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Theoretical Mechanics . Kinematics. Kinematics of particles. P ractical Lesson № 10. Problem № 1. A law of point M motion along a trajectory is given: where a =4 and. Given :. Необходимо определить:. 1) t rajectory ;. - PowerPoint PPT Presentation
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19/04/2023 1
Theoretical Mechanics.Kinematics
Practical Lesson № 10
Kinematics of particles
19/04/2023 2
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.
19/04/2023 3
-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)
19/04/2023 4
• 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
19/04/2023 5
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
19/04/2023 6
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
19/04/2023 7
• 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
19/04/2023 8
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��������������
19/04/2023 9
• 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
19/04/2023 11Figure 9
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
• When 1t 1 8 s/ , we have
x
y
x v 2 8 m s
y v 2 8 m s
*
*
, ,
, .
• When 2t 1 6 s/ , we have
x
y
x v 3 5 m s
y v 3 5 m s
*
*
, ,
, .
• When 3t 1 4 s/ , we have
x
y
x v a m s
y v a m s
*
*
,
.
4N
4N4v��������������
19/04/2023 12
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
,
.
19/04/2023 13
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
19/04/2023 14
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)
19/04/2023 15
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.
19/04/2023 16
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
19/04/2023
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
19/04/2023
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.
19/04/2023 19
Summary slide – Information about all topics studied during the lesson.
It is compiled by student him/herself!
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