TEXT BOOK PROBLESceng.tu.edu.iq/med/images/ميكانيك_هندسي2-2.pdf · ٢٨ TEXT BOOK PROBLEMS: 2/135 A car P travels along a straight road with constant speed v = 65 km/hr.

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The acceleration

Where

and

So,

TEXT BOOK PROBLES:

Ans. a) 2.4 m/s2, b)3.39 m/s2, c )5.37 m/s2

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Ans. a=1.961 m/s2

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3.3 Polar coordinate (r- ):

If a particle is located by the radial distance r [from a fixed point and an angular measurement angle started from a reference line which is the x-axes as shown in the figure below:

Where r is the position vector at A it magnitude is equal to the radial r and a direction is specified by the unit vector so the location of particle at A can be express as:

r = r …..(a)

and for the direction is a unit vector as shown in the figure below that is in the positive (+) -direction and is in the negative (-) r-direction.

Their magnitudes are:

and dividing by it can be deduced:

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Velocity and acceleration:

from equation (a) and differentiate , substitute gives:

since Then

and and

The acceleration

Substitute and

The Circular Motion:

If the motion is on a circular path r is constant in this case the velocity and acceleration can be written as:

Important note:

Note: all the equations of constant acceleration are valid only replace the acceleration ac by .

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Example:

Solution:

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Example:

A tracking radar lies in the vertical plane of the path of a rocket which is coasting in unpowered flight above the atmosphere. For the instant when =300, the tracking data give

deg./sec. The acceleration of the rocket is due only to gravitational attraction and for its particular altitude is31.4 ft./sec2vertically down. For these conditions determine the velocity of the rocket and the values of .

Solution:

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TEXT BOOK PROBLEMS:

2/135 A car P travels along a straight road with constant speed v = 65 km/hr. At the instant when the angle =600, determine the values of in m/s and in deg./s.

Ans. = 9 m/s ,

2/136 The ladder of a fire truck is designed to be extended at the constant = 0.5 m/s and to be elevated at the constant rate .As the position = 500 and l = 15 m is reached, determine the magnitudes of the velocity and acceleration of the fireman at A. Ans. v = 1.320 m/s, a= 0.0551 m/s2

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4.Relative Motion (Translating axes):

In the previous sections we discuss a single particle movement or motions so we used a fixed reference coordinate but sometimes there are more than one particles moving in one plane or in two plans, each has its own curvilinear path and we want to calculate their velocities, in this case we need to specified with respect to a fixed coordinate system with non-rotating (translating), the (x-y) coordinate to be used with the aid of vector. Remember that in the curvilinear motion there are to component of movement which they the x and its unit vector (i), and y and its unit vector (j).

In the figurer below two particles which they have a separate curvilinear motion.

The observation position is at particle B and the moved particle is A, it means that the fixed axes is in position B, so it can be say that.

and it known that

where the subscript " " means A relative to B.

Velocity and Acceleration:

A/B or VA = VB + VA/B .…(a)

A/B or aA = aB + aA/B …. (b)

j and A/B =

If the position of observation is changed from B to A as shown in the figure below

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and it known that

Where the subscript " " means B relative to A.

Velocity and Acceleration:

B/A or VB = VA + VB/A …… (c)

B/A or + ……..(d)

j and B/A =

NOTES:

1. In this case (translating) the unit vectors i and j have no derivation because their directions and magnitude are remaining unchanged.

2. , VB/A = - VA/B , and aB/A = - aA/B

3 Any axes can be used depending on the problem specifications.

4. Three ways are possible to solve the vector equations (a, b.c, d), which they the Graphical, Trigonometric, and Vector algebra.

Example:

Solution:

The observation is from A to B so

VB = VA + VB/A

a) Graphical solution.

The velocity of A is 800 km/h and its direction is on the x-axes.

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First take a suitable scale of say (1 cm = 100 km/h) and draw an 8 cm in the horizontal line from reference point (p), second draw a line with 450 from (p) represent the velocity line of B. Now from the position (o) of the horizontal line draw a line with 600 represent the relative velocity VA/B as shown below, the two line will intersect at point c. now take the scale to get the value of VB, VA/B.

PC = 7.17 cm or VB = 717 km/h

OC= 5.85 cm or VB/A = 585 km/h

b) Trigonometric solution.

Draw a vector triangle as shown

From the triangles sine law:

c) Vector algebra solution

Using unit vector i, j

VA= 800 i + 0j , VB =

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Example:

Car A is accelerating in the direction of its motion at the rate of 3 ft./sec2 car B is rounding a curve of 440 ft. radius at a constant speed of 80 mi/hr. Determine the velocity and acceleration which car B appears to have to an observer in car A if car A has reached a speed of 46 mi/hr. for the positions represented.

Solution

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TEXT BOOK PROBLEMS:

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Summary of Equations:

Kinematic Particle Motions

Rectilinear Motion Curvilinear Motion Variable acceleration

Constant acceleration Cartesian Coordinate Motion

Normal Tangential Coordinate (n, t)

Polar coordinate (r , )

=

Circular path

Circular path

= =

The Projectile Relative Motion (Translating axes)

B/A or VB = VA + VB/A B/A or +

j and B/A =

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Kinematic/Plane Rigid Bodies:

A rigid body is a solid body in which deformation is zero or so small it can be neglected. The distance between any two given points on a rigid body remains constant in time regardless of external forces exerted on it. A rigid body is usually considered as a continuous distribution of mass. We still studding the behavior of the rigid body from the kinematic point of view, it means that no study of the external force or the mass of the body effects.

Four types of motion to be investigate which they are:

a) Translation which contents the rectilinear and the curvilinear motions.

b) Rotation motion, C) Fixed axis rotation, and d) General plane motion.

As shown the figure below:

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Rotation:

a) Angular Motion Equations: In fact that all the equations derived for linear motion for the position s, velocity v, and acceleration a can be applicable for the case of angular motion if we replace them by angular position , angular velocity , and angular acceleration . So their equations can be written as:

NOTE :An integrations could be carried out to obtain the angular position

from the angular velocity equation and to obtain the angular velocity from angular acceleration .

And for rotation for constant angular acceleration ( :

At angular velocity constant the angular acceleration will be Zero. But when the angular acceleration is constant the following equations for angular displacement, angular velocity can be derived as: from equation

Or

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(a) From equation

Substitute the values of the velocity from equation (a) and integrate the above equation then

or

(b) And from equation

After integration and rearrangement, the teams it can be deduce:

NOTE: The above equations (a, b, and c) are used when the angular acceleration is constant ( .

b) Rotation About a Fixed Axis: When a rigid body rotates about a fixed axis. All points other than those on the axis moves in concentric circles about the fixed axis. Thus for the rigid body shown in the figure rotating about a fixed axis throw O

Point A is moved on a circle of radius r since it moves on a circle the equations of the circular path in curvilinear motion can be rewritten as:

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Velocity and acceleration in a vector form: Using the cross product, and the right hand rule the direction and magnitude of the angular velocity and the angular acceleration cab be evaluated where vector V can be obtaining by crossing into r

Where the vector is normal to the plane of rotation shown in the figure

And the reverse order is.

The acceleration of point A can be obtained by differentiating the cross product of V gives.

Then

As shown in the figure below

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When the gemetric of the rigid body is complecated usually the following procedure is recommended :

Since in the x-y plane motion then only and the other are zero so

And for plane motion usually the angulaur velocity is prependuclaur to the plane x-y it mean that k and its sign is depend on the right hand rule if it C.W it mean negative (-) or postive (+) if it C.C.W

Or

-

And

Since we have only and with same dirction of it sign to be negative (-) C.W and postive(+) C.C.W so,

a = = ]

The magntude of the velocity v and the acceleration a area:

and

Note: the dimensions of is in [rad], is in [rad/s],and is in [rad/s2]

Example:

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Solution:

The sign of the angular velocity and angular aceeleration to be c.w (-).

When then

The flywheel change direction when its angular velocity is momentarily zero, thus

1 rev = thus

Or N2=410x0.159= 65.3 rev

N=N1 +N2 =194.2+65.3= 259 rev

Example:

Solution:

Using the right hand rule:

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Since is C.W then it sign (-)

And also in the same direction it mean it sign (-) but it decreasing it mean the sign will be -4 so the final sign is (+) and in the Z axis so .

-1.2i +1.6j

a = =

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