Curvilinear Motion Equations of Motion

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```					Curvilinear Motion &
Equations of Motion
ENGR 221
April 16, 2003
Lecture Goals

• Normal and Tangential Coordinates
• Rectilinear Equations of Motion
Equations of motion

From Newton’s Second Law of motion

F  ma
where the force and acceleration are vector
quantities.
Equations of motion

The vectors can be broken into components similar to
equilibrium


Fx i  Fy j  Fz k  m ax i  ay j  az k        
Fx  max       & Fy  may             & Fz  maz
Equations of motion
From equilibrium equations

F 0 x

F 0 y

 F 0 z
Equations of motion
The difference between the equilibrium equations and
the equations of motion is that the right hand side is
no longer zero.

 F  ma
x         x

 F  ma
y         y

 F  ma
z        z
Procedure for solving the
Equations of Motion

1. Draw the Free-Body Diagram.
2. Write the equations of motion
3. Solve for the velocities and accelerations need
or obtain the equations for the components.
4. Solve for the forces
Example Problem – Equation of
Motion
A 80-kg skydiver falls at 85 m/s
when she opens her parachute. If
her speed is reduced to 5 m/s
during the next 60 m of fall,
determine the average force exerted
on her body by the parachute
during this interval.
Example Problem – Equation of
Motion
Draw the free-body diagram
of the skydiver.

F    y    may
Example Problem – Equation of
Motion
Rewrite the equation

F  y    may
80 kg  9.81 m/s   Fp   80 kg  ay
2

The acceleration is going to be a
constant
Fp
ay  9.81 m/s 
2

80 kg 
Example Problem – Equation of
Motion
The acceleration is defined as

dv dv dx
a   
dt dx dt
60 m           5 m/s


0
85 m/s
vdv
Example Problem – Equation of
Motion
Plug in the acceleration equation
              Fp 
5 m/s
60 m
v   2

      9.81 m/s            dx   
2
            80 kg  
0                                 2  85 m/s
60 m
              Fp  
 9.81 m/s            x             3600 m 2 /s 2
2
           80 kg    0

                       
Fp  5584.8 N
Class Problem – Rectilinear
Motion
The 2000-lb elevator cage is
brought to rest from an initial
speed of 25 ft/s in a distance
of 50 ft. Determine the
uniform deceleration and the
tension in the elevator cable
while the cage is coming to
rest.
Example Problem – Rectilinear
Motion
Two bodies A and B with masses of 25 kg
and 30 kg, respectively. During motion of
the bodies,
a) Determine the acceleration of body
A and tension in the cable
connecting the bodies.
b) Determine the acceleration of body
B if body A is replaced with a
constant force of 245 N.
Example Problem – Rectilinear
Motion
Draw the free body diagram of the boxes A and B. The
normal weight, NB and R is resistance due to friction and
the tension T.
Example
Problem –
Equation of
Motion
Find the equations of motion
for the block.

 FAy  mA aAy  T   25 kg   9.81 m/s 2    25 kg  aAy

F  Bx    mB aBx  T  R   30 kg  aBx

F  By    mB aBy  N B   30 kg   9.81 m/s 2   0
N B  294.3 N
Example
Problem –
Equation of
Motion
If there is no friction
resistance then – aAy = aBx

F   Ay    mA aAy  T   25 kg  aAy   25 kg   9.81 m/s   2

F   Bx    mB aBx  T   30 kg  aBx
 30 kg  aBx   25 kg   aBx    25 kg   9.81 m/s 2 
aBx  4.46 m/s 2  T   30 kg   4.46 m/s 2   133.8 N
Example Problem –
Equation of Motion

The force in the cable is T =245 N

245 N   30 kg  aBx
aBx  8.167 m/s 2
Class Problem –
Rectilinear Motion
Two bodies A and B with masses of 100 kg and
300 kg, respectively. During motion of the
bodies,
a) Determine the acceleration of body A and tension
in the cable connecting the bodies if the fiction is
neglected.
Example Problem – Rectilinear
motion
Twp bodies A (40 kg) and B (30 kg) are connected by a flexible
cable. The kinetic coefficient of friction between body A and the
inclined surface is mk = 0.25 and the system is released from rest.
During the motion of the body determine.

a) The acceleration of A
b) The tension in the cable
c) The velocity of body B
after 5 sec.
Example Problem – Rectilinear
motion
Draw the free body diagrams of the bodies
Example
Problem –
Rectilinear
motion
The equations are
 3
 FAx  mA aAx  2T  FA   40 kg   9.81 m/s   5    40 kg  aAx
2

 
2  4
 FAy  mA aAy  N A   40 kg   9.81 m/s   5   0
 
N A  313.92 N
 FBy  mB aBy  T   30 kg   9.81 m/s 2    30 kg  aBy
Example
Problem –
Rectilinear
motion
Use the constraints of the two length of the cable
L  2 sA  sB  constant
 2 sA  sB  0
2 s A   sB

The direction so the block are moving the upward
Example
Problem –
Rectilinear
motion
The frictional force is

FA  m A N A  0.25  313.92 N 
 78.48 N

Rearrange the equations in terms of T
Example
Problem –
Rectilinear
motion
The equations are

T   30 kg  aBy  294.3 N
2T  78.48 N  235.44 N   40 kg  aAx
Example
Problem –
Rectilinear
motion
The equations are

2   30 kg  2aAx   294.3 N   313.92 N   40 kg  aAx
274.68 N  160 kg  aAx
aAx  1.717 m/s  T  191.3 N
2
Example Problem –
Rectilinear motion

The velocity can be found by using integration of the
acceleration

vB  aBt  constant  2aAx t
  3.434 m/s   2
  5 s   17.17 m/s
Equations of motion
The difference between the linear and curvilinear
equations of motion is the defined system.

F     n    man
F     t    mat
Example Problem – Curvilinear
motion
The bob of a 2 m pendulum describes an arc of circle in
a vertical plane. If the tension in the cord is 2.5 times
the weight of the bob for the position show, find the
velocity and acceleration of the bob at the position. .
Example Problem – Curvilinear
motion
Draw the free body diagram of the mass .

+   F   t    mat
 mg sin  30o   mat
+
F n      man
 2mg  mg cos  30o   man
Example Problem – Curvilinear
motion
Solve for the acceleration .

mg sin  30o   mat  at   9.81 m/s 2  sin  30o 
at  4.90 m/s 2

2mg  mg cos  30o   man  an   9.81 m/s 2  2  cos  30o    
at  16.30 m/s 2
Example Problem – Curvilinear
motion
Compute the tangential velocity .

vt2
an            vt  an 

vt      16.30 m/s 2   2 m 
 5.71 m/s
Example Problem – Curvilinear
motion
Determine the rated speed of a highway curve of radius
 = 400 ft. banked through an angle, q = 18o. The rated
speed of the banked curve road is the speed at which a
car should travel if no lateral friction force is to be
exerted on its wheel.
Example Problem – Curvilinear
motion
Determine the equation of motion
for the car with a radius  = 400 ft.
banked through an angle, q = 18o.

F   y   0
 R cos 18o   W  0

F   n    man
 R sin 18o   man
Example Problem – Curvilinear
motion
Solve for R and find an.
mg
R
cos 18o 
 mg 
man  R sin 18   
o
 sin 18o 
 cos 18o  
            
an  g tan 18o    32.2 ft/s 2  tan 18o 
 10.46 ft/s 2
Example Problem – Curvilinear
motion
Compute the tangential velocity .
vt2
an           vt  an 

vt      10.46 ft/s 2   400 ft 
 64.7 ft/s

 60 mph 
 64.7 ft/s            44.1 mph
 88 ft/s 
Class Problem – Curvilinear
motion
An airplane is descending at an
angle, q = 20o with respect to
the horizontal when it drops a
bomb. If the altitude at the
time of release is 5000 m and
speed of the plane 750 km/hr,
determine the range
(horizontal distance traveled)
of the bomb and the elapsed
time before it strikes the
ground. Neglect air friction.
Homework (Due 4/23/03)
Problems:
15-10, 15-21,15-23, 15-57, 15-60, 15-61,
15-62, 15-63, 15-64
Bonus Problem – Curvilinear
motion
A block B of mass m may slide
freely on a frictionless arm OA,
which rotates in a horizontal
plane at a constant rate ,   q
Knowing that B is released at a
distance r0 from O and express as
a function of r (a) the
components vr of the velocity of
B along AO, (b) the magnitude
of the horizontal force F exerted
on B by the arm OA.
Bonus Problem – Curvilinear
motion
Draw the free body diagram of
the block B.
+
F  r    mar

 0  m r  rq 2   
+
 Fq  maq

 F  m rq  2rq       
Bonus Problem – Curvilinear
motion
Determine the velocity of the r component

vr  r  r  vr
dvr dr      dvr
r  vr          vr
dr dt       dr
Substitute into the equation for r double dot
 dvr     2
0  m  vr   rq 
 dr       
Bonus Problem – Curvilinear
motion
Integrate

vr   f
dvr
vr      rq   vr dvr   rq dr
2                 2

dr         o          r0

vr  q  r  r
2
0   
2 0.5
Bonus Problem – Curvilinear
motion
Insert the r and   q  q  0               to find the
force.


F  m r  0  2 q  r  r         2
0 
2 0.5
q 

 m 2q    2
r   2
r  0 
2 0.5


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