KINEMATICS OF PARTICLES
 t  v  0
11.28 a = ksin  
 at t  0
 T  s  0
v
t
dv
 t 
 t 
 k sin     dv  k  sin  dt
(a)
0
0
dt
T
T
T
T   t 
T   t  
v =  k cos   k cos   1
   T  0
 T 
v=
  t 
kT
1  cos 

  T 
v=
ds kT 
 t 

1  cos 

dt
 
 T 
S
  ds 
0
kT t
 0

 t  
1  cos T dt 
  

t
kT  T  t 
t  sin  
s=
    T  0
s=
kT  T  t 
t  sin  
    T 
T2
s=k 2

 T  t 
 t   sin  T 
 

(b) v = vmax when
i.e. when t = T
vmax = 2
kT


kT 
 t 
1  cos has max imum value 

 
T

(c) S/t=2T = k
T2
2

T  t 
t   sin  T 
  t  2 T

T2 
T  .2T 
. 2T. T  sin 

2 
  T 
 
=
k
=
k T 2 .2
2
2kT 2
S=

S / t 2T 
kT  T  t 
t  sin  
    T  1t  2T
=
S/t=2T = 2
(d) vave =
Vave =
kT 
T  2T 
2T  sin 


 
  T 
kT 2

s s t  2T  S t 0

t
2T  0
2kT 2
0


2T
kT

A small object is released from rest in a tank of oil. The downward acceleration of the object
Is (g-kv) where g is the constant of the object is (g-kv) where g is the constant acceleration
due to gravity, k is a constant which depends on the viscosity of the oil and the shape of the
object and v is the downward velocity of the object. Derive expressions for the velocity, v, and
the vertical drop, y, as function of the time, t, after release.
a=
v
t
dv
dv
 (g  kv)  
  dt (separation of variables)
dt
g  kv 0
Let (g-kv) = u  -kd0v=du

 1 g  kv du
1
 ln u gkv
t
t
g

g
k
u
k
 g  kv 
g  kv
  kt 
ln 
 e kt
g
g


V=g
(1  e  kt )
k
y
dy g
g t
 (1  e kt )   dy   (1  e kt )dt
o
v = dt k
k o
(separation of var iables !)
y=
g 1

t  (1  e  kt )

k k

Y=
g 1

t  (1  e kt )

k k

11.42
xb
xo
T
e
xA
xE
xD
xP
1
Consider rope TCD: (assume rope inextensible)
 X
  (X  X )  Const
X
P
D
P
E
 X
 X
 X
 0
X
P
D
P
E
E is fixed
 X
 (ii )
2X
P
D
 X

XB = XP + l  X
B
P
  2X
 (iii )
X
D
B


  X B  2X B  3 X
  3 X 450  675 mm / s
(a) X
A
B
2
2
2
  2X
  900 mm / s 
(b) X
D
B


  675  450  225mm / s 
(c) X A / B  X A  X
B



 X
  450  900  450mm / s  450mm / s 
(d) X  X  X  X
CD
C
D
B
D
Xt
I
XF
XE
XA
E
XD
F
Determine the relationship which
governs the velocities of the four
cylinders. Express all velocities
as positive down. How many
degrees of freedom are there?
A
II
D
III
XC
XB
C
B
Assumption: Ropes I, II & III are inextensible
Rope I: XC + 2XE = Constant(i)
Rope II: (XC-/XE) + XC + 2/XF = Constant (ii)
Rope III: (XB-XF) +XB +XA = Constant (iii)
XD
= Constant (iv)
2
XD
Substitute (iv) in (iii)  2XB + XA + XC +
 Cons tan t
4
or 4XA +8XB +4XC +XD = Constant (v)
Differentiate (v) with respect to t
Substitute (i) in (ii)  2XC + 2XF +
  8X
  4X
 X
 0
 4X
A
B
C
D
OR
4VA + 8VB + 4VC +VD = 0
Only 3 of XA, XB, XC and XD in (V) can be chosen  system has 3 degrees of freedom
A ball is dropped vertically onto a 20° incline at A; the direction of rebound forms an
angle of 40° with the vertical. Knowing that the ball next strikes the incline at B,
determine
a) the velocity of rebound at A, (b) the time required for the ball to travel from A to B
Y
40o
A
X
B
3m

20o
Motion in the x-direction
  0  X
  Const  V cos 50.t
X
0
but X  3m  t  3

V0
cos 50(i)
Motion in the Y-direction
  g  Y
 Y
  gt  V  V sin 50  gt (ii )
Y
0
0
t2
 Y  Y0  V0 sin 50.t  g (iii )
2

Geometry:
Y
X

 tan 20  Y   X tan 20  3 tan 20
t t 
(iii)
then becomes: -1.092 = V0sin50°.
-1.092 = 3 tan 50° - 4.5
(a)
3
, Y  1.092m
V0 cos 50
V0 = 4.785 m
s
2
3
9

  3

V0 cos 50 2  V0 Cos50 
(9.81)
m2
2

V

22
.
89
0
s2
V02 cos 2 50
(b)
T = 0.975secs
Calculate the minimum possible magnitude of the muzzle velocity, u, which a projectile
must have when fixed from point A to reach a target B on the same horizontal plane 12
km away.
Y

A

B
X
12 km
 Motion in the X=direction
  0  V  Const  U cos   X X  U cos .t
X
X
0
12000 = ucos.t (i)

Motion in the Y-direction
dv
  g  y  V  V  gt  V U sin   gt (ii )
Y
Y
Y0
Y
dt
u sin 
 At maximum height, VY = 0  0 usin-gt1t1=
(iii)
g
 Because of symmetry, time taken, t, to travel from A to is twice t1 (t = 2t1)
u sin  2u 2 sin  cos 
From (ii)  12000 = ucos .2

g
g
12000g
 u  u min when sin 2 is max imum (i.e. when 2  90)
u2 =
sin 2

Umin = 
12000(9.81)  343.1m / s
(negative value for u not
physically possible)
The vertical slot moves to the right with a constant speed v0 in
meters per second for an interval of motion and causes the pin
P to move along the parabolic slot x = y2/3 where x and y are
in meters. Calculate the radius of curvature  to the path for
the position where y = 2 m and calculate the tangential
acceleration at of the pin when it passes this position. Ans.  =
6.94 m, at = -0.169v02 m/s2
For y = 2m,  x = 4/3 m (since x = y2/3)
t

Y
V

y = 2m
 x = v0
x
n
an
y = 1m
X = 4/3m
  2 YY

X
3
  2 (Y
 2  YY
 )  0(sin ce X
  const ) 
X
3
4
  4Y
 Y
  3V
mX
0
3
3
4
2
Y   9 V0   9 V 2
0
16 2
32
At Y = 2m, X =
 2 Y
 2  V 2  9 V 2  25 V 2  V  5 V
V2 = X
0
0
0
0
16
16
4
 cos   Y
 sin   9 V 2 sin 
an = X
0
32
4

V
5 4
tan   X   0 V0 
 sin   4

3/ 4
5
Y
3
3
9 2 4 9 2
V0  
V0
32
5 40
V2
25 2 40
250
an 

V0 

 6.94 m/s2
2

16
36
9V0
 sin   Y
 cos    9 V 2  3   27 V 2  0.169V 2 m/s2
at  X
0
0
0
32
5 160
an 
A nozzle discharges a stream of water in the
direction shown with an initial velocity of 25 m/s.
Determine the radius of curvature of the stream (a)
as it leaves the nozzle, (b) at the maximum height of
the stream.
  Y
2/
Y
Y
At A
4
(9.81)  7.85m / s 2
5
2
V
V 2 (25) 2
but a n A  A   A  A 
A
a nA
7.85
a n A  glos  
t
at

an
g
 A  79.6m
At B
VB = (VA)X (Since ax = 0 and B is at the maximum)
Height of the stream
VB = VAcos = 25 (4/5) = 20 m/s
a n B  g  9.81m / s 2
B 

an
VB2 (20) 2

 40.8 m
a nB
9.81
As the hydraulic cylinder rotates about 0 the length, l, of
the piston rod P is controlled by the action of oil pressure
in the cylinder. If the cylinder rotates at the constant rate
 = 60 deg/sec and is decreasing at the constant rate of
152.4 mm/sec, calculate the magnitudes of the velocity
and acceleration of end B when l = 127 mag.
Solution
1
T1
B
T
1
O
r = 381 + 127 = 508 mm
r    152.4 mm / sec  cons tan t  r  0
deg
60

  60

x 2  rad / sec  cons tan t    0
sec 360
3

Vt = r  0.1524 m / sec,V   r  508 x  0.532 m / sec
3
V  V2  (01524) 2  (0.532) 2  0.553 m / sec
508 x 
ar = r  r 2  0 
 0.557 m / sec 2
1000 x 9

a = r   2r  508 x 0  2(0.1524) x  0.319 m / sec 2
3
2
a  a 2r  a 2  (0.557) 2  (0.319) 2  0.642 m / sec 2
The slotted arm is pivoted at O and carries the slider C. The
position of C in the slot is governed by the cord which is
fastened at D and remains taut. The arm turns
counterclockwise with a constant angular rate  = 4rad/sec
during an interval of its motion. The length DBC of the cord
equals R, which makes r = 0 when  = 0. Determine the
magnitude of the acceleration of the slider at the position for
which  = 30o. The distance R is 15 in. Ans. A = 489 in./sec2
Solution
1
t1
B
R
C
r

D
O
Geometry:
R
r  BC  R 
  r  BD
BC  BD  R 
BD
 R sin   r  2R sin 
2
2
2
2
Kinematics: ac = t1 r  r + 1 (2 r  r)

R
r = 2Rsin  2  r  R cos  , r  R cos  2   2 sin  2
2
2
From triangle OBD,


T  T 
2 2
ac =
 (2 r  r) 2
 = 30°  /2 = 15°, R = 15",
  4 rad
sec
 cons tan t
  0
r = 30 sin 15° = 7.76"
r  15 cos 15(4)  57.96" / sec, r  0 
a c  31.06  7.76(4) 2
15 2
(4) sin 15  31.66" / sec
2
  (2)(4)(57.96)  0
2
2
 489" / sec 2
A police car P is traveling at velocity vP
= 20j km/h when its radar gives the
following data about car A : r = 100 m,
r = -60 km/h (r decreasing), r  0,  =
0.5 rad,  = -10-3 rad/s ( decreasing),
and   0. Determine the velocity vA.
Solution:

A
Relationship between the unit vectors i, j & r1, 1
y, j
r1
VP
r



P
Note: 1 is positive
in the direction of
increasing 
1

x, i
r1 = i sin + j cos
1 = I cos - j sin
Velocity car A relative to the Policecar:
 60
V A P  r r 1  r 1 
i sin 28.65 o  j cos 28.65 o  100(10 3 ) i cos 28.65 o  j sin 28.65 o
3.6
V A P  (8.07i  14.59 j )m / s


Absolute velocity of car A: V A  V A P  V p  8.07i  14.59 j 
V A  (8.07i  9.03 j )m / s

20
j
3.6

Geometry of a space curve
Consider the following space curve along which a particle moves from A to B
m1
S
n1
A
t1
I
s
B
T(s)
T(s+s)
O

t1 =

Unit vector tangent to the curve at A is t1
lim
s 0
v dv

s ds




Unit vector perpendicular to t1 and pointing toward centre of curvature of the curve is
the principal normal n1
Binotmal unit vector m1 = = t1 x n1
-Plane of t1 & n1 : OSCULATING PLANE
-Plane of m1 & n1: NORMAL PLANE
-Plane of m1 & t1: RECTIFYING PLANE

Consider segment AB of curve is more detail
B
T1 (s)
A
t1
t1 (s+s)
Osculating plane
t1 (s+s)
 The orientation of n1 (i.e. principal normal) is determined from the following:
(i)
n1 1 t1 (ii) n1 lies in the osculating plane (iii) 1n11 = 1
dt
d
t1 .t1 = 1 
( t 1  t 1 )  2t 1 1  0
ds
ds
dt 1
 t1 1
ds
The vector
dt1

ds
Lim
t1
s 0 s
Lies in the osculating plane since
t1 =t1 =t1 (S + S) -t1 (S) lies in the same plane. Therefore, the vector
dt 1
satisfies
ds
conditions (i) and (iii) above; hence
n 1  d t 1 / ds
d t 1 / ds
Osculating Plane
C: Centre of curvature
 : radius of curvature
P: Osculating plane
X
Z
PV
A

C
x
R


Y
r
X
Consider the space curvilinear motion of a particle along the path shown. The motion at
position A may be considered to be taking place in the plane P which contains the path at
this position. This plane, often called the OSCULATING PLANE, may be defined by
point A and two adjacent points, one on either side of A, on the path curve. As these
points are brought closer to A, the plane containing the three points approaches the
limiting plane P.
The osculating plane is, by definition, the plane containing the unit tangential (t1) and unit
normal (n1) vectors. The path coordinates t and n, are used for solving most of the
problems of planar motion. The osculating plane thus remains fixed in the plane of
motion. Although it is possible to obtain the principal normal n and the direction of the
osculating plane for spatial motion, it is generally easier to define the spatial motion of a
particle using other coordinate systems, e.g. X,Y,Z or T,  , Z.