Non-Central Coordinate System

1. Non-Central Coordinate System In addition to the 1-D, 2-D, and 3-D coordinate systems discussed in previous lectures, there is another useful coordinate system known as normal/tangential coordinates. The main difference in this coordinate system is that origin is no longer fixed. Instead, this coordinate system references particle motion by the tangential velocity along its path of motion where v is the velocity magnitude and is a unit vector denoting the tangential direction of the path. As long as the motion is primarily confined to two dimensions, this coordinate system proves to be very useful in solving problems.

2. For example, given the path trajectory s, the associated directions of the tangential unit vector, which is parallel to the instantaneous velocity, and normal unit vector are shown below. However, the normal unit vector could be drawn any number of ways that is perpendicular to the tangential direction. Why is it drawn this way? What significance is this normal direction, given the radius of curvature?

3. Unit Vector Time Derivatives The time derivative of a unit vector is proportional to how quickly it changes direction. For example, suppose you have a vector rotating about an axis with angular velocity vector where is the frequency of rotation, denoted in revolutions (or radians) per second, and is the axis around which the vector is rotating. The direction change is shown mathematically above.

4. Let’s provide a model system equivalent with cylindrical coordinates to make the derivation easier. For example, in the system of a particle in a stable circular orbit around a point, the vectors can be written as follows:

5. Thus, the normal-tangential coordinate system is fundamentally different than all the previously studied coordinate systems in that the origin is not at a fixed location. To see why, consider the path of a particle shown in the diagram below… Why would we ever want to use this coordinate system, whose origin is constantly shifting?

6. Indianapolis 500 speedtrack 1/8 mile straight- away 1/4 mile turn 5/8 mile straightaway Given the average speed of a car for 1 lap is around 220 mph and the driver is moving at a constant speed, what is the acceleration around the turn?

7. In ¼ mile, the car goes through a 90 degree turn. Thus, the circumference of the circle is 1 mile, and its radius can be computed from: The acceleration of the car can be computed from: However, assuming the car has constant velocity: The acceleration becomes:

8. With advances in automotive technology, how fast can a car safely drive in the Indy 500? The catch here, is how we define “safely”. According to my definition, the word “safely” means the speed at which the car does not slip on the road.

9. In general, we know that the acceleration of the car is given by: We also have the relation for force: From this, we can arrive at two relations to solve for the two unknowns, T and N

10. Given the 2 equations and 2 unknowns: The solution is given as follows: The condition for slipping becomes:

11. Case 1: Assume no angle on the bank Assume α= 0 Qualitatively, one can see that any acceleration (braking or accelerating) reduces the maximum speed at which you can take the turn without slipping.

12. Case 2: Assume no tangential acceleration Thus, the velocity can be expressed as: Curiously, a banked curve can support potentially “infinite” velocity if the denominator equals zero: (i.e.) Given typical values for the friction coefficient, this would require that the angle of the bank exceeds 45 degrees, which would be too scary to drive on!

13. With advances in automotive technology, how fast can a car safely drive in the Indy 500? So, we have arrived at 2 qualitative conclusions… The first is that when taking a steep curve, you should maintain constant velocity to prevent slipping. The second is that the maximum speed you can take the curve depends on the friction coefficient and the bank angle. If you were on the NASCAR team, you would be well-advised to know these parameters in designing the car. In the INDY 500, the bank angle around the turns is 9o 12’. Assuming a friction coefficient of 0.3, what is the maximum speed you can take around the turn?

14. At the Indy500, the maximum speed around a turn assuming μ=0.3 is: assuming μ=0.7 assuming μ=1.8

15. Problem 1 The driver of an automobile decreases her speed at a constant rate from 45 to 30 mi/h over a distance of 750 ft along a curve of 1500-ft radius. Determine the magnitude of the total acceleration of the automobile after the automobile has traveled 500 ft along the curve.

16. Problem 1 v x vo xo The driver of an automobile decreases her speed at a constant rate from 45 to 30 mi/h over a distance of 750 ft along a curve of 1500-ft radius. Determine the magnitude of the total acceleration of the automobile after the automobile has traveled 500 ft along the curve. 1. Use tangential and normal components:These components are used when the particle travels along a circular path. The unit vector et is tangent to the path (and thus aligned with the velocity) while the unit vector en is directed along the normal to the path and always points toward its center of curvature. 2. Determine the tangential acceleration: For a constant tangential acceleration, the acceleration can be determined from v dv = at dx

17. Problem 1 v2 r The driver of an automobile decreases her speed at a constant rate from 45 to 30 mi/h over a distance of 750 ft along a curve of 1500-ft radius. Determine the magnitude of the total acceleration of the automobile after the automobile has traveled 500 ft along the curve. 3. Determine the normal acceleration: For known velocity and radius of curvature, the tangential acceleration is determine by an = 4. Determine the magnitude of the total acceleration: The magnitude of the total acceleration is given by a = at2 + an2

18. Problem 1 Solution v x v dv = at dx vo xo 44 750 v dv = at dx 66 0 1 2 45 mi/h 45 mi/h = 44 ft/s 45 mi/h = 66 ft/s at 750 ft Determine the tangential acceleration. 30 mi/h 1500 ft ( 442 - 662 ) = at ( 750 - 0 ) at = - 1.613 ft/s2

19. Problem 1 Solution 45 mi/h at = -1.613 ft/s2 500 ft v1 an v x v dv = at dx vo xo 1500 ft v1 500 v dv = -1.613 dx 66 0 1 (v12 - 662 ) = -1.613 ( 500 - 0 ) 2 The speed after the automobile traveled 500 ft. v1 = 52.4 ft/s

20. Problem 1 Solution v2 r 45 mi/h at = -1.613 ft/s2 Determine the normal acceleration. 500 ft 52.4 ft/s an = an ( 52.4 ft/s )2 an = = 1.828 ft/s2 1500 ft 1500 ft Determine the magnitude of the total acceleration. a = at2 + an2 a = ( - 1.613 ft/s2 )2 + ( 1.828 ft/s2 )2 a = 2.44 ft/s2

21. v0 A 15 ft a B 25 ft Problem 2 Knowing that the conveyor belt moves at the constant speed v0= 24 ft/s, determine the angle a for which the sand is deposited on the stockpile at B.

22. Problem 2 v0 A 15 ft a B 25 ft Knowing that the conveyor belt moves at the constant speed v0= 24 ft/s, determine the angle a for which the sand is deposited on the stockpile at B. 1. Analyzing the motion of a projectile:Consider the the vertical and the horizontal motion separately. 2. Consider the horizontal motion:When the resistance of the air can be neglected, the horizontal component of the velocity remains constant (uniform motion). The distance, velocity, and time are related by x = x0 + (vx)0t

23. Problem 2 v0 A 15 ft a B 25 ft 1 2 Knowing that the conveyor belt moves at the constant speed v0= 24 ft/s, determine the angle a for which the sand is deposited on the stockpile at B. 3. Consider the vertical motion:When the resistance of the air can be neglected, the vertical component of the acceleration is constant (uniformly accelerated motion). The distance, velocity, acceleration and time are related by y = y0 + (vy)0t - g t2

24. Problem 2 Solution y v0 x A 15 ft a B 25 ft 1 2 1 (-15 ft) = 0 + (24 ft/s) sin at - (32.2 ft/s2) t2 2 Consider the horizontal motion. x = x0 + (vx)0t x = -25 ft x0 = 0 (vx)0 = v0 cos a (vx)0 = (24 ft/s) cos a Consider the vertical motion (25 ft) = 0 + (24 ft/s) cos at y = y0 + (vy)0t - g t2 y = -15 ft y0 = 0 (vy)0 = v0 sin a (vy)0 = (24 ft/s) sin a Two equations with two unknowns, a and t.

25. Problem 2 Solution y v0 x A 1 15 ft -15 = 0 + 24 sin at - 32.2 t2 a 2 B 25 ft 25 24 t 25 24 t 25 = 0 + 24 cos at Solve the equations for a and t. eliminate a: sina = 1 - cos2a = 1 - ()2 -15 = 24 t 1 - ()2 - 16.1 t2 250.21 t4 - 1059 t2 + 850 = 0 two solutions: t = 1.048 s , a = 6.09o t = 1.729 s , a = 52.9o a = 6.09o or 52.9o