# Two Dimensional Motion How do we work with two dimensions when we consider motion? We work with vectors by working with the components! Does it matter.

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Two Dimensional Motion How do we work with two dimensions when we consider motion? We work with vectors by working with the components! Does it matter which form we use: rectangular or polar? Since vectors add nicely in rectangular, we need to work with rectangular components!

Two Dimensional Motion But if we work in components, how are the components “connected” ? By time! Time is the same for the x and for the y. For each time, there is only one x and one y.

Special Case #1: Trajectory Motion We have equations for objects falling under the influence of gravity if we neglect air resistance and if we don’t go too far from the earth’s surface: y = y o + v yo t + (1/2)a y t 2 v y = v yo + a y t

Special Case #1: Trajectory Motion If we again consider no air resistance, there should be zero acceleration in the x direction. This leads to the following equations for x: x = x o + v xo t + (1/2)*0*t 2 v x = v xo + 0*t which become simply: v x = v xo and x = x o + v x t.

Special Case #1: Trajectory Motion In effect, then, we have three equations: x = x o + v x t y = y o + v yo t + (1/2)a y t 2 v y = v yo + a y t. In these three equations we have nine quantities: x o, x, y o, y, v x, v yo, v y, a y and t. This means that we need to know 6 of these in order to solve for the other 3.

Example of Trajectory Motion Problem: If you are at the top of a building 12 meters above the ground, and you throw a ball out with a speed of 33 m/s at an angle of 40 o above the horizontal, how far away from the building will the ball land (assuming the ground is level at the base of the building) ?

Example of Trajectory Motion The first thing we recognize is that this is a trajectory problem, so we have the three equations: x = x o + v x t y = y o + v yo t + (1/2)a y t 2 v y = v yo + a y t.

Example of Trajectory Motion The second thing we recognize is that this takes place on the earth, so that a y = -9.8 m/s 2. Next let’s draw the diagram and put our information from the problem on that diagram: (go to the next slide for the diagram).

Example of Trajectory Motion x o = a y = -9.8 m/s 2 y o = v xo = t = v yo = x = y = v x = v xo v y =

Example of Trajectory Motion Let’s look at the information given in the problem and see how that specifies some of the remaining unknown quantities: Problem: If you are at the top of a building 12 meters above the ground,... We can identify this value as the initial y value, so y o = +12 meters. We can also choose to measure from the building, so this means x o = 0 meters.

Example of Trajectory Motion x o = 0 ma y = -9.8 m/s 2 y o = 12 m v xo = t = v yo = x = y = v x = v xo v y =

Example of Trajectory Motion Let’s continue to look at the information: … and you throw a ball out with a speed of 33 m/s at an angle of 40 o above the horizontal … This information refers to the initial velocity, but it is in polar form, not rectangular! So we need to use the polar to rectangular transformation equations:

Example of Trajectory Motion v yo = v o sin (  o ) = 33 m/s sin (40 o ) = 21.21 m/s, and v xo = v o cos (  o ) = 33 m/s cos (40 o ) = 25.28 m/s.

Example of Trajectory Motion x o = 0 ma y = -9.8 m/s 2 y o = 12 m v xo = 25.28 m/s t = v yo = 21.21 m/s x = y = v x = v xo v y =

Example of Trajectory Motion Now we look at the last part of the problem statement: … how far away from the building will the ball land … How far away from the building is asking the question: x = ? But we do know it lands on the ground, so this indicates that: y = 0 meters.

Example of Trajectory Motion x o = 0 ma y = -9.8 m/s 2 y o = 12 m v xo = 25.28 m/s t = v yo = 21.21 m/s x = ? y = 0 m v x = v xo v y =

Example of Trajectory Motion Notice that we have three equations, and we have three quantities still unknown: x, t and v y. One of these, x, is explicitly asked for in the problem. The other two, t and v y are not explicitly asked for but neither are they specified. Now we can use our three equations to solve for our three unknowns!

Doing the algebra x = x o + v x t y = y o + v yo t + (1/2)a y t 2 v y = v yo + a y t Substituting in the knowns, we have: x = (0 m) + (25.28 m/s * t) 0 m = (12 m) + (21.21 m/s*t) + ( 1/2 *-9.8 m/s 2 *t 2 ) v y = (21.21 m/s) + (-9.8 m/s 2 * t)

Doing the algebra x = (0 m) + (25.28 m/s * t) 0 m = (12 m) + (21.21 m/s*t) + ( 1/2 *-9.8 m/s 2 *t 2 ) v y = (21.21 m/s) + (-9.8 m/s 2 * t) Note that the first equation has two unknowns, but the second equation has only one unknown: t, but it has a square on it in one term and is found in another term also. Thus we use the quadratic equation to solve for t: in terms of the quadratic formula: a = -4.9 m/s 2, b = 21.21 m/s, and c = 12 m.

Doing the algebra 0 m = (12 m) + (21.21 m/s*t) + ( 1/2 *-9.8 m/s 2 *t 2 ) If we have ax 2 + bx + c = 0, then x = [-b +/- {b 2 - 4ac} 1/2 ] / [2a] so with a = -4.9 m/s 2, b = 21.21 m/s, and c = 12 m we get: t = [-21.21 m/s +/- {(21.21 m/s) 2 - (4*-4.9 m/s 2 *12 m)} 1/2 ] / [2*-4.9 m/s 2 ] = [- 21.21 m/s +/- {26.17 m/s}] / -9.8 m/s 2 = -0.51 s or 4.83 s.

Doing the algebra t = -0.51 s or 4.83 s. Which answer is the one we are looking for? The negative answer (-0.51 s) would indicate that it hit the ground before we threw it. This is obviously not the answer we’re looking for. In fact, the equations do not apply before we threw the ball because there were other forces acting on the ball besides gravity! Thus the correct time for the ball to hit is: t = 4.83 s.

Doing the algebra Having used the second equation to solve for t, we now use the first equation: x = (0 m) + (25.28 m/s * t) to solve for x: x = (0 m) + (25.28 m/s * 4.83 s) = 122.1 meters This does appear to be a reasonable distance for the ball to go (neglecting air resistance)! This answers the question posed. If we want, we could use the third equation to solve for v y.

Trajectories The previous example was showing how we could predict where the ball was going to go based on how we threw it. We can also determine how to throw a ball so that it hits where we want it to. This is the point of the computer homework program: Trajectories (Vol 1, #5).

Trajectories In the trajectories program, the setup is like the previous example, except in what is given and in what is asked for. Instead of giving the initial speed and angle, you are given the final x position. Thus we have x as known, but we have v o and  o (which is equivalent to v xo and v yo ) as unknown - NOT A FAIR TRADE! We now have four unknowns instead of three!

Trajectories What does it mean to have four unknowns (v xo, v yo, t and v y ) with three equations? It means that there is more than one right way of hitting the target (solving the problem). Does that correspond to what we expect to happen? Yes - we can hit the target by various combinations of v o and  o. In this case, you have one free choice (within limits).

Special Case #2 of 2-D Motion: Circular Motion What does it mean to go in a circle? The radius must be constant! This indicates that polar coordinates might be easier to use than rectangular! In order to work with vectors and motion, we need to work in rectangular coordinates to derive the results, but once the results are there, it may turn out that polar form is nicer for final equations!

Uniform Circular Motion Even though the radius is constant, the angle changes. For right now, we will consider the special case where the angle changes at a constant rate. This type of motion is called uniform circular motion. Note that since the radius doesn’t change, there should be no radial component of the velocity! (r = constant, v r = zero).

Uniform Circular Motion What about the angle,  ? In circular motion the angle does change! In uniform circular motion the angle changes at a constant rate! We call the change in angle with respect to time the angular speed, .  =  t Note that the units of angular speed,  are: radians/second.

Uniform Circular Motion If we wish to use cycles instead of radians, we define a new symbol, f: f =  * (1 cycle / 2  radians), or more simply:f =  / 2 . Note that the units of f are cycles/second. This unit also has its own name called a Hertz. Note also that we almost never use the unit of degrees/second when indicating angular speed.

Uniform Circular Motion Also note that the inverse of f, measured in cycles/sec, is what we call the period, T: T = 1/f, where T is measured in seconds/cycle; that is, it is the time for the object to complete one revolution. For the earth spinning on its axis, T is one day; for the earth orbiting around the sun, T is one year.

Uniform Circular Motion It might seem, then, that since the radius is constant and the radial speed is zero, and that since the angular speed is constant for uniform circular motion, it might seem that there is no acceleration in uniform circular motion! But that is not looking at the motion from the rectangular point of view:

Uniform Circular Motion From the rectangular point of view, the x motion goes forward, slows down and stops, and then reverses until it slows down and stops and the the whole process repeats. The same is true for the y motion. From this point of view it is easy to see that there is acceleration in both the x and y components.

Uniform Circular Motion How do we understand the acceleration from the polar point of view? Acceleration is the change in velocity. The velocity is a vector, and in polar form has a magnitude and a direction. Although the magnitude of the velocity (the speed) does not change in uniform circular motion, the direction changes! There is an acceleration when an object turns.

Uniform Circular Motion We need to consider two more aspects: the formula for this turning acceleration, and the formula for the speed in polar form. We already have the angular speed (in radians/sec), but this is not the same as velocity (in meters/second).

Uniform Circular Motion If we go back to the definition of angles measured in radians (  = s/r), we can see that the circular speed: v  =  s /  t =  (r  ) /  t = r(  /  t) = r  Let’s look at this situation from the rectangular point of view to check the above result and look at the acceleration.

Uniform Circular Motion Rectangular viewpoint Circular motion is defined by: r = constant. Uniform circular motion is defined by: d  /dt =  = constant, so d  =  dt; upon integration  o   d  = 0  t  dt, or  -  o =  t, so we have  =  o +  t. Converting polar to rectangular, we have: x = r cos(  ) = r cos(  o +  t) y = r sin(  ) = r sin(  o +  t).

Uniform Circular Motion Rectangular viewpoint x = r cos(  ) = r cos(  o +  t) y = r sin(  ) = r sin(  o +  t) Now we use: v x = dx/dt and v y = dy/dt: v x = dx(t)/dt = dx(  )/d  * d  (t)/dt = r (- sin(  ) ) *  = -  r sin(  o +  t) v y = dy(t)/dt = dy(  )/d  * d  (t)/dt = r (+cos(  ) ) *  =  r cos(  o +  t)

Uniform Circular Motion Rectangular viewpoint x = r cos(  ) = r cos(  o +  t) y = r sin(  ) = r sin(  o +  t) v x = -  r sin(  ) = -  r sin(  o +  t) v y =  r cos(  ) =  r cos(  o +  t) Now we use: a x = dv x /dt and a y = dv y /dt: a x = -    r cos(  ) = -    r cos(  o +  t) a y = -    r sin(  ) = -    r sin(  o +  t).

Uniform Circular Motion Back to Polar To convert back to polar for position, we use the inverse transformation equations: x = r cos(  ) y = r sin(  )    t r = [x 2 + y 2 ] 1/2 = [r 2 cos 2 (  ) + r 2 sin 2 (  )] 1/2 = r[cos 2 (  ) + sin 2 (  )] 1/2 = r  = inv tan[y/x] = inv tan[ r sin(  ) / r cos(  )] = inv tan[tan(  )] = .

Velocity Back to Polar To convert velocity back to polar: v x = -  r sin(  ) v y =  r cos(  )    t v = [v x 2 + v y 2 ] 1/2 = [  2 r 2 sin 2 (  ) +  2 r 2 cos 2 (  )] 1/2 =  r[sin 2 (  ) + cos 2 (  )] 1/2 =  r  v = inv tan[v y /v x ] = inv tan[  r cos(  ) / -  r sin(  )] if  >0, +cos(  )=+sin(   ); -sin(  ) =+cos(  +90 o ) then  v = inv tan (+sin(   ) /+cos(  +90 o ) =   if  <0, -cos(  )=+sin(   ); +sin(  ) =+cos(  -90 o ) then  v = inv tan (+sin(   ) /+cos(  -90 o ) =   

Velocity Back to Polar  velociy =  position   Note that the direction of the velocity,  v, is perpendicular to the direction of the position (the radius),  which means the velocity is tangent to the circle.

Acceleration Back to Polar To convert acceleration: a x = -    r cos(  ) a y = -    r sin(  )    t a = [a x 2 + a y 2 ] 1/2 = [  4 r 2 cos 2 (  ) +  4 r 2 sin 2 (  )] 1/2 =   r[cos 2 (  ) + sin 2 (  )] 1/2 =   r  a = inv tan[a y /a x ] = inv tan[-   r sin(  ) / -   r cos(  )] =  . Note that the direction of the acceleration,  a, is opposite to the direction of the position (the radius), , which means the acceleration points to the center of the circle.

Uniform Circular Motion Putting all of this together, we have in polar form: r = constant;  changes with time v r = zero;v  = r  a r  -r     a  = zero with  t, f =  and  T = 1/f.

Uniform Circular Motion Note that for solving circular motion problems, we have four equations: v  = r  a r  -r   f =  and  T = 1/f. In these four equations, there are six quantities: r, v , a r,  , f and T. Thus we need to identify two of these to solve for the remaining four.

Example of Uniform Circular Motion How fast are we moving now since we are “riding” the earth as it orbits the sun? Facts you should know: period for earth’s orbit is one year, distance to the sun (the radius of the circle) is 93 million miles. [Actually the earth goes in an ellipse, but it is very close to being a circle. For a first try, this will give very good results.]

Example (cont.) First step in reading the problem is to recognize this as uniform circular motion. Thus we have the four equations: v  = r  a r  -r   f =  and  T = 1/f. Next from the problem we are given: r = 93 million miles = 1.49 x 10 11 m; T = 365 days = 3.15 x 10 7 sec.

Example, cont. v  = r  a r  r   f =  and  T = 1/f. r = 93 million miles = 1.49 x 10 11 m; T = 365 days = 3.15 x 10 7 sec. Since we have four equations (above) and four unknowns (v   a r and f), we should be able to solve for the unknowns. Note that we drop the minus sign for a r since it merely indicates that the acceleration is towards the center.

Example, cont. v  = r  a r  r   f =  and  T = 1/f. r = 93 million miles = 1.49 x 10 11 m; T = 365 days = 3.15 x 10 7 sec. Using T = 1/f, we can find f: f = 1/T = 1/ 3.15 x 10 7 sec = 3.17 x 10 -8 Hz. Using f =  we can find  :   f = 2*3.14*  3.17 x 10 -8 Hz = 1.99 x 10 -7 rad/sec

Example, cont. v  = r  a r  r   f =  and  T = 1/f. r = 1.49 x 10 11 m; T = 3.15 x 10 7 sec. f = 3.17 x 10 -8 Hz.  1.99 x 10 -7 rad/sec Using v  = r  we can find v  : v  = r*  = 1.49 x 10 11 m * 1.99 x 10 -7 rad/sec = 29,670 m/s = 66,340 miles/hour !

Example, cont. v  = r  a r  r   f =  and  T = 1/f. r = 1.49 x 10 11 m; T = 3.15 x 10 7 sec. f = 3.17 x 10 -8 Hz.  1.99 x 10 -7 rad/sec v  = 29,670 m/s. Completing the problem, we can use a r  r   to find a r : a r = r  2 = 1.49 x 10 11 m * (1.99 x 10 -7 rad/sec) 2 =.0059 m/s 2

Example, cont. A couple of notes: we have a very high speed, but a very low acceleration in the past example. Note that this is the case because  was very small, and the acceleration involved  2. In many cases, we can have a very high acceleration if  is very high. This is the principle behind the centrifuge.

Relative Velocity How do we work with a situation in which an object moves on something that is itself moving? Examples: Flying an airplane in air that is moving (flying in a wind), running a boat on the river, walking inside an airplane that is moving.

Relative Velocity Since velocity is a vector, we get the total velocity by simply adding the velocity of the object in or on the medium to the velocity of the medium. But we must add these two velocities as vectors! That means, we must work in rectangular coordinates.

Example: Boat on a river Problem: A boat is to cross a river such that it gets to a point due East of where it starts. The boat goes 5 m/s and the river flows South at 2 m/s. We’ll assume here that the river flows at a uniform speed - not faster in the middle and slower at the edges. What direction should the boat be aimed? Also, if the river is 100 meters wide, how long will it take to cross the river?

Example: Boat on a river First we draw v R = 2 m/s a diagram: since the river will push the boat downstream we should head v b = 5 m/s somewhat upstream.  = ? w = 100 m

Example: Boat on a river The problem, then, is what is the angle,  ? The basic principle we use is vector addition. v b + v R = v Total But what do we want the two vectors to add up to (what do we want v Total to be)? What does the phrase “due East” mean here? It means that v Total should have a zero North- South component!

Example: Boat on a river Converting to rectangular components from polar, and then adding the components, gives: v b cos(  ) + v R (0) = v Total-x v b sin(  ) + v R (-1) = v Total-y = 0 m/s From the y component equation, we have an equation for  : (5 m/s) sin(  ) - (2 m/s) = 0 m/s.

Example: Boat on a river (5 m/s) sin(  ) - (2 m/s) = 0 m/s. Solving for  gives:  = Arcsin(2 m/s / 5 m/s) = 23.6 o.

Example: Boat on a river The time to cross the river depends on the width of the river - but this is purely in the x direction. This means that we need the x component of the Total speed: v b cos(  ) + v R (0) = v Total-x = (5 m/s) cos(23.6 o ) + (2 m/s)(0) = 4.58 m/s. Since v x =  x/  t, where  x = w,  t = t = w/v x = 100 m / 4.58 m/s = 21.8 seconds.

Relative Velocity A similar type of problem involves finding the angle to go across the river in the quickest time, rather than in the shortest distance (directly across). For this problem, what we want is the biggest x-speed so that the time to cross the x distance is as small as possible. This we see can only be if we head straight across!

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