Theoretical Mechanics . Kinematics

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Theoretical Mechanics.Kinematics

Practical Lesson № 10

Kinematics of particles

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2) specify the initial position when and intermediate position when

5) found for the time moment tangent and normal acceleration;

Problem № 1Given: 2

2

x a t

y a t

cos ,

sin ,

1t 1 3 c/ , ; 0t 0 c

A law of point M motion along a trajectory is given:where a=4 and Необходимо определить:

1t 1 3 c/ , ;

1t 1 3 c/ , ;

1t 1 3 c/ ,

1t 1 3 c/ , ;

1 s/ .

1) trajectory;

3) magnitude and direction of point velocity when

4) magnitude and direction of point acceleration when

6) examine the character of motion of the point M when 7) construct the velocity hodograph.

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-4 -3 -2 -1

Figure 1О

у (m)

х (м)

• The trajectory equation will be find by adding the right and left side of the equations system (1): • Then

(2) 1 2 3 a

3

2

1

y f x( )

• Since the time the trajectory of the point will be the segment of the line, for which and (see equation (1)).

• Let us construct the trajectory of the point (Fig. 1).

• It is equation of the line (Fig. 1).

t 0,

x 0 y 0

х (m) y (m) х (m) y (m)

0 a a 0 • When t=0: x=a and y=0.

2

2

x a t

y a t

cos ,

sin .

x y a y a x

a

1 Determination of the trajectory of the point• The point moves in a plane ХОY (Fig. 1), because the changes in the coordinates х

and у are given.• Its law of motion is given in coordinate form:

(1)

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• When from equation (1) get m and m.

• When from equation (1) get m and m.

• Let us show in Fig. 2 the point М0.

• Let us show in Fig. 2 the point М1.

Figure 2

О

у (m)

х (m) 1 2 3 a

a

3

2

1М0

М1

0t 0 c 0x a 0y 0

1x 2 1y 11t 1 3 c/ ,

2 Determination of the initial and intermediate position of the point

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Figure 3

• It is known that for coordinate method point velocity is determined by the projections:

О

у (m)

х (m) 1 2 3 a

a

3

2

1

• We have when x yv v i v j,

where xv x , yv y and 2 2

x yv v v .

xv x 2a t t a 2 tcos sin sin , i.е.

x

2v a a 3 14 0 866 2 7a m s

3sin , , , ,

yv y 2a t t a 2 tsin cos sin , i.е.

yv 2 7a m s, .

• The magnitude of the point velocity М1 when

will be:

1v��

xv 0

yv 0

1t 1 3 s/ ,

1t 1 3 s/ ,

М1

1t 1 3 c/ ,

2 2 2v a 2 t a 2 t 2 a 2 t 2 a 2 t

2 32 a 3 14 2a 3 14 3 8a m s

3 2

sin sin sin sin

, sin , , / .

3 Determination of the point velocity when

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Figure 4

• It is known that for coordinate method point acceleration is determined by the projections:

О

у (m)

х (m) 1 2 3 a

a

3

2

1

• We have when

x yw w i w j, ��

where x xw v x , y yw v y and 2 2x yw w w .

2xw x 2a 2 tcos , i.е.

2 2x

2w 2a 3 14 9 9a m s

3, cos , ,

2

yw y 2a 2 tcos , i.е.

2 2y

2w 2a 3 14 9 9a m s

3, cos , .

• The magnitude of the point acceleration М1 when

will be:

2 22 2w 9 9a 9 9 a 2 9 9a 14a m s, , , . • Let us show the acceleration vector of the point in Fig. 4.1w

��

1w��

1t 1 3 s/ ,

М1

1t 1 3 s/ ,

1t 1 3 s/ ,

xw 0 yw 0

4 Determination of the point acceleration when

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• As then

• Given that nw w w ,

��where

2 2nw w w .

w

��is directed along the tangent to the trajectory ( – the unit tangent vector),

nw��

is directed along the principal normal to the trajectory of the concavity( – the unit principal normal vector).n

nw w , ��

• The magnitude is defined by the formula for the plane motion

w

2 2

xx yyw

x y.

2 2

2 2

2 2 3

2

2

a 2 t 2a 2 t a 2 t 2a 2 tw

a 2 t a 2 t

2 a 2 t 2a 2 t 2a 4 t

2 a 2 t2 a 2 t

2a 4 t

2 t

sin cos sin cos

sin sin

sin cos sin

sinsin

sin.

sin

• Then

Figure 5

О

у (m)

х (m) 1 2 3 a

a

3

2

1

1t 1 3 s/ ,

М1

5 Determination of the point tangent and normal acceleration when

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Figure 6

• When we have 22

2

34 1 4a 3 142a 23w 14a m s2 33 2

, ,sin.

sin

• Let us show in Fig. 6 the component w .��

w

��

• Normal acceleration of the point when moving straight is always zero.• Let us show this analytically:

2 2

n 2 2

a 2 t 2a 2 t a 2 t 2a 2 tw 0

a 2 t a 2 t

sin cos sin cos.

sin sin

1t 1 3 s/ ,

О

у (m)

х (m) 1 2 3 a

a

3

2

1

М1

2 2 2nw w w 14a 14a м с .

• The magnitude of point acceleration М1 when

will be:1t 1 3 s/ ,

• Constructed acceleration vector is the same with the result obtained in the coordinate method, which confirms the correctness of calculations.

1w��

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• For examine the character of motion of the point M we will use the coordinate method of specifying the motion.

2

2 2

2 3

2 2 2

2 2 2 2 2 2 2

2 4

x x y yv w v w v w a sin t a cos t

a sin t a cos t a sin t a cos t

a sin t .

�� • Then

• The time intervals (0+ n/2 , ¼ + n/2, s), when it expression is positive correspond to the accelerated motion, the time intervals (¼ + n/2, (n+1)/2, s), when this expression is negative, correspond to slowness of motion.

• When we get

• Therefore, when the motion is slowed.

1t 1 3 s/ ,

2 3 2 34 32 2 0

3 2v w a sin a .

��

1t 1 3 s/ ,

6 Examine the character of motion of the point when

1t 1 3 s/ ,

19/04/2023 10Figure 8

О х*=vx (m/s) 1 2 3 a

aπ

3π

2π

π

• Velocity hodograph equation have the form (3)

x

y

x v x a 2 t

y v y a 2 t

*

*

sin ,

sin .

• In Fig. 8 the velocity hodograph of point M is shown corresponding to the equations (3). y*=vy (m/s)

-aπ -3π -2π -π

• Added the left and right sides of the equations system we find that x yv v .

7 Construction of the velocity hodograph• The velocity hodograph is curve (or straight line), which describes the end of the

velocity vector when a point moves on the trajectory if the velocity vector beginning is enshrined in the fixed point.

v

v

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О х*=vx (m/s) 1 2 3 a

• Let us show the position of points and when t = 1/8 s, t = 1/6 s, t = 1/4 s,t = 1/3 s respectively.

y*=vy (m/s)

• When 4t 1 3 s/ , we have

x

y

x v 3 5 m s

y v 3 5 m s

*

*

, ,

, .

1N ,

1v�� 1N2v

��3v��

2N3N

2N , 3N

aπ

3π

2π

π

-aπ -3π -2π -π


x

y

x v 2 8 m s

y v 2 8 m s

*

*

, ,

, .


x

y

x v 3 5 m s

y v 3 5 m s

*

*

, ,

, .


x

y

x v a m s

y v a m s

*

*

,

.

4N

4N4v��

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Given:

1) trajectory; 2) specify the initial position when and intermediate position when

5) found for the time moment tangent and normal acceleration;

1t 1s;0t 0 s

1t 1s

3) magnitude and direction of point velocity when

4) magnitude and direction of point acceleration when

6) examine the character of motion of the point M when 7) construct the velocity hodograph.

A law of point M motion along a trajectory is given:

Determine:

1t 1s;

1t 1s;

1t 1s;

Problem № 2 (Home task)

x t

3y 2

t

,

.

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Figure 1

• The point moves in a plane ХОY (Fig. 1), because the changes in the coordinates х and у are given.

• Its law of motion is given in coordinate form:

(1)

• The trajectory equation will be find, excluding from the (1) parameter t:

• Then(2)

y f x( )

t x.

3y 2

x.

• It is equation of the hyperbole (Fig. 1).

x t

3y 2

t

,

.

Further solve the problem yourself!!!

1 Determination of the trajectory of the point

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Problem № 3• Motion of the point M (Fig. 1) is given by the equation:• It is necessary to determine: velocity, acceleration and

the character of the point motion.• Let us show in Fig. 1 position of the point M at time

of 1 s.

212 3 2r t i t j k .

Figure 1

Оу (m)

х (m)М1

z (m)

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Solution of problem № 3• It is known that for coordinate method point velocity is determined by the

formula:

x y zv v i v j v k 24ti 3j.

22 2 2 2 2 2x y zv v v v 24t 3 576t 9 3 64t 1.

• The magnitude of the point M velocity will be

dr dx dy dzv v i j k

dt dt dt dt,

• Then

• It is known that for coordinate method point acceleration is determined by the formula:

x y zw w i w j w k 24i . ��

22 2 2x y zw w w w 24 24. • The magnitude of the point M acceleration will be

2 2 2 2

2 2 2 2

d r d x d y d zw w i j k

dt dt dt dt,

��

• Then

• For examine the character of motion of the point M we will use the coordinate method of specifying the motion. Then

24 24 0 0 576x x y y z zv w v w v w v w t t. ��

• As t>0, therefore, the motion is uniformly accelerated (acceleration is constant) for any time.

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Problem № 4• It is known acceleration of the point:• It is necessary to determine: - point velocity;- equation of the point motion.If it is known that

w 12ti 5k . ��

0 0v 5 1 6 r 4 3 2, , , , , . ��

17

• As

• Let us find constants from the initial conditions,namely, when : i.e.

• Substituting these data into the expressions for the velocity components we obtain

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Solution of problem № 4 (1)

2

1 2 3

dvw v 12ti 5k dt 12tidt 0 jdt 5kdt

dt

6t C i C j 5t C k.

��

0v 5 1 6, , ,��

21 1

2

3 3

6 0 C 5 C 5

C 1

5 0 C 6 C 6

.

.

. .

2v 6t 5 i 1 j 5t 6 k.

17

2x 1 y 2 z 3v 6t C v C v 5t C, , .

t 0

0 5 0 1 0 6x y zv v v( ) , ( ) , ( ) .

18

• As

• Let us find constants from the initial conditions,namely, when : i.e.

• Substituting these data in the expressions for the equation of the point motion components we obtain

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Solution of problem № 4 (2)

18

23

x 1 y 2 z 3

5tr 2t 5t C r t C r 6t C

2, , .

t 0

0 4 0 3 0 2( ) , ( ) , ( ) .x y zr r r

2

2

23

1 2 3

drv r 6t 5 i 1 j 5t 6 k dt

dt

6t 5 idt 1 jdt 5t 6 kdt

5t2t 5t C i t C j 6t C k

2.

0r 4 3 2, , ,��

31 1

2 2

2

3 3

2t 5t C 4 C 4

t C 3 C 3

5t6t C 2 C 2

2

.

.

.

2

3 5tr 2t 5t 4 i t 3 j 6t 2 k

2.

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Summary slide – Information about all topics studied during the lesson.

It is compiled by student him/herself!

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Theoretical Mechanics . Kinematics