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# 4. coordinate systems part 1 by-ghumare s m

smghumare
20 de Apr de 2020
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### 4. coordinate systems part 1 by-ghumare s m

1. Coordinate System in Curvilinear Motion Part-1 Sub- Engg. Mechanics By: Mr. Ghumare S. M. When particle moves along the curved path, a motion is called as Curvilinear motion. In curvilinear motion velocity is always tangential to the curved path.
2. Coordinate Systems The Motion of the particle in curvilinear motion is expressed using following three Coordinate systems. i.e. To find the velocity and acceleration along a curve. • Coordinate System in Curvilinear Motion: 1. Rectangular / Cartesian Coordinate Systems (x-y system) 2. Normal and Tangential Coordinate Systems / Path variables (n-t system) 3. Radial and Transverse Coordinate Systems / Polar coordinates (r - )
3. Coordinate Systems
4. Rectangular / Cartesian Coordinate Systems (x-y) Position Vector (1)r x i y j  
5. Rectangular / Cartesian Coordinate Systems (x-y) Position Vector (1)r x i y j        Differentiate Eq.(1) w.r.t. t (2)V x y dr dx dy i j dt dt dt V i V j         Resultant Velocity :     22 V x yV V  Differentiate Eq.(2) w.r.t. t (3) yx x y dvdvdv i j dt dt dt a a i a j         Direction: tan y x V V   Direction: tan y x a a  
6. Example on Rectangular Coordinate System Ex.1 The motion of a particle is defined by the equation and where in meters. t is in seconds. Determine the velocity and acceleration when t =4 seconds Given         3 3 2 2 22 2 2 0 16 16 (4) 1018 / , 18 4 18 (4) 4 4 280 / 1018 280 1055 / 15.38 V V x V x y x y t m s t t m s V V m s                2 4 6x t t  3 2 6 2y t t  andx y 4 3 2 2 2 4 6 16 6 48 x x x t t dx V x t dt d x a x t dt          & && 3 2 2 2 2 6 2 18 4 (1) 36 4 (2) y y y t t dy V y t t dt d y a y t dt                       & && Find v & a at t = 4 Sec. Put t= 4 secs in Eq. (1) and find Velocity Comp. Vx and Vy
7. Example 1 Continue…. • To find acceleration (a)         2 2 22 2 2 2 0 48 48(4) 768 / , 36 4 36(4) 4 142 / 768 142 781 / tan 10.48 x y x y y x a t m s a t m s a a a a a m s a a                
8. Example -2 Ex.2 The motion of a particle is defined by the relation and , where in meters. t is in seconds. Determine the velocity and acceleration when t =2 seconds. Given         2 2 22 2 2 0 2 5 2 2 5 9 / , 3 16 3 (2) 16 2 20 / , 9 20 21.9 tan 65.32 / 77 V x V x V y x x y x y t m s t t m s V v V m s v                     2 5x t t  3 2 8 2y t t   andx y 2 2 2 5 2 5 2 x x x x t t dx V t dt dVd x a dt dt         3 2 2 2 2 8 2 3 16 (1) 6 16 (2) y y y y t t dy V t t dt dVd y a t dt dt                   Find v & a at t = 2 Sec. Put t= 4 secs in Eq. (1) and find Velocity Comp. Vx and Vy
9. Example 1 Continue…. • To find acceleration (a) Put t = 2 Sec         2 2 22 2 2 2 0 2 2 / , 6 16 6(2) 16 4 / 2 4 4.472 / tan 63.44 x y x y y x a m s a t m s a a a a a m s a a                 
10. Rectangular Coordinate system x yV and V are given by differntiating it you have calculate a and ax y  Some times in the problem only velocity Components are given  Some times in the problem only Acceleration Components are given x yand are given by integrating given equation it you have calculate v andx y a a v
11. Tangential and Normal Comp. of Velocity and Acceleration (n-t System) Consider the motion of particle along the curved path. The velocity vector is always tangential to the path but acceleration vector is at some other angle. Let, ˆ ˆ t te =e = Unit vector in Tangential Dirrction e =e = Unit vector in Normal Dirrction n n
12. Tangential and Normal Comp. of Velocity (n-t ) Velocity ˆ ˆ t n t t n n V= V i +V j V= V e +V e Here, Tangential Comp. of Velocity (Vt)= V itself and Normal Comp. of Velocity (Vn)= 0 Resultant Velocity (V) = Velocity Components: 2 2 )t nV= (V ) +(V
13. Normal and Tangential Acceleration Compo. (n-t) Acceleration Components: ˆ ˆ t n t t n n a a i a j a a e a e     t dv a dt  2 n v a   Here, Tangential Comp. of Accl. and Normal Comp. of Acceleration Acceleration 2 2 )t nAccl. a= (a ) +(a
14. Important Points Velocity Components: Tangential Comp. of Velocity Vt =V Normal Comp. of Velocity Vn = 0 Accl. Components: Tangential Comp. of Accl. at = = Rate of Change Velocity Normal Comp. of Accl. an = = Directed towards Centre of Curvature dv dt 2 v  To find Radius of Curvature ( ) 1. an  2 2 / , / nv v a   2 2 2 1 ( / ) ( ) ( / ) 2. if Eq is expressed as dy dx y f x d y dx     
15. Important Points 1. If the magnitude of velocity / speed is constant then velocity comp. Vt =0. only Vn will present. 2. If the direction of the particle does not change i.e. Motion along straight line, then an=0, only at will present. 3. If the speed in increasing along the curve then at will be + ve and directed in the direction of motion. 4. If the speed in decreasing along the curve then at will be -ve and directed opposite to the direction of motion.   
16. Numerical Example-1 Ex.1. An outdoor track is 126m in diameter. A runner increases his speed at a constant rate from 4.2 to 7.2 m/s over a distance of 28.5m. Determine the total acceleration of the runner 2 sec after he began to increase his speed. Given, Diameter = 126m Hence Radius( )= 63m, t= 2 secs Find a=? As speed increases at constant rate, u=4.2 m/s and v=7.2 m/s S=28.5 m Using , 4.2 0.6 2 Find Velocity t = 2secs x x v = 5.4m/sectv u a t v     2 2 2 2 2 2 , 7.5 4.5 2 28.5 0.6 / x xt t t v u a s a a m s     
17. Example 1 continue…. Find normal component of Acceleration (an) 2 2 25.4 0.463 / 63 n v a m s r      Total Acceleration (a)         2 2 2 2 2 0.6 0.463 0.758 /Total Accl. t na a a a m s     
18. Numerical Example-2 Ex.2. A train runs at a speed of 120km/hr in a curved track of radius 900m. The application of brakes suddenly, causes the train to slow down at a constant rate. After 6 secs, the speed has been reduced to 72km/hr. Determine the acceleration immediately after the brakes are applied. Given, Radius ( )= 900m, t= 6 secs Find a=? Initial Speed , u= 120km/hr. =120 x = 33.33 m/s and Final velocity v= 72km/hr= 72 x = 20 m/s S=28.5 m Since the train slows down the constant rate, find (at )  5 18 5 18 220 33.33 2.222 / 6 t v u a m s t      
19. Example 2 Continue…. Immediately after the brakes are applied (an) 2 2 233.33 1.235 / 900 Here, velocity v = 33.33 m/sec n v a m s r    Total Acceleration (a)         2 2 2 2 2 2.222 1.235 2.542 /Total Accl. t na a a a m s      
20. Numerical Example-3 Ex.3. The automobile is originally at rest at x = 0, It then starts to increase its speed v according to the equation. It then starts to increase its speed v according to the equation where t in seconds and s in meters. Determine magnitude of velocity and acceleration when t =18 seconds. Find v and a = ? at t=18 secs 2 2 / 0.015 /dv dt t m s 2 2 3 1 1 3 ' 4 2 2 0.015 (3) 3 4 0 0 0 0.0 0.015 / (1) .(1) 0.015 (2) 3 0 0 0 0.015 (2 ) 3 1 Integrate Eq. (2) w.r.t.t to get disIntegrate w.r.t.t to get Velocity p. x Given t S C dv a t m s dt At t Eq t v C At t v t C S C v S                               4 4 '5 0.015 (3 ) 3 4 12x t t     
21. 4 4 2 2 ' 3 0.015 0.015(18) 12 12 . 131.22 90 .(1) 0.015 0.015 (18) .(2 ) 18sec 0.015 0 3 t 2 t=18 Sec Find displacement (S)- 131.22mts i.e. vehicle is on the Curved path. Find a at t=18secs 4.86 m/st t S See Fig m From Eq dv a t dt From Eq t s t v              3 22 .015(18) 3 : 26.16 72 n 2 26.16 m/sec To Find a 9.5m/sn v a      Example 3 Continue….
22. Also see the next video on Coordinate System Part-2 for third coordinate system i.e. Radial and Transverse Component System Thank You
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