Bristol Dynamics Reader 1 PDF

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Dynamics lecture notes for engineering classes...

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Module UFMEQT-20-1

Stress and Dynamics Dynamics – Revision Notes

October 2010

Department of Engineering Design and Mathematics University of the West of England, Bristol

Abstract This book covers some basics of physics, kinematics and mathematics. The material in this book will not be covered in the Dynamics lectures, and much of the content of the lectures will assume that the students have an understanding of this material. The students may refer to this material in order to aid understanding, and if necessary, can work through some of the problems and exercises at the end of each chapter to reinforce this knowledge.

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Contents 1 Linear Motion 1.1 Introduction . . . . . . . . . . . . . . . . . . . . 1.2 Derivation of Kinematic Equations . . . . . . . 1.3 Kinematic Equations for Constant Acceleration 1.4 Free Fall Under Gravity . . . . . . . . . . . . . 1.5 Dot Notation . . . . . . . . . . . . . . . . . . . 1.6 Examples . . . . . . . . . . . . . . . . . . . . . 1.6.1 Constant Acceleration . . . . . . . . . . 1.6.2 Free Fall Under Gravity . . . . . . . . . 1.7 Exercises . . . . . . . . . . . . . . . . . . . . . . 1.7.1 Constant Acceleration . . . . . . . . . . 1.7.2 Free Fall Under Gravity . . . . . . . . .

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7 7 7 9 10 10 11 11 12 15 15 16

2 Angular Motion 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.1 Angular Displacement . . . . . . . . . . . . . . . . . . . . . . . . 2.2.2 Angular Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.3 Angular Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Equations for Angular Motion . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Linear and Angular Motion . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Centripetal Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6.1 Constant Angualar Acceleration . . . . . . . . . . . . . . . . . . . 2.6.2 Angular and Linear Motion . . . . . . . . . . . . . . . . . . . . . 2.6.3 Centripetal Acceleration . . . . . . . . . . . . . . . . . . . . . . . 2.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7.1 Constant Angular Acceleration . . . . . . . . . . . . . . . . . . . 2.7.2 Centripetal Acceleration . . . . . . . . . . . . . . . . . . . . . . .

17 17 17 17 17 18 18 19 20 21 21 22 23 23 23 24

3 Scalars and Vectors 3.1 Introduction . . . . . . . . . . . . . . . . 3.2 Vector Representation in 2-D . . . . . . 3.3 Vector Addition and Resultant Vector . 3.4 Restultant of Two Perpendicular Vectors 3.5 Examples . . . . . . . . . . . . . . . . .

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Department of Engineering Design and Mathematics, UWE Bristol 3.6

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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4 Projectiles in 2-D Motion 4.1 Introduction . . . . . . . . . . . . . . . . . 4.2 Velocity Components . . . . . . . . . . . . 4.3 Vertical Motion (Height) . . . . . . . . . . 4.4 Horizontal Motion (Range) . . . . . . . . 4.5 Maximum Range . . . . . . . . . . . . . . 4.6 Trajectory Across a Non-Horizontal Plane 4.7 Example . . . . . . . . . . . . . . . . . . . 4.8 Exercises . . . . . . . . . . . . . . . . . . .

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5 Additional Notes 5.1 Introduction . . . . . . . . . . . . . . 5.2 Trigonometry . . . . . . . . . . . . . 5.2.1 Radians . . . . . . . . . . . . 5.2.2 The Trigonometric Functions 5.2.3 The Trigonometric Identities 5.3 Derivatives and Integrals . . . . . . . 5.3.1 Basic Differentiation Rules . . 5.3.2 Integration Table . . . . . . . 5.4 Unit Conversion . . . . . . . . . . . .

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1 Linear Motion 1.1 Introduction The following section deals with linear motion with constant acceleration. These equations can only be applied when it is known that acceleration does not vary with respect to time. • Linear motion is motion along a straight line, i.e. one-dimensional motion. • In one-dimensional motion, the direction component of vector quantities is assumed to be in the positive x -direction in an x -y plane (with the exception of constant acceleration due to gravity – see Section 1.4.) • Constant acceleration in a straight line is called uniform acceleration. • The SI units for displacement are metres (m). Metres per second (m/s or ms−1 ) are used for velocity, and metres per second per second (m/s2 or ms−2 ) for acceleration. • The variable t represents time, v represents velocity, and a represents acceleration. Displacement is generally represented by s, but sometimes x and y are used when we know that the displacement is along a certain axis.

1.2 Derivation of Kinematic Equations Constant acceleration is important as it applies to many objects in nature. For example, an object in free-fall near the Earth’s surface moves in the vertical direction with constant acceleration (neglecting air resistance). Constant acceleration is denoted as a. Acceleration is the change in velocity over the change in time: vf − vi dv = a= dt tf − ti

where the f and i subscripts denote ‘final’ and ‘initial’. For convenience, let ti = 0 and tf be any arbitrary time t. Also, let vi = v0 and vf = v1 . With this notation, the acceleration is expressed as this: v1 − v0 a= t or: v1 = v0 + at

(1.1)

When dealing with constant acceleration, velocity increases linearly with time, as shown in Figure 1.1. (The equation for this line is equation 1.1). The average velocity is just the mean value of the initial and final velocities.

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Velocity, v

v1 Slope = a0

v0

t1 Time, t

Figure 1.1: Velocity-time graph for constant acceleration The average velocity is: vav =

v0 + v1 2

(1.2)

The area under the graph in Figure 1.1 is the displacement, s:

Area = average height × base ∴ s = average velocity × time = 1 s = (v1 + v0 )t 2

(1.3) 

v0 + v1 2



t (1.4)

By substituting the equation for v1 (equation 1.1), an expression for the displacement as a function of time can be obtained: s=

1 (v0 + at + v0 )t 2

1 s = v0 t + at2 2

(1.5)

Finally, an expression that does not contain time can be obtained by rearranging equation 1.1 to isolate t, then substituting this into the equation for displacement (equation 1.4): v1 − v0 v1 = v0 + at → t = a

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1 Linear Motion 1 s = (v1 + v0 ) 2



v1 − v0 a



=

v12 − v02 2a

v12 = v 20 + 2as

(1.6)

1.3 Kinematic Equations for Constant Acceleration The following equations apply when the initial displacement is zero: s=0 Notation:

at

t=0

Initial velocity Final velocity Average velocity Constantacceleration Time elapsed Displacement

= = = = = =

v0 or u v1 or v vav a t s

Kinematic equations Alternative form Equation number v1 = v0 + at

vav =

s=

(v0 + v1 ) 2

1 (v0 + v1 )t 2

v = u + at

(u + v) 2

1. 2

1 (u + v )t 2

1. 4

vav =

s=

1.1

s = vav t

1 s = v0 t + at2 2

s = ut +

1 2 at 2

1. 5

v 21 = v02 + 2as

v 2 = u2 + 2as

1.6

Note: • When speed is increasing, acceleration is positive • When speed is decreasing (retardation or deceleration), the acceleration is negative

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1.4 Free Fall Under Gravity A particular case of constant acceleration in one dimension is when a body falls freely towards the earth. If air resistance is neglected, and the distances of fall is short, then the downward acceleration of a body when it is close to the surface of the earth is constant. This acceleration due to gravity is represented by g where: g = 9.81 m/s2 Note: It is important to consider the direction of the acceleration due to gravity. For example, when considering the upward vertical motion of a body under the effects of gravity, then: v0 = 20 m/s ↑ indicates an initial velocity of 20 m/s upwards. (So, vertical motion upwards is positive.) a = 9.81 m/s2 ↓ indicates a acceleration of 9.81 m/s2 downwards. To be consistent with the direction of the velocity, the acceleration must be written as: a = −g = −9.81 m/s2 ↑. This indicates a deceleration or retardation of 9.81 m/s 2 in the upwards direction. It generally makes sense for the positive vertical direction to be upwards. In this case, the acceleration due to gravity is: a = −g = −9.81 m/s2 It is important to remember that once an object has been released, it is moving under the influence of gravity only, regardless of its initial motion (this is defined as free fall). Once an object is in free fall, it will have an acceleration with a magnitude of g, directed downwards. Since this is a particular case of motion in one dimension, all equations defined above apply to freely falling bodies, remembering that motion is in the vertical direction, and that the value of acceleration, a = −g .

1.5 Dot Notation In this module, another form of notation used to signify motion is known as dot notation. A dot above a variable basically means the derivative with respect to time. Two dots above a variable means the second deriviative with respect to time. So, assuming displacement is defined as x :

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x˙ =

dx dt

= velocity

x¨ =

d2 x dt2

= acceleration

1 Linear Motion So, another alternative form of the equations can be seen, when: • s is replaced with x • v is replaced with x˙ • a is replaced with x¨ For example, equation 1.5 becomes: 1 x = xt ˙ + x¨t2 2

1.6 Examples 1.6.1 Constant Acceleration Question A sports car starting from rest accelerates at a rate of 5.00 m/s2 . What is the velocity of the car after it has travelled 30.0 m? Solution The first step with these problems it to check that the units you are dealing with are consistent. Here, we have an acceleration figure in m/s2 and a distance in metres, so this is fine. Next, the car is accelerating in a straight line, so we can assume that the car is travelling along the x-axis. The origin of the x-axis is at the initial location of the car, and the positive direction is to the right. Using this convention, velocity, accelerations and displacements to the right are positive. Now, there are three bits of information given in the question. These are: • The initial velocity, v0 , which is zero. • The aceleration, a, which is 5.00 m/s2 . • The displacement, s, of 30.0 m. We are required to find out v1 . Since we have v0 , a, and s, the most appropriate equation to use is equation 1.1: v12 = v 20 + 2as Substituting in the numbers, results in: v 21 = (0 m/s)2 + 2(5.00 m/s2 )(30.0 m) = 300 m2 /s2 v1 =

p

300 m2 /s2 = 17.3 m/s

The velocity of the car after 30 m is 17.3 m/s.

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Question A manufacturer of a high performance sports car claims that it can accelerate from 0 to 100.0 mph in 8.00 s. Assuming constant acceleration, determine: (a) the acceleration of the car (b) the displacement of the car in the first 8.00 s. Solution First, we have to make sure that the units are consistent. Since the final velocity is reported in miles per hour, this needs converting to m/s: 100

✘✘ ✘✘ 1609.33 metres 1✘ ✘ hour miles × × = 44.70 m/s ✘ ✘ ✘ ✘✘ hour 1 mile 3600 seconds

From, the question we have the following information: • The initial velocity, v0 is zero • The final velocity, v1 is 44.70 m/s • The time to accelerate, t is 8.00 s (a) The most appropriate equation to calculate the acceleration is therefore equation 1.1: v1 − v0 v1 = v0 + at → a = t Substituting in the values: a=

44.70 m/s − 0 m/s = 5.59 m/s2 8.0 s

(b) The displacement travelled by the car in the first 8.00 seconds can be found using equation 1.5: 1 s = (5.59 m/s2 )(8.00 s)2 = 179 m 2

1.6.2 Free Fall Under Gravity Question A golf ball is released from rest at the top of a very tall building. Neglecting air resistance, calculate the position and velocity of the ball after 1.00, 2.00 and 3.00 s. Answer We choose our coordinates so that the starting point of the ball is at the origin (y = 0 at t = 0) and remember that y is normally defined as positive in the upwards direction. We know that v0 = 0, and a = −g = −9.81 m/s2 .

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1 Linear Motion The most appropriate equations to use are equations 1.1 and 1.5: v1 = v0 + at = −9.81t 1 1 s = y = v0 t + at2 = (−9.81 m/s2 )t2 2 2 Therefore, at t = 1.00s: v1 = (−9.81 m/s2 )(1.00 s) = −9.81 m/s 1 y = (−9.81 m/s2 )(1.00 s)2 = −4.91 m 2 Likewise, at t = 2.00 s: v1 = (−9.81 m/s2 )(2.00 s) = −19.6 m/s 1 y = (−9.81 m/s2 )(2.00 s)2 = −19.6 m 2 At t = 3.00 s, v0 = −29.4 m/s and y = −44.1 m. Note: the minus sign for v0 indicates that the velocity vector is directed downwards, and the minus sign for y indicates displacement in the negative y-direction. Question A stone is thrown from the top of a building with an initial velocity of 20.0 m/s straight upward. The building is 50.0 m high, and the stone just misses the edge of the roof on its way down. Determine: (a) (b) (c) (d) (e) (f)

the time needed for the stone to reach its maximum height the maximum height the time needed for the stone to return to the level of the thrower the velocity of the stone at this instant the velocity and position of the stone at t = 5.00 s the velocity of the stone just before it hits the ground.

Answer (a) To find the time necessary for the stone to reach the maximum height, use equation 1.1: v1 − v0 v1 = v0 + at → t = a knowing that v1 = 0 at the maximum height. v0 = 20 m/s, a = −9.81 m/s2 : t=

0 m/s − 20.0 m/s = 2.04 s −9.81 m/s2

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(b) The maximum height can be determined using equation 1.5: s = v0 t +

1 2 at 2

where s is replaced by ymax , (which will be the height as measured from the thrower’s position). Substituting the value for time found in part (a) results in: ymax = (20.0 m/s)(2.04 s) +

1 (−9.81 m/s2 )(2.04 s)2 = 20.4 m 2

(c) When the stone is back at the height of the thrower, the value for displacement, y, is zero. Again, using equation 1.5 (with y replacing s): 1 y = v0 t + at2 2 and setting y = 0, a value for t can be determined: 1 0 = (20.0)t + (−9.81)t2 = t(20.0 − 4.90t) 2 This is a quadratic equation and has two solutions for t. One solution is t = 0, which corresponds to the time at which the stone starts its motion. The other solution is t = 4.08 s, which is the answer to this part. (d) The value for t found in (c) can be inserted equation 1.1 to find v1 . v1 = (20.0 m/s) + (−9.81 m/s2 )(4.08 s) = −20.0 m/s which is directed downwards. Note that the velocity of the stone when it arrives back at its original height is equal in magnitude to the stone’s initial velocity but opposite in direction. (e) To find the velocity after 5.00 s, again use equation 1.1: v1 = (20.0 m/s) + (−9.81 m/s2 )(5.00 s) = −29.1 m/s To find the position (equation 1.5): y = (20.0 m/s)(5.00 s) +

1 (−9.81 m/s2 )(5.00 s) = −22.6 m 2

which is 22.6 m below the height of the thrower. (f) To find the velocity of the stone just before it hits the ground, we use equation 1.6 since we know: v0 = 20, 0 m/s, a = −9.81 m/s2 and s = y = −50.0 m (50 metres below the thrower). Substituting these values into the equation: v 21 = (20.0 m/s)2 + 2(−9.81 m/s2 )(−50.0 m) = 1381 m2 /s2 p v1 = 1381 m2 /s2 = 37.1 m/s 14

1 Linear Motion

1.7 Exercises 1.7.1 Constant Acceleration 1. An athlete swims the length of a 50.0 m pool in 20.0 s and makes the return trip to the starting position in 22.0 s. Determine the swimmer’s average velocities in: (a) the first half of the swim (b) the second half of the swim and (c) the round trip. (Ans: (a) 2.50 m/s (b) −2.27 m/s (c) 0) 2. A speedy tortoise can run with a speed of 10.0 cm/s, and a hare can run 20 times as fast. In a race, they both start at the same time, but the hare stops to rest for 2.0 minutes. The tortoise wins by a shell (20 cm). (a) How long does the race take? (b) What is the length of the race? (Ans: (a) 126 s (b) 12.6 m) 3. Two cars travel in the same direction along a straight road, one at a constant speed of 55 mph and other at 70 mph. (a) Assuming that they start at the same point, how much sooner does the faster car arrive at a destination 10 miles away? (b) How far must the faster car travel before it has a 15 minute lead on the slower car? (Ans: (a) 2.34 minutes (b) 64.17 miles) 4. Jules Verne in 1865 proposed sending men to the Moon by firing a space capsule from a 220 metre long cannon with a final velocity of 10.97 km/s. What would have been the unrealistically large acceleration experienced by the space travellers during launch? (Ans: 273502 m/s2 ) 5. A car travelling initially at 7.0 m/s accelerates at the rate of 0.8 m/s 2 for an interval of 2.0 s. What is its velocity at the end of the acceleration? (Ans: 8.60 m/s2 ) 6. A driver in a car travelling at a speed of 60 mph sees a deer 100 m away on the road. Calculate the minimum constant acceleration that is nece...