Engineering Mechanics –II (Dynamics)/2026
Chapter Two
Kinematics of Particle
2.1. Rectilinear motion
A particle moving in straight line is said to be in a rectilinear motion. Consider a particle P
moving in straight line, Fig 2.1. The position of P at any instant of time t can be specified by
its distance s measured from some convenient reference point O fixed on the line. At time
t + Δt the particle has moved to P’ and its coordinate becomes s + Δs. The change in the
position coordinate during the interval Δt is called the displacement Δs of the particle. The
displacement would be negative if the particle moved in the negative s-direction.
-S
O
P
P’
+S
Δs
s
Fig 2.1
Velocity and Acceleration
The average velocity of the particle during the interval Δt is the displacement divided by the
time interval or ʋav= Δs/Δt. Δs, Δt becomes smaller & approaches zero in the limit, and then
the average velocity approaches the instantaneous velocity of the particle which is
𝛥𝑠
𝑑𝑠
ʋ = lim𝛥𝑡→0 𝛥𝑡 or ʋ = 𝑑𝑡 = ṡ
[2.1]
Thus the velocity is the time rate of change of the position coordinate s. The velocity is
positive or negative depending on whether on the corresponding displacement is positive or
negative.
The average acceleration of the particle during the interval Δt is the change in its velocity
divided by the time interval or αav =Δʋ/Δt. Δʋ, Δt becomes smaller & approaches zero in the
limit and then the average acceleration approaches the instantaneous acceleration of the
particle, which is
𝑑2 𝑠
𝑑ʋ
α = 𝑑𝑡 = ʋ̇ or α = 𝑑𝑡 2 = s̈
[2.2]
The acceleration is positive or negative depending on whether the velocity is increasing or
decreasing. Note that the acceleration would be positive if the particle had negative velocity
which was becoming less negative. It the particle is slowing down the particle is said to be
decelerating.
From eq. 2.1,ds = ʋdt& from eq. 2.2,dʋ = αdt. By combining eq. 2.1 & 2.2 and eliminating dt
we obtain a differential equation relating displacement, velocity and acceleration, which is
ʋdʋ = αds
Bililign F.
ECE
[2.3]
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Graphical Interpretations
The differential equations governing rectilinear motion are clarified by representing the
relationships among s, ʋ, α and t graphically.
a
b
c
Fig 2.2
Figure 2.2a is schematic representation of the variation of s with t from time t1 to time t2 for
some given rectilinear motion. By constructing the tangent line to the curve at any time t, we
obtain the slope, which is the velocity ʋ = ds/dt. Thus the velocity can be determined at all
points on the curve and plotted against the corresponding time as shown in the fig 2.2b.
Similarly, the slope dʋ/dt of the ʋ-t curve at any instant gives the acceleration at that instant,
and the α-t curve can therefore be plotted as in fig 2.2c.
From the fig. 2.2b it is obvious that the area under the ʋ-t curve during the time dt is ʋdt
which is the displacement ds. Consequently the net displacement of the particle during the
interval t1to t2is the corresponding area under the graph, which is
𝑠
𝑡
1
1
2
2
∫𝑠 𝑑𝑠 = ∫𝑡 ʋ𝑑𝑡 = s2 – s1
Similarly from fig 2.3c we see that the area under α-t curve during time dt is αdt which is dv.
Thus the net change in velocity between t1 and t2 is the corresponding area under the curve
which is
ʋ
𝑡
1
1
2
2
∫ʋ 𝑑ʋ = ∫𝑡 𝛼𝑑𝑡 = ʋ2 – ʋ1
Note: There are two additional considerations in graphical representations. (1) When the
acceleration α is plotted as a function of the position coordinate s, fig 2.3a, the area under the
curve during a displacement ds is αds which is ʋdʋ from eq. 2.3. Thus the net area under the
curve between position coordinates s1 and s2 is
ʋ
𝑠
1
1
1
2
2
2
2
∫ʋ ʋ𝑑ʋ =∫𝑠 𝛼𝑑𝑠 = 2 (ʋ 2 – ʋ 1)
(2) When the velocity ʋ is plotted as a function of the position coordinate s, fig 2.3b the slope
of the curve at any point A is 𝑑ʋ/𝑑𝑠 which is equal to α/ʋ from eq. 2.3. Rearranging the
equation gives ʋ(𝑑ʋ/𝑑𝑠) = 𝛼, the acceleration.
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ECE
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a)
b)
Fig. 2.3
Analytical Integration
We have two cases in a relation to acceleration: constant acceleration and variable
acceleration.
Constant acceleration
When acceleration is constant, dʋ = αdt and ʋdʋ = αds can be integrated directly. For
simplicity, s = s0, ʋ = ʋ0 and t = 0 at the beginning of the interval, then at any time t the
integrated equations become
ʋ
𝑡
∫ʋ 𝑑ʋ = 𝛼 ∫0 𝑑𝑡 or ʋ2 = ʋ0 + 𝛼𝑡
0
ʋ
𝑠
0
0
2
2
∫ʋ ʋ𝑑ʋ = 𝛼 ∫𝑠 𝑑𝑠 or ʋ = ʋ 0 + 2𝛼(s – s0)
Substitution of the integrated expression for ʋ into ds = ʋdt and integrating with respect to t
gives
𝑠
𝑡
∫𝑠 𝑑𝑠 = ∫0 (ʋ0 + 𝛼𝑡)𝑑𝑡
or
0
1
s = s0 +ʋ0t + 2 𝛼t2
Variable acceleration
a) Acceleration given as a function of time, 𝛼 = f(t). In this case we use dʋ = αdt
ʋ
𝑡
∫ʋ 𝑑ʋ =∫0 𝑓(𝑡)𝑑𝑡
0
or
𝑡
ʋ = ʋ0 +∫0 𝑓(𝑡)𝑑𝑡
From this integral expression as ʋ is a function of t the position coordinate s is obtained by
integrating ds = ʋdt while keeping ʋ in the integral.
𝑠
𝑡
∫𝑠 𝑑𝑠 =∫0 ʋ𝑑𝑡
0
or
𝑡
s = s0 +∫0 ʋ𝑑𝑡
b) Acceleration given as a function of velocity, 𝛼 = f(v). In this case we use dʋ = αdt but
we need to rearrange it into dt = dʋ/α and integrating
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ECE
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𝑡
ʋ 𝑑ʋ
∫0 𝑑𝑡 = ∫ʋ 𝛼
ʋ 𝑑ʋ
t = ∫ʋ 𝑓(𝑣)
or
0
0
This equation gives t as a function of ʋ and it can be rearranged to get ʋ as a function of t.
We can also use ʋdʋ = αds. Again it needs to be rearranged into ds = ʋdʋ/α and
integrating
𝑠
ʋ ʋ𝑑ʋ
0
0
∫𝑠 𝑑𝑠 = ∫ʋ
ʋ ʋ𝑑ʋ
s = s0 + ∫ʋ 𝑓(𝑣)
or
𝛼
0
c) Acceleration given as a function of displacement, α = f(s). In this case we use
ʋdʋ = αds and integrating
ʋ
𝑠
0
0
𝑠
ʋ2 = ʋ20 + 2 ∫𝑠 𝑓(𝑠)𝑑𝑠
∫ʋ ʋ𝑑ʋ =∫𝑠 𝑓(𝑠)𝑑𝑠or
0
Next we solve for ʋ to get ʋ = h(s) a function of s. Now we can use ds = ʋdt but it needs to be
rearranged into dt = ds/ʋ and integrating
𝑡
𝑠 𝑑𝑠
∫0 𝑑𝑡 = ∫𝑠
0 ʋ
or
𝑠 𝑑𝑠
t = ∫𝑠 ℎ(𝑠)
0
This equation gives t as a function of s. Finally, it can be rearranged to obtain s as a function
of t.
Example 2.1
The position coordinate of a particle which is confined to move along straight line is given by
s = 2t3 – 24t + 6 where s is measured in meters from convenient origin and t is in seconds.
Determine,
(a) The position of the particle when t = 0 and t = 5 s
(b) The time required for the particle to reach a velocity of 72 m/s from its initial condition at
t = 0,
(c) The acceleration of the particle when v = 30 m/s, and
(d) The net displacement of the particle during the interval from t = 1s to t = 4s.
(e) The total distance covered by the particle during the first 5 seconds.
Bililign F.
ECE
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2.2. Plane curvilinear Motion
Curvilinear motion occurs when a particle moves along a curved path. In plane curvilinear
motion, since the path is often described in two dimensions, vector analysis will be used to
formulate the particle's position, velocity, and acceleration.
2.2.1. Position
Consider a particle located at a point on a plane curve
defined by the path function s(t), Fig 2.6a.The position of
the particle, measured from a fixed point O, will be
designated by the position vector r = r(t). Notice that both
the magnitude and direction of this vector will change as
the particle moves along the curve.
2.2.2. Displacement
Suppose that during a small time interval ∆t the particle
moves a distance ∆s along the curve to a new position,
defined by r' = r + ∆r, Fig. 2.6b.The displacement ∆r
represents the change in the particle's position and is
determined by vector subtraction; i.e. ∆r= r' - r.
2.2.3. Velocity
During the time ∆t, the average velocity of the particle is
∆𝐫
vav = ∆𝑡
The instantaneous velocity is determined from this equation
by letting ∆t→0, and consequently the direction of ∆r
approaches the tangent to the curve. Hence,
v = lim∆𝑡→0 (∆𝐫/∆𝑡)or
𝒅𝒓
v = 𝑑𝑡
Since dr will be tangent to the curve, the direction of v is
also tangent to the curve, Fig. 2.6c.
Fig 2.6
The magnitude of v, which is called the speed, is obtained by realizing that the length of the
straight line segment ∆𝐫 in Fig. 2.6b approaches the arc length ∆s as ∆t→0, we have
v = lim∆𝑡→0 (∆r/∆𝑡)= lim∆𝑡→0 (∆s/∆𝑡) or
𝑑𝑠
v = 𝑑𝑡
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Thus, the speed can be obtained by differentiating the path function s with respect to time.
2.2.4. Acceleration
If the particle has a velocity v at time t and a velocity
v' = v + ∆v at t + ∆t, Fig. 2.7, then the average acceleration
of the particle during the time interval ∆t, is
∆𝐯
aav = ∆𝑡
Fig 2.7
Where ∆v = v' - v. To study this time rate of change, the two velocity vectors in Fig. 2.7 are
plotted in Fig. 2.8 such that their tails are located at the fixed point O and their arrowheads
touch points on a curve. This curve is called a hodograph, and when constructed, it describes
the locus of points for the arrowhead of the velocity vector in the same manner as the path s
describes the locus of points for the arrowhead of the position vector, Fig. 2.6a.
To obtain the instantaneous acceleration, let ∆t→0, in the above equation. In the limit ∆v will
approaches the tangent to the hodograph, and so a = lim∆𝑡→0 (∆𝐯/∆𝑡) or
a=
𝐝𝐯
𝑑𝑡
Fig 2.8
The direction of the acceleration is neither tangential nor normal to the path curve.
Fig 2.9
2.2.5. Coordinate systems
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ECE
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Three types of coordinate systems are commonly utilized to analyse curvilinear motion.
a) Rectangular coordinate system (x-y)
b) Normal-Tangential coordinate system (n-t) and
c) Polar coordinate system (r-θ)
2.2.5.1.
Rectangular coordinate system (x-y)
The position of the particle r, the velocity
v and the acceleration aof the particle are
described by their x- and y- components.
r = xi + yj
𝑑𝒓
𝑑
𝑑𝑥
𝑑𝑦
v = 𝑑𝑡 = 𝑑𝑡(xi + yj) = 𝑑𝑡 i + 𝑑𝑡 j = ẋi + ẏj
v = vxi +vyj similarly
𝑑𝐯
𝑑
a= 𝑑𝑡 = 𝑑𝑡(vxi +vyj) = v̇xi +v̇yj = ẍi + ÿj
a = axi + ayj
Fig.2.10
The magnitude of r = √x 2 + y 2 and the angle which r makes with the horizontal is
tan θ = y/x. The magnitude of v = √vx 2 + vy 2 and the angle which v makes with the
horizontal is tan θ = vy/vx. The magnitude of a = √𝑎x 2 + 𝑎y 2 and the angle which a makes
with the horizontal is tan θ = ay/ax.
Example 2.3
A particle which moves in two-dimensional motion has coordinates in millimeters which vary
1
with the time t in seconds according to x = 3t2 – 4t and y = 4t2 – 3t3. Determine the magnitude
of the position r, velocity v, acceleration aand the angle which these vectors make with the
horizontal x-axis when t = 2 s.
Example 2.4
The curvilinear motion of the particle is defined by vx = 50 – 16t and y = 100 – 4t2, where vx
in meters per second, y in meters and t in seconds. It is also known that x = 0 when t = 0.
Determine the velocity and acceleration of the particle when y = 0
Projectile motion
A projectile is an object that is thrown in air by a force. The action of the force ceases after
the projectile is launched. The path a projectile follows in air is called trajectory. In projectile
motion, we neglect the effect of aerodynamic resistances and we take a constant downward
acceleration of g. With these assumptions a rectangular coordinate system is useful for
trajectory analysis.
Bililign F.
ECE
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Fig 2.11
ax = 0 and ay = – g
The initial velocity is v0 or u and it has two components (u x &uy) since the object is thrown at
an angle θ to the horizontal.
ux = u cos θ and uy = u sin θ or u = u cos θ i + u sin θ j
The horizontal motion is uniform motion with the velocity ux = vx. Thus, the horizontal
distance at any time t is
x = x0 + uxt.
In the vertical motion velocity is varying with a constant deceleration g, hence
vy = uy – gt,
v2y = u2y – 2g (y – y0),
1
y = y0 + uyt – 2gt2
For fig 2.11, we can describe three terms: time of flight, maximum height and range & we
can determine the equation of the path.
The time from the initial point to the apex of the trajectory is tup or thalf. At this point vy is
zero. Thus,
vy = uy – gt = 0
𝑢𝑦
tup = 𝑔 =
𝑢 sin 𝜃
𝑔
The total time or time of flight tf = tup + tdown. For the case fig 2.11 tup = tdown, thus
2𝑢𝑦
tf = 2tup = 𝑔 =
Bililign F.
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2𝑢 sin 𝜃
𝑔
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The height from the initial point to the apex of the trajectory is called maximum height (hmax).
It is the maximum possible height during upward movement and calculated using tup.
Substituting tup into the y-equation and letting y0 = 0 at the initial condition to get hmax
uy
1
uy
𝑢𝑦 2
𝑢𝑦 2
hmax = uy( g ) – 2g( g )2
hmax = 𝑔 – 2𝑔
𝑢𝑦 2
(𝑢 sin 𝜃)2
hmax = 2𝑔 =
2𝑔
The total horizontal distance is called the Range (R). It is x at time tf. Thus,
R = x = x0 + uxtf and setting x0 = 0
R = u cos 𝜃
R=
2𝑢 cos 𝜃 𝑢 sin 𝜃
𝑔
R=
2𝑢 sin 𝜃
𝑔
=
2𝑢2 cos 𝜃 sin 𝜃
𝑔
𝑢2 sin 2𝜃
𝑔
Note: the equation for R and tf is only valid for case of fig 2.11.
1
The path of the projectile is parabolic and determined from y = y0 + uyt – 2gt2 and x = x0 + uxt.
If x0 & y0 are 0 at t = 0 and solving for t from x, t = x/ux. Substituting for t in the y-equation,
𝑥
1
𝑥
y = uyu – 2g(u )2
x
x
𝑔
y = (tan 𝜃)x – (2(u cos θ)2)x2
y = (tan 𝜃)x – (
𝑔 sec2 𝜃
2u2
)x2
Example 2.5
A projectile is fired from the edge of a
150m cliff with an initial velocity of
180m/s at an angle of 30° with the
horizontal. Neglecting air resistance, find
(a) the horizontal distance from the gun to
the point where the projectile strikes the
ground, (b) the greatest elevation above
the ground reached by the projectile.
Bililign F.
ECE
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Example 2.6
With what minimum horizontal velocity u
can a boy throw a rock at A and have it just
clear the obstruction at B?
2.2.5.2.
Normal and tangential Coordinates (n-t)
The n- and t-coordinates are considered to move along the path with the particle as shown in
the fig.2.12 where the particle moves from A to B to C. The positive direction for n at any
position is always taken toward the center of cuevature of the path. As shown in the fig.2.12
the positive n-direction shifts from one side of the path to the other if the curvature changes
direction. The positive t-direction is in the direction of the motion.
Fig.2.12
We now use the coordinates n ant t to describe the velocity v and acceleration a by
introducing unit vectors en in the n-direction and et in the t-direction as shown in the fig.2.13
for the position of the particle at point A on its path. During the differential time interval dt,
the particle moves a differential distance ds along the curve from A to A'. With the radius of
curvature of the path at this position designated by ρ, we see that ds = ρdβ, where βis in
radians. It is unneccessary to consider the differential change in ρ between A and A' because a
higher-order term would be introduced which disappears in the limit. Thus the magnitude of
the velocity can be written v = ds/dt = ρdβ/dt, and in vector form
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ECE
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(a)
(b)
Fig.2.13
v = vet = ρ𝛽̇ et
[2.4]
The acceleration of a of the particle is a = dv/dt, and is a vector which reflects both the
change in magnitude and the change in direction of v. Thus the differentiation of v is done by
applying the ordinary rule for the differentiation of the product of a scslsr and a vector and
we get
𝑑𝐯
a = 𝑑𝑡 =
𝑑(𝒗𝐞𝑡 )
𝑑𝑡
= vėt + v̇et
[2.5]
where the unit vector et has nonzero derivative because its direction is changing with the
direction of the curve.
To find ėtwe analyze the change in et during a differential increament of motion as the
particle moves from A to A' in fig.2.13a. The vector et correspondingly changes to e't, and the
vector difference det is shown in fig.2.13b. The vector det in the limit has a magnitude equal
to the length of the arc |et|dβ = dβ obtained by swinging the unit vector et through the angle
dβ expressed in radians. The direction of det is given by en. Thus we can write det = endβ.
Dividing by dt gives det/dt = (dβ/dt)en which can be written as
ėt = 𝛽̇̇en
[2.6]
substituting eq.2.6 and β̇ from the relation v = ρβ̇ eq.2.5 becomes
a=
𝑣2
𝜌
en + v̇et or a = anen + atet
[2.7]
2
𝑣
where an = 𝜌 = ρ𝛽̇̇2 = v𝛽̇̇ and at = v̇ = s̈ The magnitude of the acceleration is
a = √a𝑛 2 + a𝑡 2 .
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ECE
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If the equation of the curve is given as y = f(x), from calculus the radius of curvature ρ at any
point can be calculated as
ρ=
[1+ (𝑑𝑦/𝑑𝑥)2 ]3/2
|𝑑2 𝑦/𝑑𝑥 2 |
Circular Motion
Circular motion is special
case of plane curvilinear
motion where the radius
of curvature ρ becomes
the constant radius r of
the circle and the angle β
is replaced by the angle θ
measured
from
any
convenient refference to
OP, fig.2.14. The velocity
and
acceleration
components
for
the
particle P in a circular
motion become
Fig.2.14
v = r𝜃̇
an = v2/r = r𝜃̇2 = v𝜃̇
[2.8]
at = v̇ = r𝜃̈
Example 2.7
A test car starts from rest on a horizontal
circular track of 80 m radius and increases
its speed at a uiform rate to reach 100
km/hr in 10 seconds. Determine the
magitude of the total acceleration of the
car 8 seconds after the start.
Bililign F.
ECE
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Example 2.8
If the car decelerates uniformly along the curved road from 25 m/s at A to 15 m/s at C, and
the total acceleration at A is 2 m/s2 determine (a) the radius of curvature at A. (b) the
acceleration of the car at B.
2.2.5.3.
Polar Coordinates (r-θ)
In polar coordinate system, the particle is located by the radial distance r from a fixed point
and by angular measurement θ to the radial line. Polar coordinates are particularly useful
when a motion is constrained through the control of radial distance and an angular position or
when an unconstrained motion is observed by measurements of a radial distance and an
angular position.
Figure 2.15a shows the polar coordinates r and θ which locate a particle travelling on a
curved path. An arbitrary fixed line, such as x-axis, is used as arefference for the
measurement of θ. Unit vectors er and eθ are established in the positive r- and θ-directions
respectively. The position vector r to the particle at A has a magnitude squal to the radial
distance r and a direction specified by the unit vector er. Thus the vector natation for locaton
of a particle at A is
r = rer
Fig.2.15
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ECE
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Velocity
We are now ready to differentiate r = rer with respect to time using the rule for differentiating
the product of a scalar and a vector because the direction of eris significantly changing. Thus
v = 𝒓̇ = 𝑟̇ er + rėr
Now we differentiate er in exactky the same way as et but β is replaced by θ and the direction
en is replaced by eθ i.e. ėr = θ̇eθ
Similarly, for eθ only the direction en is replaced by negative θ-direction(– eθ) i.e. ėθ = – θ̇er
Thus,
v = 𝑟̇ er + r𝜃̇eθ or v = vrer + vθeθ
[2.8]
where vr = 𝑟̇ , vθ = r𝜃̇ and the magnitude of v = √𝑣𝑟 2 + 𝑣𝜃 2
Acceleration
Now we differentiate v with respect to time to get
a = v̇ =
𝒅
(𝑟̇ er + r 𝜃̇eθ)
𝒅𝒕
a = (𝑟̈ er + 𝑟̇ ėr) + (𝑟̇ 𝜃̇eθ + r𝜃̈eθ + r𝜃̇ėθ)
Substitution of ėr = 𝜃̇̇eθ and ėθ = - 𝜃̇̇er gives
a = (𝑟̈ er + 𝑟̇ 𝜃̇eθ) + (𝑟̇ 𝜃̇eθ + r𝜃̈eθ – r𝜃̇ 𝜃̇er)
Rearranging
a = (𝑟̈ – r𝜃̇2)er + (r𝜃̇̈ + 2𝑟̇ 𝜃̇ )eθ or a = arer + aθeθ
[2.9]
where ar = 𝑟̈ – r𝜃̇2, aθ = r𝜃̇̈ + 2𝑟̇ 𝜃̇ and the magnitude of a = √a𝑟 2 + a𝜃 2
Circular Motion
For circular motion r is constant thus,
vr = 0
vθ = r𝜃̇
ar = – r𝜃̇2
aθ = r𝜃̈
Example 2.9
Rotation of the radially slotted arm is governed by θ = 0.02t3+ 0.2t, where θ is in radians and
t is in seconds. Simultaneously,the power screw in the arm engages the slider B and controls
its distance from O according to r = 0.04t2 + 0.2, where r is in meters and t in seconds.
Determine the magnitude of velocity and acceleration of the slider for the instant when t = 3s.
Bililign F.
ECE
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Example 2.10
As it passes the position
shown, the particle P has
constant speed v = 100
m/s along the straight line
shown. Determine the
corresponding values of
𝑟̇ , 𝜃̇, 𝑟̈ and 𝜃.̈
2.3. Relative Motion (Translating(Non-Rotating) Axes)
In the previous sections we have described particle motion using coordinates referred to fixed
reference axes. Now we use a moving reference axes with respect to another fixed reference
axes.
Now consider two particles A and B which
may have separate curvilinear motions in a
given plane. We arbitrarily attach the
origin of a translating axes x-y to particle B
and observe the motion of A as measured
relative to the frame x-y is rA/B = xi + yj,
where the subscribt notation “A/B” means
“A relative to B” or “A with respect to B.”
Fig.2.16
Bililign F.
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The absolute position of B is defined by a vector rB measured from the origin of the fixed
axes X-Y. The absolute position of A is determined by the vector rA = rB + rA/B. Rearranging
gives
rA/B = rA – rB
Now we take the first and second derrivative to get the corresponding relative velocity and
acceleration of the particle respectively. Thus,
ṙA/B = ṙA – ṙB or vA/B = vA – vB
[2.10]
r̈ A/B = r̈ A – r̈ B or aA/B = aA – aB
[1.11]
Example 2.11
Passenger in the jet
transport A flying east at a
speed of 800 km/hr
observe a second jet plane
B that passes under the
transport in horizontal
flight. Although the nose
of B is pointed in the 450
northeast direction, plane
B
appears
to
the
passengers in A to be
moving away from the
transport at the 600 angle
as shown. Determine the
true velocity of B.
Example 2.12
Rapid-transit trains A and
B travel on parallel tracks.
Train A has a speed of 80
km/hr and is slowing
down at the rate of 2 m/s2,
while train B has a
constant speed of 40
km/hr. Determine the
velocity and acceleration
Bililign F.
of train B relative to train A.
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2.4. Constrained Motion of Connected Particles
Sometime the motion of particles are interrelated because of the constraints imposed by
interconnecting members. In such case it is neccessary to account for these constraints in
order to determine the respective motion of the particle.
One Degree of Freedom
Consider very simple system of two interconnected particles A and B as shown in the
fig.2.17. the motio of B is clearly the same as that of the center of its pulley, so we can
establish coordinates y and x measured from a convenient fixed datum. The total length of the
cable is
𝜋𝑟
L = x + 22 + 2y + 𝜋r1 + b
With L, r2, r1and b all constant, the first and second time derivatives of the equation give
0 = ẋ + 2ẏ
or
0 = vA + 2vB
0 = ẍ + 2ÿ
or
0 = aA + 2aB
Fig.2.17
The system in the above figure has one degree of freedom since only one variable, either x or
y , needed to specify the positions of all the parts of the system.
Two Degree of Freedom
The system with two degree of freedom is shown in the fig.2.18. Here the positions of the
lower cylinder and pulley C depend on the separate specifications of the two coordinates yA
and yB.The length of the cables attached to cylinders A and B can be written, respectively as
LA = yA + 2yD + constants
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LB = yB + yC + (yC – yD) + constants
And their first & second time derivatives are
0 = ẏA + 2ẏD and
0 = ẏB + ẏC + ẏC – ẏD
0 = ÿA + 2ÿD and
0 = ÿB + ÿC + ÿC – ÿD
Eliminating ẏD and ÿD gives
ẏA + 2ẏB + 4ẏC = 0
or
vA + 2vB + 4vC = 0
ÿA + 2ÿB + 4ÿC = 0
or
aA + 2aB + 4aC = 0
Fig.2.18
Example 2.13
If the velocity ẋ of block
A up the incline is
increasing at the rate of
0.044 m/s each second,
determine
the
acceleration of B.
Bililign F.
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Dynamics (Meng1062)//2021
Example 2.14
At a certain instant, cylinder A has a downward velocity of 0.8 m/s and an up ward
acceleration of 2 m/s2. Determine the corresponding velocity and acceleration of cylinder B.
Example 2.15
For a given value of y, determine the upward velocity of A interms of the down ward velocity
of B. Neglect the dimeter of the pulley.
Keyru S.
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