Motion, Forces and Energy PHY101 Dr. Tim Richardson Semester 1 (Block 1) 2005

1. Motion, Forces and Energy PHY101 Dr. Tim Richardson Semester 1 (Block 1) 2005 What does the University motto mean?

2. Objectives of PHY101 To gain a deep understanding of the behaviour of moving bodies and the causes of their motion. Course Format Formal Lectures (3 per week for 6 weeks). Tutorials (small group sessions with a tutor aimed at cementing the lecture material). Problems Classes in which you have an opportunity to develop skills necessary to tackle challenging physics problems (FRIDAYS 11:10 – 12:00) Recommended Book & Material See First Year Noticeboard Various websites advertised throughout the course.

3. Motion, Forces and Energy PHY101 Dr. Tim Richardson Semester 1 (Block 1) 2005 Part 1a: Scale and All That Part 1b: Motion in 1D and 2D Part 2: That Guy Newton Part 3: Circles and Resistance Part 4: Work – A Strange Concept Part 5: Potential Energy & Conservation Part 6: Snooker (Collisions) Part 7: Rotational Kinematics Part 8: Not-So-Simple Harmonic Motion Part 9: That Sinking Feeling

4. Part 1a: Scale and All That Standards of Length, Mass and Time The metre (m) and the length scale The kilogram (kg) – platinum / iridium alloy cylinder Time – Cesium 133 atoms Units and Dimensions Scale: From a Quark to a Universe

5. Units and Related Matters Seriously, units and dimensions are important. Without units, a value of a particular quantity may be meaningless. It is truly amazing how often physics students omit units when quoting a value of a quantity. Get into the habit of using units at all times – it is one of the trademarks of a good and professional physicist! FORGETTING UNITS IS A PUNISHABLE OFFENCE! Who’s the scientist in the picture?

6. Introductory Concepts in Motion Displacement – the change in position of an object, Dx (vector quantity) Dx = xf – xi where the subscripts f and i refer to final and initial positions respectively. It is important to understand the difference between displacement and distance (purely scalar): Running one lap of a running track gives you a displacement of 0 m but you have travelled a distance of 400 m. Who’s the athlete on the right?

7. Velocity and speed Average velocity of a particle v is given by: v = Dx / Dt where Dt is the time interval. Average velocity is a vector quantity whereas average speed is a scalar quantity (distance / Dt). Again we can use the example of a 400 m athlete who completes one lap; her average speed might be around 11 ms-1 but her average velocity is formally zero. Instantaneous velocity The instantaneous velocity vx is the limiting value of the above ratio as Dt approaches zero. In calculus notation, this limit is called the derivative of x with respect to t, written as dx.dt: vx = limDx/Dt = dx/dt Dt~0 ~ 10-3 ms-1 ~ 104 ms-1

8. Example: • A particle moves along the x axis such that its displacement x is given by • x = -4t + 2t2 . • Find the instantaneous velocity of the particle at t = 2.5 s and t = 3.8 s. • We could plot this function and measure the gradient of the curve at the above • time values (do this in your own time), but using basic calculus is easier: • v = dx/dt = 4t – 4 • Now substituting the values for t yield the velocity: • For t = 2.5 s, v = 6.0 ms-1 and for t = 3.8 s, v = 11.2 ms-1. • See how close your graphical estimate comes the above values.

9. Acceleration We can define average acceleration as a = Dvx / Dt but more importantly Instantaneous acceleration, ax as: ax = lim Dvx / Dt = dv/dt Dt~0 Now we can link displacement to acceleration using calculus: ax = dv/dt = d (dx/dt) = d2x / dt2 dt For our example where x = -4t + 2t2 , we can now see that the acceleration is single-valued and equal to + 4 ms-2 (think about the meaning of this).

10. Problems to try: • An athlete runs along part of a track with a displacement given by • x = 5t + 4. Find (i) her change in position over the first 3 seconds, • (ii) her velocity at t=2s and t=3s and (iii) her acceleration. • 2 A particle’s displacement varies with time as x = 4t3 - 3.5t2 + t + 9. • Find (i) the change in position of the particle between t=4 s and t=6s, • (ii) its instantaneous velocity at t=3.2 s and (iii) its instantaneous • acceleration at t=4.1s. • (iv) Does the particle ever stop in the time interval 0 < t < 0.5 s? • (NB I sometimes like “trick” questions.)

11. Part 1b: Motion in 1- and 2-D. If the acceleration of a particle varies in time, its motion can be complex and difficult to analyse. However, one-dimensional motion in which the acceleration is a constant value is straightforward to deal with. We saw earlier that we can define acceleration as a where: a = v – u(1) t where v is the final velocity and u is the initial velocity for the time interval, t. We can rearrange this expression to yield our first KINEMATIC EQUATION: v = u + at (KE 1) This equation allows us to relate velocity to acceleration and time.

12. Using x for the change in position, we can find our second kinematic equation: Average velocity = u + v(2) 2 and displacement is average velocity x time, so x = (u + u + at). t 2 So x = ut + ½ at2 (KE2) Finally we use equations (2) and (KE1) to yield our 3rd kinematic equation: x = ( u + v ) ( v – u ) = v2 - u2 2 a 2a Therefore v2 = u2 + 2ax (KE3)

13. Deriving the Kinematic Equations from Calculus We know that a = dv/dt and that v = dx/dt: So dv = a dt or v = a dt In the case of constant acceleration, v = at + c, but when t=0, c is just the initial velocity u, so we derive v = u + at . Also we can write dx = v dt or x = v dt But from above we know that v = u + at, so x = (u + at) dt = u dt + at dt Finally therefore we have x = ut + ½ at2 What was the nickname of the athlete on the right?

14. Freely Falling Objects We’ll be talking about gravity at great length later in the course, but for now we just need to remember that in the absence of air resistance, all bodies fall at the same rate under the influence of the Earth’ gravitational field. We use +g for objects falling and –g for rising objects decelerating. vy = v0y+ g t y = v0y t + ½ g t2 vy2 = v0y2+ 2 g y What year was the “hammer and feather” experiment performed on the Moon?”

15. The Coin in the Well Experiment We use h = v0y t + ½ g t2 to find the depth of the well, counting how long it takes for the coin to splash into the water. Since the coin is dropped from rest, we Know that v0y = 0 so that: h = ½ g t2 If for example it takes 4.3 s for the coin to fall, then the depth of the well would be 90.6 m. What’s wrong with this diagram?

16. 2 h / g Reaction Time Game A further refinement on this theme gives us a “reaction time” testing game. All you need is a 30 cm ruler (unless you’re a real dork in which case you might need a 1 metre ruler) and a partner. Ask your partner to drop the ruler through your forefinger and thumb without telling when he/she is going to do it. As soon as you see the ruler falling, try to catch it. Then use h = ½ g t2 to calculate your reaction time. Usually students’ reaction times are in the range 0.15 – 0.3 s. t = If we assume that the reaction time for pressing a car brake pedal is similar to that found above, then we can work out the distance travelled at various speeds before we even begin to brake! eg. t ~ 0.2s, so that: Speed (ms-1) (mph) Distance (m) 13 30 2.6 22 50 4.4 31 70 6.2 39 90 7.8 http://www.chss.montclair.edu/psychology/museum/mrt.html

17. Projectile Motion • Projectile motion is quite simple to analyse if we make two assumptions: • The free-fall acceleration due to gravity g is constant over the range of motion • The effect of air resistance is negligible. • We’ll start by showing that the projectile path is parabolic. We know that • ay = -g and ax = 0:

18. See Board Work