Theoretical Mechanics . Kinematics

1. Theoretical Mechanics.Kinematics Kinematics of particles Practical Lesson № 10

2. Problem№ 1 A law of point M motion along a trajectory is given: wherea=4 and Given: Необходимо определить: 1) trajectory; 2) specify the initial position whenandintermediate position when 3) magnitude anddirection ofpoint velocity when 4) magnitude anddirection ofpoint acceleration when 5) found for the time moment tangent and normalacceleration; 6) examine the character of motion of the pointM when 7) construct the velocity hodograph.

3. 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) у (m) a • The trajectory equation • will be findby adding the right and left side of the equationssystem (1): 3 2 1 • Then • (2) х (м) О -4 -3 -2 -1 • It is equation of theline(Fig. 1). 1 2 3 a Figure1 • Whent=0: x=aandy=0. • Since the timethetrajectory of the point will be thesegment of the line, for whichand(see equation (1)). • Let us construct the trajectory of the point (Fig. 1).

4. 2 Determination of theinitial and intermediate position of the point • Whenfrom equation (1) getm andm. • Let us showinFig. 2 thepointМ0. • When from equation(1) getmandm. • Let us showinFig. 2 the pointМ1. у (m) М1 a 3 2 1 М0 О 1 2 3 a х(m) Figure2

5. 3 Determination of thepoint velocitywhen • It is known that for coordinate method point velocity is determined by the projections: • where • and • We have when у (m) • i.е. М1 a 3 2 1 • i.е. О х(m) 1 2 3 a • The magnitude of the point velocityМ1when • will be: Figure3

6. 4Determination of thepoint accelerationwhen • It is known that for coordinate method point acceleration is determined by the projections: • and • where • We havewhen у (m) • i.е. М1 • i.е. a 3 2 1 • The magnitude of the point accelerationМ1when • will be: О 1 2 3 a х(m) Figure4 • Let us show the acceleration vector of the pointinFig.4.

7. 5Determination of the point tangent and normal accelerationwhen • Given that • where • is directed along the tangent to the trajectory ( – the unit tangent vector), • is directed along the principal normal to the trajectory of the concavity( – the unit principal normal vector). у (m) • As then • The magnitudeis defined by the formula for the plane motion М1 • Then a 3 2 1 О х(m) 1 2 3 a Figure5

8. When • we have • Normal acceleration of the point when moving straight is always zero. • Let us show this analytically: у (m) • Let us show in Fig. 6 the component • The magnitude of point accelerationМ1when • will be: М1 a 3 2 1 • Constructed acceleration vector is the same with the result obtained in the coordinate method, which confirms the correctness of calculations. О х(m) 1 2 3 a Figure6

9. 6 Examine the character of motion of the pointwhen • For examine the character of motion of the pointM we will use the coordinate method of specifying the motion. • 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.

10. 7 Construction of the velocity hodograph • The velocity hodographiscurve (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. • Velocity hodograph equation have the form • (3) • Added the left and right sides of the equations system we find that • InFig. 8 the velocity hodograph of point M is shown corresponding to the equations(3). y*=vy(m/s) aπ 3π 2π π О -aπ-3π-2π-π 1 2 3 a х*=vx(m/s) Figure8

11. Let us show the position of points and whent = 1/8s, t = 1/6s, t = 1/4 s,t = 1/3s respectively. • When • we have • When • we have • When • we have • When • we have y*=vy(m/s) aπ 3π 2π π -aπ-3π-2π-π О 1 2 3 a х*=vx(m/s) Figure9

12. Problem № 2 (Home task) A law of point M motion along a trajectory is given: Given: Determine: 1) trajectory; 2) specify the initial position whenandintermediate position when 3) magnitude anddirection ofpoint velocity when 4) magnitude anddirection ofpoint acceleration when 5) found for the time moment tangent and normalacceleration; 6) examine the character of motion of the pointM when 7) construct the velocity hodograph.

13. 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) • The trajectory equation will be find, excluding from the (1) parametert: • Then • (2) Figure1 • It is equation of thehyperbole (Fig. 1). • Further solve the problem yourself!!!

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. z(m) О у (m) х(m) М1 Figure1

15. Solution of problem№ 3 • It is known that for coordinate method point velocity is determined by the formula: • Then • The magnitude of the point M velocity will be • It is known that for coordinate method point acceleration is determined by the formula: • Then • The magnitude of the point M acceleration will be • For examine the character of motion of the pointM we will use the coordinate method of specifying the motion. Then • Ast>0, therefore, the motion is uniformly accelerated (acceleration is constant) for any time.

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

17. Solution of problem№ 4 (1) • 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 17

18. Solution of problem№ 4 (2) • 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 18

19. Summary slide – Information about all topics studied during the lesson. It is compiled by student him/herself!