Lecture 6 V4

DYNAMICS

DR. MOHD KHIR BIN MOHD NORUniversiti Tun Hussein Onn Malaysia

BDA 20103 – LECTURE 6

Dynamics - LECTURE 6 2

General Plane Motion

mF a

P GI m d M a

G GI M





Dynamics – LECTURE 6 7














Principle of energy conservation

The principle of work and energy applies to a rigid body system

1 2c

g eU T V V

1 2 : Work of all external non conservative forces: Total conservative gravitational potential energy

: Total conservative elastic potential energy: Total kinetic Energy

c

g

e

UV

VT

1 2cU Fdr 1 2

cU Md


Energy involved

2 2 2 22 1 2 1

1 12 2

T m v v I

2 1( )gV mg h h (Rotation)


EXAMPLE 1The bar shown has a mass of 10-kg and is subjected to a couple moment of M = 50 N.m and a force of P = 80 N, which is always applied perpendicular to the end of the bar. Also, the spring has an unstretched length of 0.5 m and remains in the vertical position due to the roller guide at B. determine the total work done by all the forces acting on the bar when it has rotated downward from θ = 0 to θ° = 90°.


EXAMPLE 1

SolutionFirst the free-body diagram of the bar is drawn in order to account for all the forces that act on it.

View Free Body Diagram


EXAMPLE 1

Weight W. Since the weight 10(9.81) = 98.1 N is

displaced downward 1.5 m, the work is

Couple Moment M. The couple moment rotates through an angle

of θ = π/2 rad. Hence,

JUW 2.147)5.1(1.98

JUM 5.78)2/(50


EXAMPLE 1

Spring Force Fs When θ = 0°the spring is stretched (0.75 – 0.5) =

0.25 m, and when θ = 90°, the stretched is (2 + 0.75) – 0.5 = 2.25 m. Thus,

By inspection the spring does negative work on the bar since Fs acts in the opposite direction to displacement.

JUs 0.75)25.0)(30(21)25.2)(30(

21 22


EXAMPLE 1

Force P As the bar moves downward, the force is

displaced through a distance of (π/2)(3) = 4.172 m.

The work is positive.JUP 0.377)172.4(80


EXAMPLE 1

Pin Reactions Forces Ax and Ay do no work since they are

not displaced.

Total Work The work of all forces when the bar is

displaced is thus JU 5280.3770.755.782.147


EXAMPLE 2

The 10-kg rod AB is confined so that its ends move in the horizontal and vertical slots. The spring has a stiffness of k = 800 N/m and is unstretched when θ = 0°. Determine the angular velocity of AB when θ = 0°, if the rod is released from rest when θ = 30°.


EXAMPLE 2

Potential Energy The two diagrams of the rod,

when it is located at its initial and final positions as shown

The datum, used to measure the gravitational potential energy, is placed in line with the rod when θ = 0°.


EXAMPLE 2

When the rod is in position 1, the center of gravity G is located below the datum so that the gravitational potential energy is negative.

(positive) elastic potential energy is stored in the spring, since it is stretched a distance of s1 = (0.4 sin 30°) m, thus

J

ksWyV

19.6

)30sin4.0)(800(21)30sin2.0(1.98

21

2

2111


EXAMPLE 2

When the rod is in position 2, the potential energy of the rod is zero, since the spring is unstretched, s2 = 0, and the center of gravity G is located at the datum. Thus,

Kinetic Energy The rod is released from rest from position 1,

thus (vG)1 = 0 and ω1 = 0, and

02 V

01 T


EXAMPLE 2

In position 2, the angular velocity is ω2 and the rod’s mass center has a velocity of (vG)2. Thus,

Using kinematics, (vG)2 can be related to ω2 as shown

22

222

22

222

])4.0)(10(121[

21))(10(

21

21)(

21

G

GG

v

IvmT


EXAMPLE 2

At the instant considered, the instantaneous center of zero velocity (IC) for the rod is at point A; hence (vG)2 =(rG/IC)ω2 = (0.2)ω2

Substituting into the previous expression and simplfying, we get

222 267.0 T


EXAMPLE 2

Conservation of Energy

srad

VTVT

/82.40267.019.60

2

22

2211


See you …….In the FINAL EXAMINATION

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Lecture 6 V4