Presentation on theme: "Non-Central Coordinate System"— Presentation transcript:
1Non-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 motionwhere 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.
2For 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?
3Unit 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 vectorwhere 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.
4Let’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:
5Thus, 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?
6Indianapolis 500 speedtrack 1/8 milestraight-away1/4 mileturn5/8 mile straightawayGiven 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?
7In ¼ mile, the car goes through a 90 degree turn 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:
8With 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.
9In 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
10Given the 2 equations and 2 unknowns: The solution is given as follows:The condition for slipping becomes:
11Case 1: Assume no angle on the bank Qualitatively, one can see that any acceleration (braking or accelerating) reduces the maximum speed at which you can take the turn without slipping.
12Case 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!
13With 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?
14At the Indy500, the maximum speed around a turn assuming μ=0.3 is:
15Problem 1 The driver of an automobile decreases her speed at a constant rate from 45 to 30 mi/h over a distance of 750 ftalong a curve of 1500-ft radius. Determine the magnitudeof the total acceleration of the automobile after theautomobile has traveled 500 ft along the curve.
16Problem 1The driver of an automobile decreases her speed at aconstant rate from 45 to 30 mi/h over a distance of 750 ftalong a curve of 1500-ft radius. Determine the magnitudeof the total acceleration of the automobile after theautomobile has traveled 500 ft along the curve.1. Use tangential and normal components: These componentsare used when the particle travels along a circular path. Theunit vector et is tangent to the path (and thus aligned with thevelocity) while the unit vector en is directed along the normal tothe path and always points toward its center of curvature.2. Determine the tangential acceleration: For a constanttangential acceleration, the acceleration can be determined fromv dv = at dxvovxox
17Problem 1The driver of an automobile decreases her speed at aconstant rate from 45 to 30 mi/h over a distance of 750 ftalong a curve of 1500-ft radius. Determine the magnitudeof the total acceleration of the automobile after theautomobile has traveled 500 ft along the curve.3. Determine the normal acceleration: For known velocity andradius of curvature, the tangential acceleration is determine byan =4. Determine the magnitude of the total acceleration: Themagnitude of the total acceleration is given bya = at2 + an2v2r
1845 mi/h = 44 ft/s 45 mi/h = 66 ft/s v dv = at dx v dv = at dx 1 Problem 1 Solution45 mi/h45 mi/h = 44 ft/s45 mi/h = 66 ft/sat750 ftDetermine the tangential acceleration.30 mi/hvovxoxv dv = at dx1500 ft6644750v dv = at dx21( ) = at ( )at = ft/s2
19v dv = at dx v dv = -1.613 dx 1 (v12 - 662 ) = -1.613 ( 500 - 0 ) 2 Problem 1 Solutionat = ft/s21500 ft45 mi/hv1an500 ftThe speed after the automobiletraveled 500 ft.vovxoxv dv = at dx66v1500v dv = dx21(v ) = ( )v1 = 52.4 ft/s
20v2 an = r ( 52.4 ft/s )2 an = = 1.828 ft/s2 1500 ft a = at2 + an2 Problem 1 Solution45 mi/hat = ft/s2Determine the normal acceleration.500 ftv2r52.4 ft/sanan =( 52.4 ft/s )2an = = ft/s21500 ft1500 ftDetermine the magnitude of the total acceleration.a = at2 + an2a = ( ft/s2 )2 + ( ft/s2 )2a = 2.44 ft/s2
21Problem 2 A a v0 B Knowing that the conveyor belt 15 ftA25 ftav0BKnowing that the conveyor beltmoves at the constant speedv0 = 24 ft/s, determine the angle afor which the sand is deposited onthe stockpile at B.
22Knowing that the conveyor belt moves at the constant speed Problem 215 ftA25 ftav0BKnowing that the conveyor beltmoves at the constant speedv0 = 24 ft/s, determine the angle afor which the sand is deposited onthe stockpile at B.1. Analyzing the motion of a projectile: Consider the the verticaland the horizontal motion separately.2. Consider the horizontal motion: When the resistance of the aircan be neglected, the horizontal component of the velocityremains constant (uniform motion). The distance, velocity, andtime are related byx = x0 + (vx)0 t
23Knowing that the conveyor belt moves at the constant speed Problem 215 ftA25 ftav0BKnowing that the conveyor beltmoves at the constant speedv0 = 24 ft/s, determine the angle afor which the sand is deposited onthe stockpile at B.3. Consider the vertical motion: When the resistance of the aircan be neglected, the vertical component of the acceleration isconstant (uniformly accelerated motion). The distance, velocity,acceleration and time are related byy = y0 + (vy)0 t g t212
24(-15 ft) = 0 + (24 ft/s) sin a t - (32.2 ft/s2) t2 1 2 Problem 2 Solution15 ftA25 ftav0BxyConsider the horizontal motion.x = x0 + (vx)0 tx = -25 ftx0 = 0(vx)0 = v0 cos a(vx)0 = (24 ft/s) cos aConsider the vertical motion(25 ft) = 0 + (24 ft/s) cos a t12y = y0 + (vy)0 t g t2y = -15 fty0 = 0(vy)0 = v0 sin a(vy)0 = (24 ft/s) sin aTwo equations withtwo unknowns, a and t.(-15 ft) = 0 + (24 ft/s) sin a t (32.2 ft/s2) t212
25Problem 2 Solution15 ftA25 ftav0Bxy25 = cos a t-15 = sin a t t212Solve the equations for a and t.2524 teliminate a: sina = cos2a = ( )22524 t-15 = 24 t ( ) t2t t = 0two solutions:t = s , a = 6.09ot = s , a = 52.9oa = 6.09o or 52.9o