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PLANAR KINETİCS OF RIJID BODY

DYNAMICS – Lecture Notes / Mehmet Zor

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Force and Acceleration

4.1.a

4.1

(F=ma, M=Iα)

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4.1.1 Deriving the relationships between external loads and accelerations in the plane motion of rigid bodies:

According to Newton's 2nd law, when more than one force acts on a material point, let's recall the relationship between the total resultant force, mass and acceleration:

The acceleration of the particle is in the direction of the resultant force.

(2.2.a)

Our goal now is to obtain the force and moment equation for a rigid body in terms of accelerations.

We can think of a rigid body as a combination of infinitely many material points with mass dm that are stuck together. (Figure 4.2.b)

(a)

A rigid body subjected to external loads

Differential particles forming a rigid body.

(b)

External and internal forces acting on material point i.

(c)

Figure 4.2

4.1 Planar Kinetics of Rigid Body / Force and Acceleration

Figure 4.1

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4.1 Planar Kinetics of Rigid Body / Force and Acceleration

resultant internal force

resultant force

Forces acting on particle i:

acceleration of particle i :

According to Newton's 2nd law:

Points i and P are points of the same object.

From the absolute acceleration equation 3.5:

(3.5)

According to the law of action and reaction, all internal forces balance each other and therefore the total internal force is zero.

If the center of gravity is taken as G instead of P:

(4.1)

or simply:

x

i

x

y

G

y

x

y

i

P

y

Figure 4.3

Figure 4.5

Figure 4.4.a

Figure 4.4.b

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4.1 Planar Kinetics of Rigid Body / Force and Acceleration

If we write the moment of the resultant force acting on particle i about point P:

Since the internal forces balance each other, the sum of their moments about point P is zero::

Since the total moment about point P will be the sum of the total moments of each particle i:

According to equation 4.2, the total moment about point P is only the sum of the moments of the external loads.

(4.2)

..>>

The explicit expression of equation 4.2 is:

x

dm

x

y

i

P

Figure 4.6

Figure 4.4.a

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4.1 Planar Kinetics of Rigid Body / Force and Acceleration

(4.3)

For particule i :

Vector multiplication of a vector with itself is zero.

In planar motion, the moment vector is perpendicular to the plane of motion.

(3.5)

Now we will try to express the total moment about point P in terms of accelerations

dm

x

y

i

P

Figure 4.6

y

Figure 4.4.a

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4.1 Planar Kinetics of Rigid Body / Force and Acceleration

x

dm

x

y

i

P

Figure 4.6

(4.3)

(4.4)

It is the mass moment of inertia about the axis perpendicular to the plane passing through P.

x

dm

i

G

P

y

In this case, equation 4.4 will become:

(4.5)

I_G is the mass moment of inertia about the axis perpendicular to the plane passing through the center of gravity G.

Let's rewrite equation 4.3:

Equations 4.4 and 4.5 give the expression of the moments of external forces with respect to a point in terms of accelerations. Now we will examine these equations a little more and understand the concept of kinetic moment..>

Figure 4.7

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4.1 Planar Kinetics of Rigid Body / Force and Acceleration

From the Parallel Axis Theorem :

4.1.2 Kinetic Moment:

If we substitute equations (I), (II) and (III) into equation 4.4:

From equation 3.5:

We place the axis set at any point P.

(I)

(II)

(I)

(II)

(4.4)

(III)

(4.6)

x

y

P

G

Figure 4.8.b

x

y

P

G

Figure 4.8.a

(see Statics-lecture notes - chapter 8)

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4.1.2 Solution of Kinetic Problems According to D’alembert’s Principle

External Forces System

Effective Forces System

(4.1)

(4.7.a)

2 Independent Vector Equations

(4.7.b)

(4.8)

(4.9)

(Pozitive (+)directions are the same on both sides of the equation)

y

x

3 Independent Scalar Equations

A summary and application of the rigid body kinetic equations derived so far is D’alembert’s Principle. According to this principle, the system of external forces acting on a rigid body at a time t is equal to the system of effective forces. The movement is in the x-y plane and the axis of rotation is the z axis, which is perpendicular to the plane of movement. G: The center of gravity of the object.

or

Figure 4.9.b

G

O

(4.5)

(4.10)

y

x

4.1 Planar Kinetics of Rigid Body / Force and Acceleration

Figure 4.9.a

G

O

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4.1.3 Rigid Body Systems:

External Forces System

Effective Forces System

In Plane Motion, 3 Scalar Equations are written:

: D’Alembert’s principle can be applied to systems consisting of more than one rigid body.

G₁

G₂

G_n

d₁

O

The signs were written by paying attention to the rotation directions in the figure.

G₁

G₂

G_n

O

e₂

e₁

1-)

2-)

3-)

2 Vectorial Equations

1-)

2-)

(4.12)

(4.13)

(4.14)

(4.15)

(4.16)

4.1 Planar Kinetics of Rigid Body / Force and Acceleration

Figure 4.10.a

Figure 4.10.b

*(Example 4.1.3 and Question 4.1.2 are related to rigid systems.)

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(m: mass of object)

y

x

G

Rod

G

Sphere

G

Disk

G

z

Cylindir

(4.11.a)

(4.11.b)

(4.11.c)

(4.11.d)

4.1.4 Mass Moment of Inertia

4.1 Planar Kinetics of Rigid Body / Force and Acceleration

Figure 4.11.a

Figure 4.11.b

Figure 4.11.c

Figure 4.11.d

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Example 4.1.1

G

.

1m

1.8m

51m

1m

y

x

A

B

Solution logic : The car will slow down with constant acceleration. If we find its acceleration, we can find the distance S covered from the timeless velocity formula, which is valid for linear motion with constant acceleration. If the value of S is less than 51 meters, the turtle will not be crushed.

initial velocity:

4.1 Planar Kinetics of Rigid Body / Force and Acceleration

Figure 4.4

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External Forces System

Effective Forces System

G

W

+x

Turtles cannot be crushed ☺

G

Since there is no movement in the y direction:

Since the car does not rotate on the z-axis:

(I)

(II)

(III)

from equatios (I), (II) and (III) we can find:

from timeless velocity equations (equ. 1.9):

A practical information: A practical tip: In order not to make a sign error in the calculations, we always choose the direction of the unknown values (even if we guess correctly) in the + direction and we can use the result exactly as it is, together with its sign, in the calculations.

m=1 ton=1000kg, W=mg=10000N

1m

4.1 Planar Kinetics of Rigid Body / Force and Acceleration

1m

1.8m

Figure 4.5.a

Figure 4.5.b

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Example 4.1.2

P = 120N

0.4m

G

B

A person holding a disk weighing 80N and having a radius of 0.4m in space suddenly withdraws his hand and at that instant a pulling force of 120N P is applied vertically to the rope wrapped around the disk. Accordingly, at this instant, calculate

a-) the angular acceleration of the disk and

b-) the acceleration of the rope

External Forces System

Effective Forces System

+y

P = 120N

G

B

W = 80N

0.4m

4.1 Planar Kinetics of Rigid Body / Force and Acceleration

G

+x

Solution:

Let's calculate the mass moment of inertia with respect to the axis passing through G, perpendicular to the plane:

a-)

Figure 4.6

Figure 4.7.a

Figure 4.7.b

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b-)

0.4m

G

B

Since G and B are two points of the same object, equation 3.5 can be applied between them.

The acceleration of the rope is equal to the acceleration component of point B in the y-direction:

B is the common point between the disk and the rope.

So we need to find the acceleration of B vectorially.

Kinematic Analysis

Since it is in the +y direction, its vector expression is:

(For the instant t=0)

4.1 Planar Kinetics of Rigid Body / Force and Acceleration

Figure 4.8

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Question 4.1.1 (*)

4.1 Planar Kinetics of Rigid Body / Force and Acceleration

Figure 4.9

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Example 4.1.3

Weights A and B were hung on the pulley system, which was supported at the center O and could rotate freely around the perpendicular axis passing through this point, and the system was left in motion. Accordingly, at any time t, calculate

a-) the angular acceleration of the pulleys,

b-) the translational accelerations of the weights A and B,

c-) the reaction force from the bed to the pulley,

d-) the forces in the ropes.

100 N

50 N

0.8m

0.4m

A

B

O

x

y

Total Weight of the pulleys : W_o= 49N,

Total radius of inertia of the pulleys: : k = 1m

Gravity acceleration : g=9.8m/s²

4.1 Planar Kinetics of Rigid Body / Force and Acceleration

Solution..>>

Figure 4.11

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Solution:

Total mass moment of inertia of the pulleys with respect to the center of gravity O :

100 N

50 N

r₁

A

B

O

r₂

x

y

c

d

1

2

The system is a rigid body system consisting of 2 masses and 2 pulleys.

Since the pulleys are mounted to each other at point O, they can be considered as a single object and have the same angular acceleration (∝) value.

Masses A and B move in the y-direction and their accelerations are equal to the y-components (or tangential components) of the accelerations of the points (c and d) common to the pulleys at that instant.

O and c are points of the same object. The acceleration of point c at any time t is:

We can now write the tangential acceleration component we need as a scalar:

Similarly for point d;

from the equation 3.5 :

4.1 Planar Kinetics of Rigid Body / Force and Acceleration

Figure 4.12

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For the whole system

3 independent scalar equations

Effective Forces System

(1)

(2)

(3)

From equ. (3):

From equ. (1):

From equ.(2):

External Forces System

(We found the questions asked in options a, b and c.)

Now let's expand these equations :

Attention: The rope forces for the entire system are internal forces and therefore are not shown in the external forces system. Since we separate the system from the fixed joint at O, the Rx, Ry forces are shown. An external force is always shown in each part we separate from the system. (For better understanding, examine the Free Body Diagrams topic explained in Statics.)

100 N

50 N

A

B

O

x

y

c

d

1

2

0,8m

0,4m

4.1 Planar Kinetics of Rigid Body / Force and Acceleration

A

B

O

0,8m

0,4m

x

y

c

d

1

2

Figure 4.13.a

Figure 4.13.b

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d-) To find the forces in the ropes, we need to examine objects A and B separately.

External Forces Sys.

Effective Forces Sys.

control

A

B

A

B

x

y

O

0,8m

0,4m

x

y

O

0,8m

0,4m

External Forces Sys.

Effective Forces Sys.

External Forces Sys.

Effective Forces Sys.

If we examine the pulley system separately

Same results as we found before

4.1 Planar Kinetics of Rigid Body / Force and Acceleration

+y

Figure 4.14.a

Figure 4.14.b

Figure 4.15.a

Figure 4.15.b

Figure 4.16.a

Figure 4.16.b

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A pendulum was created by connecting a sphere of 15kg mass and 100mm radius to the OA rod of 10kg mass. An external moment of M= 50Nm is applied to the rod in the clockwise direction. If 𝜃=45⁰ and the angular velocity of the pendulum is 𝜔=3𝑟𝑎𝑑/𝑠 at the position shown,

a-) find the angular acceleration of the pendulum,

b-) find the reaction forces at the fixed joint O.

Question 4.1.2 (*)

4.1 Planar Kinetics of Rigid Body / Force and Acceleration

Answer

Figure 4.17