{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

summary_ch1-4

# summary_ch1-4 - Getting ready for the Midterm Physics 111...

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Getting ready for the Midterm Physics 111 DISCLAIMER: It is not guaranteed. that the midterm questions are taken exclusively from the material discribed in this note. Although I tried to be careful. there might be some typos. In that case consult the textbook for the correct answer! I hope this helps a little bit in your preparation for the midterm. Klaus Honscheid Uniformly Accelerated Motion Constant acceleration: a = const We derived the following four equations: V = V0 + at x=x0 + vot+ 1/2at2 v2 =v02+2a(x-xo) v- : (v + v0)f2 (average velocity) Solving problems: Write downthe quantities you know and then those you want to know! Apply the right equation. Does your result make sense? Are the units correct? Typical questions: A car going 80 km/h is stopped over a distance of 50 m. What is the average acceleration. Falling Bodies: Acceleration due to gravity g = 9.80 m/sz. This acceleration is always directed towards the center of the earth. Watch out for minus signs depending on which direction (for the displacement y) you deﬁne positive! If y is deﬁned to be positive in the upward direction (away from the center of the earth!) a minus sign is introduced in our equations: (In this class we use g as a symbol for the absolute value of the acceleration due to gravity. So g is always 9.80 III/52.) y=yo+vot 41'2th x-t, v-t, a-t diagrams: The relation between displacement, velocity or acceleration vs. time can be displayed in diagrams. In uniformly accelerated motion the acceleration is constant (ie litre parallel the horizontal t axis), the velocity changes linearly with time (ie a straight line with a slope is the acceleration is not 0) and the displacement changes with the square of the time (parabola) Vectors Physical quantities such as force and velocity that have both, magnitude and dirmtion are represented by vectors. Once a coordinate system is deﬁned, we can talk about the components of a vector in the x and the y direction. If the angle between the vector and the x—axis is a, then the length of each component is given by dx = d cos(a) dy = d sin(0t) where d is the magnitude (length) of the vector. Using the Pythagorean theorem we can write: d2 : dXZ + dyl Two vector are added by adding the corresponding components: c=a +b => cx=ax +bx andCy=ay +by Also: tip—to-tail method, trigonometric functions Kinematics in 2 dimensions In problems that involve two or more velocities pointng into different directions we have to use vector addition to determine the resulting velocity Typical problems: A boot crossing a river“. time to reach the other side, displacement along the shore line etc. Or an airplane ﬂying in some direction and wind blowing from a different direction. Problem solving: Make a drawing. Indicate the velocity vectors and the displacement vectors but be sure not to mix them up! Use trigonometry to deterime the unkowns in the problem. Projectile Motion The modem in the horizontal and the vertical direction are independent. We use our kinematics equation separately for the vertical and horizontal components. In our problems the acceleration due to gravity is the only acceleration. Which means we have constant motion (ie v = const) in the horizontal direction and the motion of a falling body in the vertical direction. [fy is taken upward positive,we have: Horizontal (x) Vertical (y) Vx = Vox Vy = Voy ' gt X=X0+V0xl y=yO+Vle-11Qgt2 Vyz = Voyz ' 2 g(y ' Yo) Problem solving: Make a drawing. Deﬁne your coordinate system. This deﬁnes the starting point of the motion (x0, yo) and the positive y - direction with respect to the direction of the acceleration due to gravity. Note: It is not required to place the origin of the coordinate system at the origin of the motion (ie x0 = 0, yo = O) . However, should you choose a different deﬁnition, you must include the correct x0 and yo into you calculation! Range: starting at ground level, the range is given by R = vczlg sin(20t) => longest range for a = 450 Dynamics: Newton’s lst law: Every object continues in its state of rest or of uniform speed in a straight line unless it is compelled to change that state by a net force acting on it. Newton’s 2nd Law: The acceleration of an object is directly proportional to the net force acting on it and is inversely proportional to its mass. The direction of the acceleration is in the direction of the applied net force. Fnet = m 3 (This is a vector equation) Newton’s 3rd Law: Whenever an object exerts a force on a second object, the second exerts an equal and opposite force on the ﬁrst. Note: Action and reaction forces act on different objects Typical Problems: Combination of kinematics and dynamics. Tension in strings, free body diagrams, overall force etc. Example: A bullet (mass = 0.01 kg) is accelerated in a gun (distance Ax = 0.5m) to a speed of 400 this. What is the average force excerted onto the bullet? Solution: given: mass, v, v0, Ax wanted: F 2nd law: F = m a so we need to know the acceleration: (remember kinematics) v2 = v02 + 2 a Ax a = v2/(2Ax) a = 160000 m/s2 and then it follows for the force F = ma F = 1600 N Free Body Diagrams: Many problems in dynamics can be solved with the help of free body diagrams. These diagrams are nothing special, they are just a convenient way to determine the net force acting on an object. Here is the procedure, step by step: Step 1: For each object in the problem draw a free body diagram. This is just a simple drawing symbolizing the object and all the forces acting ON this object (Weight, normal force, tension, pull or push force etc). Step 2: Determine the net force acting on each object in the problem. This gives you one equation per object! In a 2 dim. problem, deﬁne a coordinate system and determine the x component and the y component of the net force separately (so you have two equations per object in this case) Step 3: Apply Newton‘s 2nd law: Fnet = ma (2 dim: for x and y component separately) for each object in the problem. Step 4: Use these equations to solve for the unknowns in the problem. This might involve some mathematical transformations and you might have to solve a set of coupled equations. Example: What is the tension in the cables connecting the train cars? What is the overall acceleration. We assume, all cars have the same mass M and the train is pulled by a horizontal force F. F” r 2 {3:27 7;”: “‘2‘ fg-T'éjp ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 4

summary_ch1-4 - Getting ready for the Midterm Physics 111...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online