑𝑠
2ℎ
𝑑𝜃
= ( 2 )(2𝜃)( )
𝑑𝑡
𝛽
𝑑𝑡
The velocity equation would be:
𝑉=
4ℎῳ
𝜃
𝛽2
b. Velocity Equation of the follower from C to B and β/2 ≤ θ ≤ β
𝜃 2
𝐹𝑟𝑜𝑚: 𝑠 = ℎ [1 − 2 (1 − ) ]
𝛽
Derivation:
𝑑𝑠
𝜃 2 𝑑𝜃
𝑉=
= [ℎ − 2ℎ (1 − ) ]
𝑑𝑡
𝛽
𝑑𝑡
𝑑𝜃
𝜃 𝑑𝜃
𝑉 = ℎ ( ) − 2ℎ(1 − )2
𝑑𝑡
𝛽 𝑑𝑡
𝜃 −(𝛽)(1) − (𝜃)(0)
𝑉 = 2ℎ((2)(1 − )(
)ῳ
𝛽
𝛽2
𝑉=
4ℎῳ
𝜃
(1 − )
𝛽
𝛽
And the Maximum velocity of the follower, it occurs at point C where θ = β/2
𝑉𝑐 = 𝑉𝑚𝑎𝑥 =
4ℎῳ 𝛽
2ℎῳ
( )=
2
𝛽
2
𝛽
ACCELERATION EQUATION
a. For the acceleration equation, the follower distance from A to C and 0 ≤ θ ≤ β/2 where, dθ / dt =
ῳ = angular velocity of the cam in radians/seconds.
𝑎=
𝑑𝑉 4 ℎ 𝜃 𝑑𝜃
= 2 ( )
𝑑𝑡
𝛽
𝑑𝑡
The acceleration equation would be:
ῳ
𝑎 = 4 ℎ ( )2
𝛽
b. Acceleration of the follower from C to B and β/2 ≤ θ ≤ β (Deceleration Equation)
ῳ
𝑎 = −4 ℎ( )2
𝛽
JERK EQUATION
𝐽=
𝑑𝑎
=0
𝑑𝑡
For this topic, here are the example diagrams for displacement, velocity, acceleration and jerk.
The follower dwells from 0 to 30 degrees, rises with a constant positive acceleration from 30 to 90
degrees and continues to rise with constant negative acceleration from 90 to 150 degrees and dwells
from 150 to 210 degrees and returns with a constant negative acceleration from 210 to 270 degrees and
completes the return with constant positive acceleration from 270 to 330 degrees and lastly it dwells
from 330 to 360 degrees.