This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full 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(xxo)
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 xt, vt, at 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—totail 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+Vle11Qgt2 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‘ fgT'éjp ...
View
Full Document
 Spring '11
 UNKNOWN

Click to edit the document details