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2. physics

Remember EVERY ANSWER needs a number unit and direction.
Cross check the equations on this document with equations given in your
notes and in your book!
Rectilinear Motion – Motion in one direction
If the acceleration is constant we can use Kinematic Equations
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***NOTE these equations are only good if the acceleration is constant. If the acceleration is not
constant then you will have to use the equations. ( Remember these are how the kinematic equations
were derived). Separation and integration is important as well as determining the correct bounds for
integration. The information given in the problem will lead you to what equation to use
Pulleys –
When you are giving a pulley problem the first thing you should do is dimension the pulley system. You
should be able to determine the relationships between velocities and the accelerations before you even
look at what the problem is asking. After the pulley have been dimensioned use kinematics to complete
the problems
How to dimension a pulley system
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Create a datum line. This datum should be connected to a fixed point.
One datum line should be indicated for every direction a particle is moving in the system. EVERY
particle that is moving in that direction should be dimension from the same point
The positive direction of the problem is dimensioned FROM the datum line TO the particle.
Trace the length of each rope in a system. Each rope in a system has its own equation. ( 2 ropes
2 equations; 3 ropes 3 equations etc)
Take the time derivate to get velocity. Take the time derivative again to get the acceleration.
The derivative of any constant length in the system is equal to 0.
Relative Motion –
If two particles undergo independent motion , then these motion can be related to each other. You can
use I,j,k components to relate these motions
To find the angle between vectors you can always put the vectors tail to head. Then theta will be the
angle between these two vectors and can be found by
X component
theta
Y component
Projectiles
Break up the problem into horizontal and vertical motion. The velocity in the X direction is constant
. The main equation to use for the X direction Is
. In the Y direction all of
the kinematic equations are valid. The gravity in projectile problems is 9.81m/s^2 or 32.2ft/s^2 . WATCH
THE SIGNS. The negative sign is the following equations so therefore you would not put a –g term in
these equations.
( )
( )
(
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)
The magnitude of the Velocity V can be broken up into components
Also when the projectile is as high as it can go, the velocity in the y direction is = 0.
Normal and Tangential Components
When a particle is moving along a curved path, it has both a normal and tangential component of
acceleration. The normal component of acceleration always points back towards the center of the curve
it is moving on. The tangential component always acts tangent in the direction of motion. These are
perpendicular to each other. Another equation for the normal acceleration is
. V is the velocity
and p is the radius of curvature of the path the particle is moving on. The tangential acceleration is
equal to these equations of acceleration
The radius of curvature is equal to
[
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if the path that the particle is moving on is given as a y
is function of x y=f(x). Don’t use this equation unless the path is given as a function of x.
Newtons Law of Motion
1st thing Draw a Free Body Diagram and a Kinematic Diagram
The Free Body Diagram shows all of the forces that are on the system. You use the Free Body Diagram
∑
for the ∑
. The kinematic side of the diagram identifies all the accelerations and the
masses of the system. From your diagrams you should be able to fill out the Newtons law equation. If
the accelerations are unknown assume they are in the positive direction
Sometimes it is easier to express the problem in terms of N-t coordinates ( particle moving over a curved
path) The sum of the forces in the normal, and tangential directions are equal to the mass * acceleration
in those directions. Also the Forces in the binormal direction are = 0.
X-Y Coordinates
∑
∑
∑
N-t Coordinated
∑
∑
∑
Planar Kinematics of a Rigid BodyWhen dealing with rigid bodies, bodies can undergo different types of motion (Translation, Rotation
about a Fixed Axis, and General Planar Motion)
Translation- The body just moves and does not change its orientation
Rotation about a Fixed Axis- The body rotates about some fixed axis. All points except points on the axis
of rotation move on a circular path
General Planar motion- The body undergoes translation and rotation about a fixed axis.
With translation all points have the same velocities and accelerations
Rotation about a fixed axis- If the angular acceleration is constant and only if it is constant we can derive
equations just like rectilinear motion. Here w is the angular velocity, alpha is the angular acceleration,
and theta is the angular position
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)
***NOTE these equations are only good if the angular acceleration is constant. If the angular
acceleration is not constant then you will have to use the equations. ( Remember these are how the
kinematic equations were derived). Separation and integration is important as well as determining the
correct bounds for integration. The information given in the problem will lead you to what equation to
use
Some other equations that are of use for these problems is
S is the length that point has traveled, and r is the distance to the fixed axis. These equations work wwell
for simple geometries but if the geometry is difficult to visualize then vector equations should be used
Here the bold face indicates that these equations are vector equations. The X indicates a cross product.
Notice that the only term in the acceleration equation that is not a vector is the angular velocity. THIS IS
IMPORTANT.
General Planar Motion- Velocities
This is pretty much saying that any motion that has a rotation and velocity can be broken up into a
translation and then a rotation about a point. The relative velocity method must use vectors. This means
the terms have i,j,k components.
This equations says the velocity of point B is equal to the velocity of point A (Translation) + the angular
velocity crossed with the position vector of b with respect to a (Rotation about a Fixed Axis). The
subscripts are important. With respect to A means you start at A and go to B. If you had decided to start
at B and go to A the two velocity terms must be switched. If the velocities are unknown assume they
are positive. If the angular velocity is not known assume it is positive. The positive direction is found
with the right hand rule.
If you are looking at a bar or something that is rotating about a fixed axis then the above equation will
simplify. Lets say point A is the fixed axis, then point A will not have a velocity so the equation becomes
Also another trick with these problems involve a rolling disk or cylinder. If the disk is rolling, then the
point of contact the disk has with a surface takes on the velocity of that surface. For example, lets say
the cylinder is rolling on the ground, then the velocity of the point of the ground is equal to 0. If the
ground was moving, lets say the ground was a conveyor belt and was moving at 2m/s, then the point of
the cylinder touching the ground would have a velocity of 2m/s.
Also the velocity of the mid-point of the wheel, cylinder, disk, whatever it is going to be completely in
one direction. Again for example lets say the wheel is rolling on the ground to the right, then the velocity
of the center of the wheel is going to be only in the x direction towards the right.
Instantaneous Center of Zero Velocity
Know the different ways of obtaining the instantaneous center. From the is point, the velocity of any
point of a rigid body can be obtained. The equation that we used is the
. This is a scalar
equation so there is NO ijk components. THIS IS VERY IMPORTANT.
How to determine the instantaneous center. These are the 4 ways to determine the instantaneous
center. I will expand on the second method of nonparallel vectors. The distance of the point with
respect to the instantaneous center is ALWAYS perpendicular to the velocity. So after the velocity of the
two vectors is determined you can draw a perpendicular line THROUGH the point. Don’t draw the line at
the head of your velocity vector. The point of intersection of these two lines is the IC. NEVER ASSUME A
RIGHT TRIANGLE. This was a big mistake on the mid term exam. Determine all of the angles through
geometry. IF the triangle turns out to be a right triangle you can use SOHCAHTOA. IF the triangle is not a
right triangle then you should use the law of sines.
. After you determine the
distance you need you can use the equation
velocity is the same for both of these equations.
and
. Notice that the angular
Relative Motion Acceleration
Relative Acceleration equations also need I,j,k components because these equations use vector
notation.
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)
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)
Notice that all of the terms in this equation are vectors except for the angular velocity which is a scalar
quantity. However it is multiplied by the position vector which makes the normal velocity term a vector.
If the accelerations are unknown, assume a positive normal and a positive tangential accelerations. If
the angular acceleration is not known assume it is positive. This direction is determined by the right
hand rule. The acceleration of A (Aa) can also have a normal and tangential component of acceleration if
it is moving on a circular path. (We did one or two long examples of this)
A couple of key points for a disk that rolls without slipping. From the velocities we know that the center
of it rolls with the velocity
. The acceleration of it center is also going to be in a straight line and
be
.
Planar Kinetics of a Rigid Body
Moment of Inertia- lookup in the back of a table. Parallel Axis theorem is very powerful. This says that
the moment of inertia of an object about some parallel axis (A) is equal to the moment of inertia about
the center of mass + the mass of the object times the distance between these two axes squared.
Another important equation is if you are given the radius of gyration (kg).
Composite bodies – You can figure out the moment of inertias by subtracting two parts from each other
or adding moment of inertias together For example a circle with a hole in it. The equation for composite
∑
bodies is
where IG is the center of mass for each of the composite parts
Mass = density*Volume
Newtons Law of Motion Translation
Translation means no rotation so the sum of the moments is equal to 0. Also the acceleration in these
equations are equal to the acceleration of the center of mass. It is also possible to take the moments
about a different point other than the center of mass of take the sum of the moments about the kinetic
moments. X-Y coordinates are easier when the object is not rotating on a curved path. If the object is
rotating on a curved path then N-t coordinates should be used.
X-Y Coordinates
∑
∑
∑
∑
∑
∑
N-t Coordinated
Rotation about a Fixed Axis
If the rotation of an object is rotating about a fixed axis then the following set of Newtons equations can
be used. Don’t forget in these problems that pin reaction forces have to be accounted for. In the
following equations rG is the distance from the fixed axis to the center of mass
N-t Coordinated
∑
∑
∑
Or
∑
General Planar Motion
If the body is experiencing general planar motion, the set of Newtons equations to use is. Typically an
application of general planar motion will have 4 unknowns and 3 equations.
X-Y Coordinate
∑
∑
∑
In order to solve these problems it is important to first assume that the body rotates without slipping.
Since this rotates without slipping the acceleration of the center of mass
. This becomes your
fourth equation. The MOST IMPORTANT PART OF THIS PROBLEM is that you CANNOT assume that the
frictional force
. DON’T DO THAT. After you figure out the magnitude of the frictional force
from the above equations then you have to perform a check. The check is
. If this is true then
the assumption that no slipping occurs is correct you are done with the problem. If this is not true then
the object slips and you have to redo the problem with the
equation.
Also if the point on the object you are trying to solve for moves in a circular path (center of gravity of a
unbalanced mass) then you have to use relative acceleration equations
Work and Energy - Particles
Need to know the forces that do work on a system. These forces are a variable force (VF), a constant
force (CF), a weight of an object(W), and spring force(K). These work of forces are shown in the
equations below. Some ideas to remember. The work done by the Weight only occurs if it changes
height. If the object doesn’t change height then no work is done. Also if the particle moves down, the
work done is positive( the delta y would be negative) and if the particle moves up the work done is
positive. ALSO don’t forget the negative sign in front of everything for the work of a spring. S is the
distance that the spring is stretched past its unstretched distance. Also the normal force does no work
because the force is ALWAYS perpendicular to the objects direction of motion
∫
Principle of Work and Energy States that the initial kinetic energy plus all of the work on the system is
equal to the final kinetic energy.
∑
Power is the amount of work over time and is the dot product of the force and the velocity
This can be rewritten as
and theta is the angle between the force and the velocity. So it
could be the velocity acting in the direction of the force or the force acting in the direction of the
velocity.
The efficiency of a machine is
Conservative forces The two conservative forces we are interested in are the gravitational potential
energy and the spring potential energy. SO the total potential energy of a system is equal to the
gravitational potential energy + the spring potential energy.
The gravitational potential energy = mgh and depends on the placement of the datum line. If you
assume the datum line is a line of 0 potential energy anything that is below the datum line is going to
have negative potential energy and anything that is above is going to have positive potential energy The
spring potential energy
. S is the distance that the spring is stretched. If the spring is
unstretched it has 0 potential energy.
If all of the forces in the system are conservative we can apply the conservation of energy equation
This says that the initial kinetic energy plus the initial potential energy is equal to the final kinetic energy
plus the final potential energy. If a something is at rest then the kinetic energy is 0 and that term in the
equation will drop out. It is a good strategy to draw the system at the two positions that you are looking
at and dimension a datum line if possible so at least one of these positions will have 0 potential energy (
datum line going through it )
Work and Energy – Rigid Bodies
Just like particles, the work of rigid bodies of different forces do different types of work. These forces
are a variable force (VF), a constant force (CF), a weight of an object(W), and spring force(K). These
work of forces are shown in the equations below. Additionally the rigid body since it has a moment of
inertia will have the work of a couple moment which can be constant(CM) or variable (VM)
∫
∫
Rigid bodies also follow the principle of work and energy the initial kinetic energy plus all of the work
on the system is equal to the final kinetic energy. However T is going to be different depending on what
type of motion the object is undergoing.
∑
If the body is undergoing pure translation then
With V being the Velocity of the Center of
Mass
If the body is undergoing Rotation about a Fixed Axis then
. The
difference in these two equations is the axis you are taking the moment of inertia about.
If the body is undergoing general plane motion then
Using methods of instantaneous centers you can relate the velocity of a body with its angular velocity
and
When you use look at potential energy and conservation of energies for a rigid body, then the change in
height for gravitational potential energy is about the center of mass. This is important because
sometimes the center of mass doesn’t move as far as part of the object does.
Linear Impulse and Momentum
Linear momentum (L)
L = mv
The impulse in a system is the effect of the force over the time the force is acting on the system.
The vector equation
∑∫
can be broken up into 3 scalar equations for the
x,y,and z components
∑∫
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)
∑∫
∑∫
If the sum of the external particles acting on a system is 0 then the conservation of linear momentum
equation can be used. The final mass can be added together if the particles become attached or
subtracted if the particles leave each other ( gun firing a bullet)
∑
∑
Impacts
Establish a normal and tangential coordinate system. The tangential coordinate system is along the
plane of impact and does not see any of the impact. The normal system is along the impact. Draw two
diagram one just before impact and one just after impact. Assume positive velocities if they are not
known. From this and conservation of momentum along the normal direction
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( )
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( )
(
And
)
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(
e is the coefficient of restitution and has a value between 0 and 1. 0 means perfectly plastic collision and
1 means perfectly elastic collision.
From the conservation of momentum of the particles you will get that the initial velocity in the
tangential direction is the same as the final velocity in the tangential direction
Conservation Angular Momentum
∑∫
If the system has no moments of forces then the above equation simplifies to
Then
where d is the distance from axis that momentum is conserved
Linear Impulse and Momentum- Rigid Bodies
Translation Linear Momentum
Rotation about a Fixed Axis
General Planar Motion and d is the distance from the point of rotation to the center of mass
∑∫
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)
∑∫
∑∫
Finally the linear momentum is conserved if there is no external impulsive forces acting on the body.
Conservation of angular momentum applies about a fixed point . We can say that this happens when the
time the force is acting on the body is very short. So if the time is very short then the impulse is approx..
0. And the conservation of angular momentum will apply.