240 likes | 536 Views
普通物理. 吳桂光 東海大學物理系 kkng@thu.edu.tw Office: ST223, Tel: 3467 Office hour: 10 -11am (Tue, Fri). Test book: Fundamentals of Physics, 7th Ed., Halliday/ Resnick/ Walker, John Wiley & Sons, Inc. Course content:
E N D
普通物理 吳桂光 東海大學物理系 kkng@thu.edu.tw Office: ST223, Tel: 3467 Office hour: 10 -11am (Tue, Fri) Test book: Fundamentals of Physics, 7th Ed., Halliday/ Resnick/ Walker, John Wiley & Sons, Inc.. Course content: Force, linear and rotation motion, energy, waves, thermodynamics 熱力學 (temperature, first and second law, …), electronics and electromagnetism 電磁學(electric and magnetic fields), optics (electromagnetic wave, diffraction, interference, …). Part 1-4 of Halliday.
Lecture 1: Motion along a straight line (chapter 2 of Halliday) • Kinematics (運動學): physics of motions • We will study: • Motion along a straight line only (one dimension), horizontal, vertical, … • How does the moving object speed up, slow down, stop, or reverse direction? How is time involved in the change? • The moving object is a particle (or can be thought as a particle).
Understand the rules of kinematics, we will be able to solve simple problems like this one at the end of this lecture. Position and displacement (位移)
Average velocity (速度) and average speed (速率) vavg is equal to savg, except vavg has direction. Average velocity and speed can be found a graph of x against t, which is equal to the slope from the initial to the final position.
Instantaneous velocity (or simply velocity) which is the slope of x(t) curve. Therefore one can obtain a v(t) curve from a x(t) curve. Question: can we do the reverse, obtain a x(t) curve from a v(t) curve? Answer: almost, but we don’t know the initial position x0. If we take the slope of vel. vs. time curve, we will get the acceleration curve.
Question (a) –, (b) +, (c) 0, (d) –
Acceleration (加速率) second derivative of x(t). If the unit of velocity is m/s, what is the unit of acceleration? m/s2 The sign of acceleration follows the change of velocity, no matter the speed is increasing or decreasing. (a) + (b) - (c) - (d) +
Constant acceleration: a special case There are many situations that the acceleration is, or approximately is, constant. If it is the case, average acceleration = instantaneous acceleration: fig. (c) or fig. (b) since integrate this out gives (or follow Halliday): fig. (a) From these two equations, one can derive the following equations:
Totally there are five variables, x-x0, v, t, a, v0. Each equations have four variables (the fifth is irrelevant), if three variables are given, the fourth can be found by solving one of the equations. Eqn. 2-16 m/s2
Free fall (自由下落) acceleration A free falling object accelerates downward constantly (air friction is neglected). This acceleration is a=-g=-9.8m/s2. This is the same for all masses, densities and shapes. So the above equations are valid.
available information: y=48m, v0=0, g=9.8m/s2 from 2-15:
Note the sign we used for y and g. In fact we can set downward direction to be positive or negative, what is important is all signs of variables have to be consistent. again from 2-15, for t=1s
from 2-11, or from 2-16, from 2-11, for t=1s,
Available information: v0=12m/s, a=-g=-9.8m/s2 Be careful of the signs! At max. height v=0. from 2-11,
from 2-15 again, one can use 2-16 to get the same answer.
from 2-15, which one is correct? Both are correct, one is for going up and the other is for going down.
Question (a) D, (b) E Try the following at home: Questions: 3, 5 Problems: 11, 16, 36, 37, 51, 54