UNDERSTAND MOTION IN ONE AND TWO DIMENSIONS

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1 SUBAREA I. COMPETENCY 1.0 UNDERSTAND MOTION IN ONE AND TWO DIMENSIONS MECHANICS Skill 1.1 Calculating displacement, aerage elocity, instantaneous elocity, and acceleration in a gien frame of reference Kinematics is the part of mechanics that seeks to understand the motion of objects, particularly the relationship between position, elocity, acceleration and time The aboe figure represents an object and its displacement along one linear dimension. First we will define the releant terms: 1. Position or Distance is usually represented by the ariable x. It is measured relatie to some fixed point or datum called the origin in linear units, meters, for example.. Displacement is defined as the change in position or distance which an object has moed and is represented by the ariables D, d or Δ x. Displacement is a ector with a magnitude and a direction. 3. Velocity is a ector quantity usually denoted with a V or and defined as the rate of change of position. Typically units are distance/time, m/s for example. Since elocity is a ector, if an object changes the direction in which it is moing it changes its elocity een if the speed (the scalar quantity that is the magnitude of the elocity ector) remains unchanged. i) Aerage elocity: = /. The ratio, Aerage here denotes that this quantity is defined oer a period Δ t. ii) Instantaneous elocity is the elocity of an object at a particular moment in time. Conceptually, this can be imagined as the extreme case when Δ t is infinitely small. 4. Acceleration represented by a is defined as the rate of change of elocity and the units are m/. Both an aerage and an instantaneous acceleration can be defined similarly to elocity.

2 From these definitions we deelop the kinematic equations. In the following, subscript i denotes initial and subscript f denotes final alues for a time period. Acceleration is assumed to be constant with time. = at (1) f i + 1 d t at = i + () = + ad (3) f i i + d = In two dimensions the same relationships apply, but each dimension must treated separately. f t (4) Skill 1. Soling problems inoling displacement, time, elocity, and constant acceleration Simple problems inoling distance, displacement, speed, elocity, and constant acceleration can be soled by applying the kinematics equations from the proceeding section. The following steps should be employed to simplify a problem and apply the proper equations: 1. Create a simple diagram of the physical situation.. Ascribe a ariable to each piece of information gien. 3. List the unknown information in ariable form. 4. Write down the relationships between ariables in equation form. 5. Substitute known alues into the equations and use algebra to sole for the unknowns. 6. Check your answer to ensure that it is reasonable. Example: A man in a truck is stopped at a traffic light. When the light turns green, he accelerates at a constant rate of 10 m/s. a) How fast is he going when he has gone 100 m? b) How fast is he going after 4 seconds? c) How far does he trael in 0 a=10 m/s seconds? Solution: I We first construct a diagram of the situation. In this example, the diagram is ery simple, only showing the truck accelerating at the gien rate. Next we define ariables for the known quantities (these are noted in the diagram): a=10 m/s ; i =0 m/s

3 Now we will analyze each part of the problem, continuing with the process outlined aboe. For part a), we hae one additional known ariable: d=100 m The unknowns are: f (the elocity after the truck has traeled 100m) Equation (3) will allow us to sole for f, using the known ariables: f = i + ad m s ( 0m / s) + (10m / s )(100m) 000 f = = m f = 45 s We use this same process to sole part b). We hae one additional known ariable: t=4 s The unknowns are: f (the elocity after the truck has traeled for 4 seconds) Thus, we can use equation (1) to sole for f : f = i + at f = 0m / s + (10m / s )(4s) = 40m / s For part c), we hae one additional known ariable: t= 0 s The unknowns are: d (the distance after the truck has traeled for 0 seconds) Equation () will allow us to sole this problem: 1 d = i t + at 1 d = (0m / s)(0s) + (10m / s )(0s) = 000m Finally, we consider whether these solutions seem physically reasonable. In this simple problem, we can easily say that they do.

4 Skill 1.3 Interpreting algebraically and graphically relationships among position, elocity, acceleration, and time Algebraic relationships are introduced in Skill 1.1. The relationship between time, position or distance, elocity and acceleration can be understood conceptually by looking at a graphical representation of each as a function of time. Simply, the elocity is the slope of the position s. time graph and the acceleration is the slope of the elocity s. time graph. If you are familiar with calculus then you know that this relationship can be generalized: elocity is the first deriatie and acceleration the second deriatie of position. Here are three examples: Initial position<0 Initial position =0 Initial position>0 x x x Initial elocity = 0 Initial elocity >0 Initial elocity <0 a a a a<0 a>0 a=0 t t t There are three things to notice: 1) In each case acceleration is constant. This isn t always the case, but a simplification for this illustration. ) A non-zero acceleration produces a position cure that is a parabola.

5 3) In each case the initial elocity and position are specified separately. The acceleration cure gies the shape of the elocity cure, but not the initial alue and the elocity cure gies the shape of the position cure but not the initial position. Skill 1.4 Analyzing problems inoling motion in two dimensions In the preious section, we discussed the relationships between distance, elocity, acceleration and time and the four simple equations that relate these quantities when acceleration is constant (e.g. in cases such as graity). In two dimensions the same relationships apply, but each dimension must be treated separately. The most common example of an object moing in two dimensions is a projectile. A projectile is an object upon which the only force acting is graity. Some examples: i) An object dropped from rest. ii) An object thrown ertically upwards at an angle iii) A canon ball. Once a projectile has been put in motion (say, by a canon or hand) the only force acting it is graity, which near the surface of the earth implies it experiences a=g=9.8m/s. This is most easily considered with an example such as the case of a bullet shot horizontally from a standard height at the same moment that a bullet is dropped from exactly the same height. Which will hit the ground first? If we assume wind resistance is negligible, then the acceleration due to graity is our only acceleration on either bullet and we must conclude that they will hit the ground at the same time. The horizontal motion of the bullet is not affected by the downward acceleration. Example: I shoot a projectile at 1000 m/s from a perfectly horizontal barrel exactly 1 m aboe the ground. How far does it trael before hitting the ground? Solution: First figure out how long it takes to hit the ground by analyzing the motion in the ertical direction. In the ertical direction, the initial elocity is zero so we can rearrange kinematic equation from the preious section to gie: d t =. Since our displacement is 1 m and a=g=9.8m/s, t=0.45 s. a

6 Now use the time to hitting the ground from the preious calculation to calculate how far it will trael horizontally. Here the elocity is 1000m/s and there is no acceleration. So we simple multiply elocity with time to get the distance of 450m. Motion on an arc can also be considered from the iew point of the kinematic equations. As pointed out earlier, displacement, elocity and acceleration are all ector quantities, i.e. they hae magnitude (the speed is the magnitude of the elocity ector) and direction. This means that if one dries in a circle at constant speed one still experiences an acceleration that changes the direction. We can define a couple of parameters for objects moing on circular paths and see how they relate to the kinematic equations. 1. Tangential speed: The tangent to a circle or arc is a line that intersects the arc at exactly one point. If you were driing in a circle and instantaneously moed the steering wheel back to straight, the line you would follow would be the tangent to the circle at the point where you moed the wheel. The tangential speed then is the instantaneous magnitude of the elocity ector as one moes around the circle.. Tangential acceleration: The tangential acceleration is the component of acceleration that would change the tangential speed and this can be treated as a linear acceleration if one imagines that the circular path is unrolled and made linear. 3. Centripetal acceleration: Centripetal acceleration corresponds to the constant change in the direction of the elocity ector necessary to maintain a circular path. Always acting toward the center of the circle, centripetal acceleration has a magnitude proportional to the tangential speed squared diided by the radius of the path. Centripetal acceleration pulls toward the center of the circle and changes the direction of the total elocity ector. Tangential Speed= the magnitude of the elocity ector. A tangential acceleration changes the tangential speed.

7 Uniform circular motion describes the motion of an object as it moes in a circular path at constant speed. There are many eeryday examples of this behaior though we may not recognize them if the object does not complete a full circle. For example, a car rounding a cure (that is an arc of a circle) often exhibits uniform circular motion. The following diagram and ariable definitions will help us to analyze uniform circular motion. Aboe we see that the mass is traeling a path with constant radius (r) from some center point (x 0, y 0 ). By defining a ariable (θ) that is a function of time (t) and is the angle between the mass s present position and original position on the circular path, we can write the following equations for the mass s position in a Cartesian plane. x = r cos(θ) + x 0 y = r sin(θ) + y 0

8 Next obsere that, because we are discussing uniform circular motion, the magnitude of the mass s elocity () is constant. Howeer, the elocity s direction is always tangent to the circle and so always changing. We know that a changing elocity means that the mass must hae a positie acceleration. This acceleration is directed toward the center of the circular path and is always perpendicular to the elocity, as shown below: This is known as centripetal acceleration and is mathematically expressed as: a = r = 4π r t where t is the period of the motion or the time taken for the mass to trael once around the circle. The force (F) experienced by the mass (m) is known as centripetal force and is always directed towards the center of the circular path. It has constant magnitude gien by the following equation: F = ma = m r Skill 1.5 Analyzing properties of ectors and soling problems inoling ector quantities analytically and graphically Vector space is a collection of objects that hae magnitude and direction. They may hae mathematical operations, such as addition, subtraction, and scaling, applied to them. Vectors are usually displayed in boldface or with an arrow aboe the letter. They are usually shown in graphs or other diagrams as arrows. The length of the arrow represents the magnitude of the ector while the direction in which the arrow points shows the ector direction.

9 To add two ectors graphically, the base of the second ector is drawn from the point of the first ector as shown below with ectors A and B. The sum of the ectors is drawn as a dashed line, from the base of the first ector to the tip of the second. As illustrated, the order in which the ectors are connected is not significant as the endpoint is the same graphically whether A connects to B or B connects to A. This principle is sometimes called the parallelogram rule. If more than two ectors are to be combined, additional ectors are simply drawn in accordingly with the sum ector connecting the base of the first to the tip of the final ector. Subtraction of two ectors can be geometrically defined as follows. To subtract A from B, place the ends of A and B at the same point and then draw an arrow from the tip of A to the tip of B. That arrow represents the ector B-A, as illustrated below: To add two ectors without drawing them, the ectors must be broken down into their orthogonal components using sine, cosine, and tangent functions. Add both x components to get the total x component of the sum ector, then add both y components to get the y component of the sum ector. Use the Pythagorean Theorem and the three trigonometric functions to the get the size and direction of the final ector.

10 Example: Here is a diagram showing the x and y-components of a ector D1: Notice that the x-component D1x is adjacent to the angle of 34 degrees. Thus D1x=36m (cos34) =9.8m The y-component is opposite to the angle of 34 degrees. Thus D1y =36m (sin34) = 0.1m A second ector D is broken up into its components in the diagram below using the same techniques. We find that Dy=9.0m and Dx=-18.5m. Dx D=0.6m 64 Dy Next we add the x components and the y components to get DTotal x =11.3 m and DTotal y =9.1 m

11 Now we hae to use the Pythagorean theorem to get the total magnitude of the final ector. And the arctangent function to find the direction. As shown in the diagram below. DTotal=31.m tan θ= DTotal y / DTotal x = 9.1m / 11.3 =.6 θ=69 degrees Vector multiplication (dot and cross product) The dot product is also known as the scalar product. This is because the dot product of two ectors is not a ector, but a scalar (i.e., a real number without an associated direction). The definition of the dot product of the two ectors a and b is: a b = n i= 1 a b i i = a1 b1 + ab a b The following is an example calculation of the dot product of two ectors: [1 3-5] [4 - -] = (1)(4) + (3)(-) + (-5)(-) = 8 Note that the product is a simple scalar quantity, not a ector. The dot product is commutatie and distributie. Unlike the dot product, the cross product does return another ector. The ector returned by the cross product is orthogonal to the two original ectors. The cross product is defined as: a x b = n a b sin θ where n is a unit ector perpendicular to both a and b and θ is the angle between a and b. In practice, the cross product can be calculated as explained below: Gien the orthogonal unit ectors i,,j, and k, the ector a and b can be expressed: Then we can calculate that a = a 1 i + a j + a 3 k b = b 1 i + b j + b 3 k a x b =i(a b 3 )+j(a 3 b 1 )+k(a 1 b )-i(a 3 b )-j(a 1 b 3 )-k(a b 1 ) The cross product is anticommutatie (that is, a x b= - b x a) and distributie oer n n