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FORMULAS Circular motion
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1D Motion
2D & 3D Motion
Laws of Motion
Work Energy Power
Circular Motion In the diagram v is the tangential velocity of the object. a is the
Gravitation centripetal (acting towards the center of the circle) acceleration
Properties of Matter and F is the centripetal force. r is the radius of the circle and m
Electricity is mass of the object.
How to solve problems
a = v2 / r
SOLVED EXAMPLES
F = ma = mv2/r

Center of mass

Consider that there are a number of particles forming a system,

such that, when an external force is applied, the complete
system moves as on. Now, if we are able to replace all these
particles by just one particle. And, this new particle is placed in
such a manner and has such a mass, that when acted on by an
external force, it moves exactly in the same manner as the
system it replaced. Then, we have found the center of mass of
the system. ( This sounds more complex than it is. So don't
worry! read on)

Center of mass of a two particle system

Let us now consider two particles. Let them have masses m1 and m2. Let r1(t) and
r2(t) be the position vectors of the two particles at the instant of time t. (Note: On a
web page it is difficult to use the usual vector notation. So I will use bold fonts to
depict vectors). The position vector Rc.m.(t), giving the center of mass of this system
is given by the formula.

Rc.m.(t) = {m1r1(t) + m2r2(t)} / (m1 + m2)

from the above it is obvious that for a two particle system the center of mass always
lies between the two particles. It also lies on the line joining the two particles.

Center of mass of a large number of particles

For a large number of particles, say 'n' particles the formula becomes.

Rc.m.(t) = {m1r1(t) + m2r2(t) + ----- mn rn(t)} / (m1 + m2 + ------ mn)

Center of mass of a rigid body

The position of the center of mass of a rigid body is a fixed point. This point may or
may not lie in the body. Where is lies depends upon the shape of the body. For
example of we take the case of a ring. The center of mass will lie at its geometrical
center. Obviously this point does not lie in the ring.

Center of mass of some rigid bodies

Body Center of mass

Rod Mid point of rod

Ring Center of ring

SphereCenter of sphere

Angular velocity

Average angular velocity = Angular displacement / time interval

So if ? is the average angular velocity, ? the angular displacement and t the time
interval. We can say

? = ??/?t

So we can write the instantaneous angular velocity as d?/dt

Linear velocity

Linear velocity = Radius of the circular path x Angular velocity

v = ?r

Angular acceleration

Average angular acceleration = Change in angular acceleration / Time interval

a = ??/?t

Instantaneous angular acceleration

a = d2? / dt2

Linear acceleration

Linear acceleration = Radius of the circular path x Angular acceleration

a = ra

A body that travels an equal distances in equal amounts of time along a circular path has a

constant speed but not constant velocity. This is because velocity is a vector and thus it has

magnitude as well as direction

The velocity of P is directed along the tangent at P. The speed remains constant but the

velocity has changed. We know that if the velocity changes with time then the ball on the

string is also accelerating.

The Radian
angles can be measured in radians as well as degrees. The angle in radians is defined by. If s

= r then θ=1 rad. Therefore, 1 rad is the angle subtend at the center of a circle by an arc

equal in length to the radius. When s =2π r then θ=2 π radians=360°. Therefore, 1 rad =

360°/2 π=57.3°

θ = s/r(1)

Since we are often dealing with angles and trigonomic functions, a useful approximation that

is often used is that for small angles of &theta, sin θ and tan θ = θ where θ is measured in

radians.

&theta = sin &theta = tan θ for small angles where theta is measured in radians.

Angular Velocity
The speed of a body moving in a circle can be specified either by its speed along the tangent

at any instant (linear speed) or by the angular velocity. This is the angle swept out in unit

time by the radius joining the body to the centre. It is measured in [rad s-1]

Further, the angular velocity is the time for the particle to travel around once divided by the

period. ω = 2 π/T

ω = θ/t(2)

DERIVATION

Consider a body moving uniformly from A to B in time t so that OA rotates through a small

angle θ.

The angular velocity, ω of the body about O is ω=θ/t.

If arc AB has length s and v is the constant speed of the body, v=s/t

But s=r θ.
So, v=r θ/t and ω =θ/t,

therefore v=rω

v = r/ω(3)

Acceleration
The expression for the acceleration of an object moving in circular motion of radius r moving

at a constant speedv is derived as follows. If it travels from A to B in a short interval of time

δt then, since speed = distance x time, arc AB = vδt.

Also by the definition of angle in radians, arc AB = r δθ=v δt.

So, v δt/r = δθ.

The vectors vA and vB represent the velocities at these points. The change in velocity between

A and B obtained by subracting vA from vB.

Change in velocity = vA and vB. Parallagram rule or triangle rule. The resultant is effectively

XZ. Since one vector (-vA) is perpendicular to OA and the other vB is perpendicular to OA and

the other vB is perpendicular to OB, therefore ∠ XYZ = ∠ AOB =δθ

If δt is small, the δθ is also small. Then the length of XZ in the figure will be almost the same

as X'Y' in b) X'Z'=rδθ but also δθ=vδt/r. Therefore, X'Z'=v2/r δt.

a=change in velocity/time interval = (v2δt/r)/δt=v2/r. Therefore a=v2/r also a=ω2r.

a = v2/r(4)
a = r ω2(5)

The direction of the acceleration is toward the centre O as can be seen if δt is made so that A

and B all but coincide; XZ is then perpendicular to vA (or vB) is along AO or BO. We say the

body has a centripetal acceleration (i.e. centre seeking). Does a body moving uniformly in a

circle have constant acceleration?

Centripetal Force
Since a body moving in a circle (or circular arc) is accelerating, it follows that from Newton's

2nd law that there must be a force acting on it to cause the acceleration. This force will also

be directed toward the centre and is called the centripetal force. It causes the body to deviate
from the straight line motion it would naturally follow. The magnitude of the centripetal force

is given by:

F = ma = mv2/r(6)

Where m is the mass of the body and v is the speed in the circular path of radius r. If the

angular velocity of the body is ω we can also say since v=rω.

F = ma = mrω2(7)

Other examples of circular motion will be discussed in the following sections. In all cases it is

important to appreciate that the forces acting on the body must provide a resultant force of

magnitude mv2/r toward the centre.

EXPERIMENTAL MEASUREMENT OF CENTRIPETAL FORCE

The turntable is rotated by the electric motor causing the truck of known mass m to move

out from the centre of the turntable. Opposing the motion of the truck is a spring which

provides the centripetal force required to keep the truck in position. As the speed of the

turntable is increased the spring extends until the truck reaches the stop at the end of the

rails. When the truck just hits the truck the extension of the spring is of known length. The

velocity of the turntable can be found by measuring the time it takes for the turntable make

ten revolutions and dividing by ten. The velocity it then 2π/t. We know the m and the radius

and therefore the extension of the spring. We can also measure the force required to produce

the same extension of the same using a spring balance and compare the two forces.

ROUNDING A BEND

If a car is travelling round a circular bend with uniform speed on a horizontal road, the

resultant force acting on it must be directed to the centre. ie. it must be the centripetal force.

This force arise from the interaction of the car with the air and the road. The direction of the

force exerted by the air will more or less oppose the instantaneous direction of motion. The

other more important horizontal force is the frictional force exerted inwards by the road on

the tyres of the car. The resultant of these two forces is the centripetal force. If the

centripetal force is less than the force wanting to pull the car out, a skid will result. Safe

cornering that does not rely on friction is achieved by banking the road. The problem is to

find the angle θ at which the bend should be banked so the centripetal force acting on the car

arises entirely from a component of the normal force, N of the road.

Treating the car as a particle and resolving N vertically and horizontally we have, since N sinθ

is the centripetal force. N sinθ=mv2/r where m and v are the mass, speed and r is the radius

of the bend respectively. Also the car is assumed to remain in the same horizontal plane

therefore no horizontal acceleration therefore, N cosθ=mg. Dividing both equations, tan

θ=v2/rg

A bend in a railway track is also banked. In this case, so that at a certain speed no lateral

thrust has to be exerted by the outer rail on the flanges of the wheels of the train, otherwise

the rails are strained. The horizontal component of the normal forces of the rails on the train

provide the centripetal force.

AIRCRAFT

Figure. A banking aircraft uses the horizontal component of the lift force to provide the
centripetal force for turning.

An aircraft in straight, level flight experiences a lift force perpendicular to the surface of its

wings which balances its weight, mg. To make a turn, the ailerons are operated so that the
aircraft banks and the horizontal component of the lift supplies the necessary centripetal

force to make the aircraft turn. The lift force is always perpendicular to the wings and so the

aircraft's weight has to be supported by the vertical component of the lift. Since this is

slightly less than the weight of the aircraft, the aircraft will lose altitude unless the lift is

increased by increasing the speed or pulling the nose of the aircraft up.

THE ROTOR

The rotor is often seen at a fairground. A cylinder of around 4 meters in diameter is set

spinning. The passengers stand inside the cylinder against the wall as it spins. The rotational

motion causes the passengers to be pinned to the wall of the rotor. When the rotor is

spinning fast enough, the floor of the cylinder drops down and the passengers remain pinned

to the walls. The forces acting on a passenger of m are shown. N is the normal force and is

the centripetal force needed to keep him moving in a circle. Hence if r is the radius of the

rotor and v the speed of the passengers then N=mv2/r. If F is the frictional force acting

upwards between the passenger and the rotor wall and since there is no vertical motion of

the passenger. F=mg. If μ is the coefficient of limiting friction between the passenger and the

wall we have F=μN. Therefore, μN=mg, μ=mg/N=gr/v2. Note that this is independent of the

mass m.

A typical value of μ between clothing and the rotor wall (canvas lined) is about 0.40 so if

r=2m. v=5ms-1. What would be the angular velocity?

Since v=rω, &omega=7/2 =3.5 rad s-1. Revolutions per minute = (3.5 rads-1x60 sec)/2π

rad=210/2π=105/π

LOOPING THE LOOP

Consider a bucket of water when it is at the top of the loop A. If the weight of the water mg

is less than mv2/r, the normal force N of the bottom of the bucket on the water. However, if

the bucket is swung more slowly than mg will mv2/r and the residual part of the weight

causes the water to leave the bucket.

Centrifuge and Washing Machines

Clothes get damaged in the washingmachine because the clothes experience a high force of

gravity
Space Elevator
One application of centripetal force and circular motion is found in the space elevator lift. It

was proposed by Russian scientist, Yuri Artsutanov in 1960 as a possible way of getting into

space. It works on the principle that at a object orbiting the Earth above the equator with a

period of 24 hours will remain in the same position above the Earth (see geostationary

satelites). If a rope were to be lowered from the object, then an elevator car could climb up

the rope into space without the need for a rocket. Since the rope has mass the additional

weight of the rope would cause the satellite holding it to be pulled back to Earth. However, a

counter weight orbiting the Earth a greater distance than the geostationary height would be

travelling faster to stay in the same position over the equator the centripetal force of the

counterweight can be balanced against the weight of the rope. The result is that the rope

stays in place and the space elevator becomes possible.

Even though the theory is sound, there are significant technical challenges yet to be

overcome before the space lift becomes a reality. The first is to do with the material

properties of the rope. Given that the height of the geostationary orbit is some 3.58 x 104 km

above the equator, any rope would have to be extremely strong to support its own weight. At

present there is no known material that would be able to support its own mass at this length

however, carbon-nanotubes have been suggested as a possible material for such ropes

because of their low mass per unit length and extreme strength.

Other problems such as space debris also need to be overcome.

Splung .com physics

MECHANICS

From Wikipedia, the free encyclopedia

Jump to: navigation, search

Not to be confused with angular frequency.

Angular velocity describes the speed of rotation and the orientation of the
instantaneous axis about which the rotation occurs. The direction of the angular
velocity vector will be along the axis of rotation; in this case (counter-clockwise
rotation) the vector points up.

In physics, the angular velocity is a vector quantity (more precisely, a pseudovector) which
specifies the angular speed of an object and the axis about which the object is rotating. The SI
unit of angular velocity is radians per second, although it may be measured in other units
such as degrees per second, revolutions per second, degrees per hour, etc. When measured in
cycles or rotations per unit time (e.g. revolutions per minute), it is often called the rotational
velocity and its magnitude the rotational speed. Angular velocity is usually represented by the
symbol omega (Ω or ω). The direction of the angular velocity vector is perpendicular to the
plane of rotation, in a direction which is usually specified by the right-hand rule.[1]

Contents
[hide]

• 1 The angular velocity of a particle

o 1.1 Particle in two dimensions
o 1.2 Particle in three dimensions
• 2 Angular velocity for a moving frame
o 2.1 With Euler angles
o 2.2 With rotation matrices
o 2.3 With angular displacement tensors
o 2.4 Coordinate-free description
o 2.5 Angular velocity as a vector field
• 3 See also
• 4 References

• 5 External links

[edit] The angular velocity of a particle

[EDIT] PARTICLE IN TWO DIMENSIONS
The angular velocity of the particle at P with respect to the origin O is
determined by the perpendicular component of the velocity vector v.

The angular velocity of a particle in a 2-dimensional plane is the easiest to understand. As

shown in the figure on the right (typically expressing the angular measures ɸ and θ in
radians), if we draw a line from the origin (O) to the particle (P), then the velocity vector (v)
of the particle will have a component along the radius (radial component, v∥) and a
component perpendicular to the radius (cross-radial component, v ). However, it must be
remembered that the velocity vector can be also decomposed into tangential and normal
components.

A radial motion produces no change in the direction of the particle relative to the origin, so
for purposes of finding the angular velocity the parallel (radial) component can be ignored.
Therefore, the rotation is completely produced by the tangential motion (like that of a particle
moving along a circumference), and the angular velocity is completely determined by the
perpendicular (tangential) component.

It can be seen that the rate of change of the angular position of the particle is related to the
cross-radial velocity by:[1]

Utilizing θ, the angle between vectors v∥ and v, or equivalently as the angle between vectors r
and v, gives:

Combining the above two equations and defining the angular velocity as ω=dΦ/dt yields:
In two dimensions the angular velocity is a single number which has no direction. A single
number which has no direction is either a scalar or a pseudoscalar, the difference being that a
scalar does not change its sign when the x and y axes are exchanged (or inverted), while a
pseudoscalar does. The angle as well as the angular velocity is a pseudoscalar. The positive
direction of rotation is taken, by convention, to be in the direction towards the y axis from the
x axis. If the axes are inverted, but the sense of a rotation does not, then the sign of the angle
of rotation, and therefore the angular velocity as well, will change.

It is important to note that the pseudoscalar angular velocity of a particle depends upon the
choice of the origin.

[EDIT] PARTICLE IN THREE DIMENSIONS

Classical mechanics

Newton's Second Law

History of classical
mechanics · Timeline of
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 D'Alembert's principle

[show]Core topics

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v • d • e

In three dimensions, the angular velocity becomes a bit more complicated. The angular
velocity in this case is generally thought of as a vector, or more precisely, a pseudovector. It
now has not only a magnitude, but a direction as well. The magnitude is the angular speed,
and the direction describes the axis of rotation. The right-hand rule indicates the positive
direction of the angular velocity pseudovector.

Just as in the two dimensional case, a particle will have a component of its velocity along the
radius from the origin to the particle, and another component perpendicular to that radius.
The combination of the origin point and the perpendicular component of the velocity defines
a plane of rotation in which the behavior of the particle (for that instant) appears just as it
does in the two dimensional case. The axis of rotation is then a line normal to this plane, and
this axis defined the direction of the angular velocity pseudovector, while the magnitude is
the same as the pseudoscalar value found in the 2-dimensional case. Define a unit vector
which points in the direction of the angular velocity pseudovector. The angular velocity may
be written in a manner similar to that for two dimensions:

which, by the definition of the cross product, can be written:

[edit] Angular velocity for a moving frame

Euler's rotation theorem states that, in any instant of time t there always exists a momentary
or instantaneous axis of rotation. Therefore, any transversal section of the body by a plane
perpendicular to this axis has to behave as a two dimensional rotation. Thus, the magnitude of
the angular velocity vector at a given time t is the derivative at that time of the angle rotated
about the momentary rotation axis:

The angular velocity vector lies on the momentary rotation axis, and its direction is given by
the eigenvector of the derivative of the rotation matrix at time t.

[EDIT] WITH EULER ANGLES

Diagram showing Euler frame in green

The components of the angular velocity vector were first calculated by Leonhard Euler using
his Euler angles and an intermediate frame made out of the intermediate frames of the
construction:

• One axis of the reference frame (the precession axis)

• The line of nodes of the moving frame respect the reference frame
(nutation axis)
• One axis of the moving frame (the intrinsic rotation axis)

Euler proved that the projections of the angular velocity vector over these three axes was the
derivative of its associated angle (which is equivalent to decompose the instant rotation in
three instantaneous Euler rotations). Therefore[2]:

This basis is not orthonormal and it is difficult to use, but now the velocity vector can be
changed to the fixed frame or to the moving frame with just a change of bases. For example,
changing to the mobile frame:

where IJK are unit vectors for the frame fixed in the moving body.

[EDIT] WITH ROTATION MATRICES

See also: axes conventions
Position of point P located in the rigid body (shown in blue). Ri is the position with
respect to the lab frame, centered at O and ri is the position with respect to the
rigid body frame, centered at O' . The origin of the rigid body frame is at vector
position R from the lab frame.

The same equations for the angular speed can be obtained using rotation matrices instead of
Euler angles.

To obtain the equations it is convenient to image a rigid body attached to the frames and
consider a coordinate system that is fixed with respect to the rigid body. Then we will study
the coordinate transformations between this coordinate and the fixed "laboratory" system.

As shown in the figure on the right, the lab system's origin is at point O, the rigid body
system origin is at O' and the vector from O to O' is R. A particle (i) in the rigid body is
located at point P and the vector position of this particle is Ri in the lab frame, and at position
ri in the body frame. It is seen that the position of the particle can be written:

The defining characteristic of a rigid body is that the distance between any two points in a
rigid body is unchanging in time. This means that the length of the vector is unchanging.
By Euler's rotation theorem, we may replace the vector with where is a 3x3
rotation matrix and is the position of the particle at some fixed point in time, say t=0. This
replacement is useful, because now it is only the rotation matrix which is changing in time
and not the reference vector , as the rigid body rotates about point O'. Also, since the three
columns of the rotation matrix represent the three versors of a reference frame rotating
together with the rigid body, any rotation about any axis becomes now visible, while the
vector would not rotate if the rotation axis were parallel to it, and hence it would only
describe a rotation about an axis perpendicular to it (i.e., it would not see the component of
the angular velocity vector parallel to it, and would only allow the computation of the
component perpendicular to it). The position of the particle is now written as:

Taking the time derivative yields the velocity of the particle:

where Vi is the velocity of the particle (in the lab frame) and V is the velocity of O' (the
origin of the rigid body frame). Since is a rotation matrix its inverse is its transpose. So we
substitute :

or

where is a 3x3 matrix called the angular velocity tensor. It is possible to prove
that T is a skew symmetric matrix, by taking the time derivative of :

Applying the formula (AB)T = BTAT:

Thus, T is the negative of its transpose, which implies it is a skew symmetric matrix. Since it
is also a 3x3 matrix, we can take its dual to get a 3 dimensional vector, called the angular
velocity pseudovector, ω:

Substituting ω for T into the above velocity expression, and replacing matrix multiplication
by an equivalent cross product:
It can be seen that the velocity of a point in a rigid body can be divided into two terms - the
velocity of a reference point fixed in the rigid body plus the cross product term involving the
angular velocity of the particle with respect to the reference point. This angular velocity is the
"spin" angular velocity of the rigid body as opposed to the angular velocity of the reference
point O' about the origin O.

Then we can prove that the angular velocity is independent from the choice of origin, or, the
angular velocity is an intrinsic property of the spinning rigid body.

Proving the independence of angular velocity from choice of origin

See the graph to the right: The origin of lab frame is O, while O1 and O2 are two fixed points
on the rigid body, whose velocity is and respectively. Suppose the angular velocity with
respect to O1 and O2 is and respectively. Since point P and O2 have only one velocity,

The above two yields that

Since the point P (and thus ) is arbitrary, it follows that

If the reference point is the instantaneous axis of rotation the expression of velocity of a point
in the rigid body will have just the angular velocity term. This is because the velocity of
instantaneous axis of rotation is zero. An example of instantaneous axis of rotation is the
hinge of a door. Another example is the point of contact of a pure rolling spherical rigid
body.
[EDIT] WITH ANGULAR DISPLACEMENT TENSORS

In general, the angular velocity in an n-dimensional space is the time derivative of the angular
displacement tensor which is a second rank skew-symmetric tensor. This tensor will have
n(n-1)/2 independent components and this number is the dimension of the Lie algebra of the
Lie group of rotations of an n-dimensional inner product space.[3]

In three dimensions angular velocity can be represented by a vector because second rank
tensors are dual to vectors in three dimensions. The tensor can be defined as a matrix T(t)
such that:

with:

So the cross product

can be expressed as a matrix multiplication.

[EDIT] COORDINATE-FREE DESCRIPTION

At any instant, t, the angular velocity tensor is a linear map between the position vectors
and their velocity vectors of a rigid body rotating around the origin:

where we omitted the t parameter, and regard and as elements of the same 3-dimensional
Euclidean vector space V.

The relation between this linear map and the angular velocity pseudovector ω is the
following.

Because of T is the derivative of an orthogonal transformation, the

bilinear form is skew-symmetric. (Here stands for the scalar product). So we can apply the
fact of exterior algebra that there is a unique linear form L on Λ2V that

,
where is the wedge product of and .

Taking the dual vector L* of L we get

Introducing ω: = * L * , as the Hodge dual of L* , and apply further Hodge dual identities
we arrive at

where

by definition.

Because is an arbitrary vector, from nondegeneracy of scalar product follows

[EDIT] ANGULAR VELOCITY AS A VECTOR FIELD

For angular velocity tensor maps velocities to positions, it is a vector field. In particular, this
vector field is a Killing vector field belonging to an element of the Lie algebra so(3) of the 3-
dimensional rotation group, SO(3).

[edit] See also

• Angular frequency
• Angular acceleration
• Angular momentum
• Areal velocity
• Isometry
• Lie algebra
• Orthogonal group
• Rigid body dynamics
• Rotation group

[edit] References
1. ^ a b Hibbeler, Russell C. (2009). Engineering Mechanics. Upper Saddle
River, New Jersey: Pearson Prentice Hall. pp. 314, 153.
ISBN 9780136077916. http://books.google.com/books?id=tOFRjXB-
XvMC&pg=PA314&dq=angular+velocity&rview=1.(EM1)
2. ^ K.S.HEDRIH: Leonhard Euler (1707-1783) and rigid body dynamics [1]
3. ^ Rotations and Angular Momentum on the Classical Mechanics page of
the website of John Baez, especially Questions 1 and 2.
 4.
5.
6. It would be easy just to say that all you have to do is plug the values into a formula for
determining angular velocity and linear velocity, but each question will have its own
challenges.....the most direct path to the answer will be to determine what units you
are asked to have your answer in and set up your question accordingly. With that said,
there are some general formulas that may assist you.
7. Linear Velocity....as the name implies....deals with speed in a straight line, the units
are often km/hr or m/s or mph (miles per hour).

8. Angular Velocity....as the name implies...deals with speed through an angle, or in a

circular motion, the units are often radians per second or degrees per second.

9. To relate the two types of velocity to each other, we can use the following
relationship:

where r = radius
10.
11.
12. Example 1
13. Linear Velocity
14. Driving to Moose Jaw took Bob 40 minutes to travel the 70 kms. Determine his speed
in km / hr.
15.

16.
17. Example 2
18. Angular Velocity
19. A merry-go-round makes one revolution every four seconds. Determine it's angular
velocity in radians /second.
 20.

21.
22. Example 3
23. a) A car's tire revolves 400 times per minute. Determine the angular velocity in
rads /minute.

24.
25. b) Determine the angular velocity in degrees /second.
26.

27.
28. Example 4
29. A ferris wheel with a radius of 6m rotates once every 20 seconds.Determine the linear
speed of a rider in meters/second.
 30.

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#1
01-13-2006, 08:55 AM
unregistered Posts: n/a
Guest
Linear velocity to angular velocity
Hello, does anyone know how to convert linear velocity to angular velocity?

Specifically, cm/s to rad/s or rpm?

THANK YOU

unregistered

#2
01-13-2006, 09:21 AM
ncstagger Posts: n/a
Guest
Re: Linear velocity to angular velocity
My problem is to find the angular velocity in rpm's that corresponds to the linear velocity
of 15cm/s.

Please see if my method is sound here...

I measured the diameter of my circle and it is 7.46cm.

This gives a circumference of 23.42cm (pi*diameter).

Knowing that 1 revolution (circumference) is equal to 6.28 radians I can now have a
conversion factor of 6.28 radians = 23.42 cm.
So I simply multiply the 15 cm/s by 6.28rad/23.42cm and obtain 4.03 radians/s.

This is now a common conversion to rpm and results in 38.5 rpm.

Corrections/input welcome!

ncstagger

#3
05-19-2006, 08:56 PM
sexy_caylee_91 Posts: n/a
Guest
Linear velocity
What is the defination to Linear velocity?

sexy_caylee_91

#4
05-21-2006, 03:37 PM

Mrs X Join Date: Feb 2006

Location: New Zealand
can't count, can't spell! Posts: 1,700
Rep Power: 7

Re: Linear velocity to angular velocity

Quote:

Originally Posted by ncstagger

My problem is to find the angular velocity in rpm's that corresponds to the linear
velocity of 15cm/s.

Please see if my method is sound here...

I measured the diameter of my circle and it is 7.46cm.

This gives a circumference of 23.42cm (pi*diameter).

Knowing that 1 revolution (circumference) is equal to 6.28 radians I can now have a
conversion factor of 6.28 radians = 23.42 cm.

So I simply multiply the 15 cm/s by 6.28rad/23.42cm and obtain 4.03 radians/s.

This is now a common conversion to rpm and results in 38.5 rpm.

Corrections/input welcome!
This sounds about right. My understanding of rheometry is that the linear velocity is
measured at an exact tangent to the outside of the disk. At 15cm/s, it takes 1.56s to
complete 1 revolution, which is 38.46rpm.

i.e. 23.42cm/15cmpersec = 1.56seconds per revolution.

and 60seconds/1.56seconds = 38.5revolutions per minute.

For these sorts of experiments, it always pays to have a look at what is actually happening
to your sample, and have a think about what the approximate answer should be. If you
get an answer that is close, then you are probably on the right track.

Mrs X

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#5
10-21-2006, 04:52 PM
Unregistered Posts: n/a
Guest
Re: Linear velocity
Quote:

Originally Posted by sexy_caylee_91

What is the defination to Linear velocity?
Linear velocity is velocity along a line

Unregistered

#6
05-15-2007, 07:19 PM
jzorrill Posts: n/a
Guest
Re: Linear velocity of the air
ssuming the inside diameter of the colum is 4 inches, calculate the lenear velocity of the
air at the flooding point

jzorrill

#7
11-14-2007, 05:29 PM
Buba bigz Posts: n/a
Guest
Re: Linear velocity to angular velocity
what is linear valocity ???????(definition)

Buba bigz

#8
11-26-2007, 10:28 PM
Unregistered Posts: n/a
Guest
Re: Linear velocity
What is linear velocity? Can someone please give me the definition?

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 #9
11-27-2007, 01:28 PM
Unregistered Posts: n/a
Guest
Re: Linear velocity to angular velocity
linear velocity is basically the velocity of an object traveling in a straight line, where as
angular velocity is the velocity of a obfect moving in a circular path. To convert velocity to
angular velocity you use the equation v=rω, where 'v' is linear velocity and 'ω' is angular
velocity and r is the radius of the circluar path

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#10
09-07-2008, 08:15 PM
Moe Posts: n/a
Guest
Re: Linear velocity to angular velocity
Quote:

Originally Posted by ncstagger

My problem is to find the angular velocity in rpm's that corresponds to the linear
velocity of 15cm/s.

Please see if my method is sound here...

I measured the diameter of my circle and it is 7.46cm.

This gives a circumference of 23.42cm (pi*diameter).

Knowing that 1 revolution (circumference) is equal to 6.28 radians I can now have a
conversion factor of 6.28 radians = 23.42 cm.

So I simply multiply the 15 cm/s by 6.28rad/23.42cm and obtain 4.03 radians/s.

This is now a common conversion to rpm and results in 38.5 rpm.

Corrections/input welcome!
Hi, whoever you are...you really saved mt neck because I have been stuck on this one
problem for the past hour and a half. Thanks to you, you have helped me defeat this
stupidly easy equation god bless you
For other uses, see Revolutions per minute (disambiguation).

"RPM" redirects here. For other uses, see RPM (disambiguation).

Revolutions per minute (abbreviated rpm, RPM, r/min, or r·min−1) is a unit of frequency
of rotation: the number of full rotations completed in one minute around a fixed axis. It is
used as a measure of rotational speed of a mechanical component.

Standards organizations generally recommend the symbol r/min, which is more consistent
with the general use of unit symbols. This is not enforced as an international standard. In
French for example, tr/mn (tours par minute) is commonly used, and the German unit reads
U/min (Umdrehungen pro Minute) or 1/min (= 1 min−1).
The corresponding unit in the International System of Units (SI) is hertz (symbol Hz) or s−1
(1/second). Revolutions per minute is converted to hertz through division by 60. Conversion
from hertz to rpm is by multiplication by 60.

1 rpm = 1/min = 1/(60s) = 1/60 Hertz ≈ 0.01667 Hz

Another related unit is the SI unit for angular velocity, radian per second (rad·s−1):

1 rpm = 2π rad·min−1 = 2π/60 rad·s−1 = 0.1047 rad·s−1 ≈ 1/10 rad·s−1

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Formulas of Motion

Linear and angular acceleration, velocity and displacement

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LINEAR MOTION FORMULAS

Velocity can be expressed as (velocity is constant):

v=s/t (1a)

where

v = velocity (m/s, ft/s)

s = linear displacement (m, ft)

t = time (s)

Velocity can be expressed as (acceleration is constant):

v = v0 + a t (1b)

where

v0 = initial linear velocity (m/s, ft/s)

a = acceleration (m/s2, ft/s2)

Linear displacement can be expressed as (acceleration is constant):

s = v0 t + 1/2 a t2 (1c)

Combining 1a and 1c to express the final velocity

v = (v02 + 2 a s)1/2 (1d)

Velocity can be expressed as (velocity is variable)

v = ds / dt (1f)

where

ds = change of displacement (m, ft)

dt = change in time (s)

Acceleration can be expressed as

a = dv / dt (1g)

where

dv = change in velocity (m/s, ft/s)

ANGULAR MOTION FORMULAS

Angular velocity can be expressed as (angular velocity = constant):

ω=θ/t (2a)

where

ω= angular velocity (rad/s)

θ = angular displacement (rad)

t = time (s)

Angular velocity can be expressed as (angular acceleration is constant):

ω = ωo + α t (2b)

where
ωo = angular velocity at time zero (rad/s)

α = angular acceleration (rad/s2)

Angular displacement can be expressed as (angular acceleration is constant):

θ = ωo t + 1/2 α t2 (2c)

Combining 2a and 2c:

ω = (ωo2 + 2 α θ)1/2

Angular acceleration can be expressed as:

α = dω / dt = d2θ / dt2 (2d)

where

dθ = change of angular displacement (rad)

dt = change in time (s)

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Classical mechanics

Newton's Second Law

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Equations of motion are equations that describe the behavior of a system (e.g., the motion of
a particle under the influence of a force) as a function of time.[1] Sometimes the term refers to
the differential equations that the system satisfies (e.g., Newton's second law or Euler–
Lagrange equations), and sometimes to the solutions to those equations.

Contents
[hide]

• 1 Equations of uniformly accelerated linear

motion
• 2 Classic version
o 2.1 Examples
o 2.2 Extension
• 3 Equations of circular motion
• 4 Derivation
o 4.1 Equation 2
o 4.2 Equation 3
o 4.3 Equation 4
• 5 See also
• 6 External links

• 7 References

[edit] Equations of uniformly accelerated

linear motion
The equations that apply to bodies moving linearly (in one dimension) with constant
acceleration are often referred to as "SUVAT" equations where the five variables are
represented by those letters (s = displacement, u = initial velocity, v = final velocity, a =
acceleration, t = time); the five letters may be shown in a different order.

The body is considered between two instants in time: one initial point and one current (or
final) point. Problems in kinematics may deal with more than two instants, and several
applications of the equations are then required. If a is constant, a differential, a dt, may be
integrated over an interval from 0 to Δt (Δt = t − ti), to obtain a linear relationship for
velocity. Integration of the velocity yields a quadratic relationship for position at the end of
the interval.

where...

is the body's initial velocity

is the body's initial position

and its current state is described by:

, The velocity at the end of the

interval
 , the position at the end of the
interval (displacement)

, the time interval between the

initial and current states

, the constant acceleration, or in

the case of bodies moving under
the influence of gravity, g.

Note that each of the equations contains four of the five variables. Thus, in this situation it is
sufficient to know three out of the five variables to calculate the remaining two.

[edit] Classic version

The equations below (often informally known as the "suvat"[2] equations) are often written in
the following form:[3]

By substituting (1) into (2), we can get (3), (4) and (5). (6) can be constructed by rearranging
(1).

where

s = the distance between initial and final positions (displacement)

(sometimes denoted R or x)

u = the initial velocity (speed in a given direction)

v = the final velocity

a = the constant acceleration

t = the time taken to move from the initial state to the final state

[EDIT] EXAMPLES

Many examples in kinematics involve projectiles, for example a ball thrown upwards into the
air.
Given initial speed u, one can calculate how high the ball will travel before it begins to fall.

The acceleration is local acceleration of gravity g. At this point one must remember that
while these quantities appear to be scalars, the direction of displacement, speed and
acceleration is important. They could in fact be considered as uni-directional vectors.
Choosing s to measure up from the ground, the acceleration a must be in fact −g, since the
force of gravity acts downwards and therefore also the acceleration on the ball due to it.

At the highest point, the ball will be at rest: therefore v = 0. Using the fifth equation, we have:

Substituting and cancelling minus signs gives:

[EDIT] EXTENSION

More complex versions of these equations can include s0 for the initial position of the body,
and v0 for u for consistency.

[edit] Equations of circular motion

The analogues of the above equations can be written for rotation:
where:

is the angular acceleration

is the angular velocity

is the angular displacement

is the initial angular velocity.

t is the time taken to move from the initial state to the final state

[edit] Derivation
These equations assume constant acceleration and non-relativistic velocities.

[EDIT] EQUATION 2

By definition:

Hence:

[EDIT] EQUATION 3

Using equation 2, substitute t with above:

[EDIT] EQUATION 4

Using equation 1 to substitute u in equation 2 gives:

12.3. Torque

Figure 12.7. Motion of particle P in the x-y plane.

A particle with mass m moves in the x-y plane (see Figure 12.7). A single force F acts on the
particle and the angle between the force and the position vector is [phi]. Per definition, the
torque exerted by this force on the mass, with respect to the origin of our coordinate system,
is given by

and

where r[invtee] is called the arm of the force F with respect to the origin. According to the
definition of the vector product, the vector [tau] lies parallel to the z-axis, and its direction
(either up or down) can be determined using the right-hand rule. Torque defined in this way
has meaning only with respect to a specified origin. The direction of the torque is always at
right angles to the plane formed by the vectors r and F. The torque is zero if r = 0 m, F = 0 N
or r is parallel or anti-parallel to F.

12.4. Angular Momentum

The angular momentum L of particle P in Figure 12.7, with respect to the origin, is defined as

This definition implies that if the particle is moving directly away from the origin, or directly
towards it, the angular momentum associated with this motion is zero. A particle will have a
different angular momentum if the origin is chosen at a different location. A particle moving
in a circle will have an angular momentum (with respect to the center of the circle) equal to
Again we notice the similarity between the definition of linear momentum and the definition
of angular momentum.

A particle can have angular momentum even if it does not move in a circle. For example,
Figure 12.8 shows the location and the direction of the momentum of particle P. The angular
momentum of particle P, with respect to the origin, is given by

Figure 12.8. Angular momentum of particle P.

The change in the angular momentum of the particle can be obtained by differentiating the
equation for l with respect to time

We conclude that

This equation shows that if the net torque acting on the particle is zero, its angular
momentum will be constant.

Example Problem 12-3

Figure 12.9 shows object P in free fall. The object starts from rest at the position indicated in
Figure 12.9. What is its angular momentum, with respect to the origin, as function of time ?

The velocity of object P, as function of time, is given by

The angular momentum of object P is given by

Therefore

which is equal to the torque of the gravitational force with respect to the origin.

Figure 12.9. Free fall and angular momentum

Figure 12.10. Action - reaction pair.

If we look at a system of particles, the total angular momentum L of the system is the vector
sum of the angular momenta of each of the individual particles:

The change in the total angular momentum L is related to the change in the angular
momentum of the individual particles
Some of the torques are internal, some are external. The internal torques come in pairs, and
the vector sum of these is zero. This is illustrated in Figure 12.10. Figure 12.10 shows the
particles A and B which interact via a central force. Newton's third law states that forces
come in pairs: if B exerts a force FAB on A, than A will exert a force FBA on B. FAB and FBA are
related as follows

The torque exerted by each of these forces, with respect to the origin, can be easily calculated

and

Clearly, these two torques add up to zero

The net torque for each action-reaction pair, with respect to the origin, is equal to zero.

We conclude that

This equation is another way of expressing Newton's second law in angular quantities.

Frequency and Period

You’ve encountered frequency and period when dealing with springs and simple harmonic
motion, and you will encounter them again in the chapter on waves. These terms are also relevant
to rotational motion, and SAT II Physics has been known to test the relation between angular
velocity and angular frequency and period.
Angular Frequency
Angular frequency, f, is defined as the number of circular revolutions in a given time interval.
It is commonly measured in units of Hertz (Hz), where 1 Hz = 1 s–1. For example, the second hand
on a clock completes one revolution every 60 seconds and therefore has an angular frequency of
1
/60 Hz.
The relationship between frequency and angular velocity is:

For example, the second hand of a clock has an angular velocity of s.

Plugging that value into the equation above, we get

which we already determined to be the frequency of the second hand of a clock.

Angular Period
Angular period, T, is defined as the time required to complete one revolution and is related to
frequency by the equation:

Since we know that the frequency of the second hand is 1/60 Hz, we can quickly see that the period
of the second hand is 60 s. It takes 60 seconds for the second hand to complete a revolution, so
the period of the second hand is 60 seconds. Period and angular velocity are related by the
equation

EXAMPLE

The Earth makes a complete rotation around the sun once every 365.25 days. What is the Earth’s
angular velocity?
The question tells us that the Earth has a period of T = 365.25 days. If we plug this value into the
equation relating period and angular velocity, we find:

Note, however, that this equation only gives us the Earth’s angular velocity in terms of radians
per day. In terms of radians per second, the correct answer is:

Relation of Angular Variables to Linear Variables

At any given moment, a rotating particle has an instantaneous linear velocity and an
instantaneous linear acceleration. For instance, a particle P that is rotating counterclockwise will
have an instantaneous velocity in the positive y direction at the moment it is at the positive x-
axis. In general, a rotating particle has an instantaneous velocity that is tangent to the circle
described by its rotation and an instantaneous acceleration that points toward the center of the
circle.

On SAT II Physics, you may be called upon to determine a particle’s linear velocity or acceleration
given its angular velocity or acceleration, or vice versa. Let’s take a look at how this is done.
Distance
We saw earlier that the angular position, , of a rotating particle is related to the length of the
arc, l, between the particle’s present position and the positive x-axis by the equation = l/r, or l =
r. Similarly, for any angular displacement, , we can say that the length, l, of the arc made by a
particle undergoing that displacement is
Note that the length of the arc gives us a particle’s distance traveled rather than its displacement,
since displacement is a vector quantity measuring only the straight-line distance between two
points, and not the length of the route traveled between those two points.
Velocity and Acceleration
Given the relationship we have determined between arc distance traveled, l, and angular
displacement, , we can now find expressions to relate linear and angular velocity and
acceleration.
We can express the instantaneous linear velocity of a rotating particle as v = l/t, where l is the
distance traveled along the arc. From this formula, we can derive a formula relating linear and
angular velocity:

In turn, we can express linear acceleration as a = v/t, giving us this formula relating linear and
angular acceleration:

EXAMPLE

The radius of the Earth is approximately m. What is the instantaneous velocity of a point on
the surface of the Earth at the equator?

We know that the period of the Earth’s rotation is 24 hours, or seconds. From the
equation relating period, T, to angular velocity, , we can find the angular velocity of the Earth:

Now that we know the Earth’s angular velocity, we simply plug that value into the equation for
linear velocity:

They may not notice it, but people living at the equator are moving faster than the speed of
sound.
Equations of Rotational Kinematics
In Chapter 2 we defined the kinematic equations for bodies moving at constant acceleration. As
we have seen, there are very clear rotational counterparts for linear displacement, velocity, and
acceleration, so we are able to develop an analogous set of five equations for solving problems in
rotational kinematics:

In these equations, is the object’s initial angular velocity at its initial position, .
Any questions on SAT II Physics that call upon your knowledge of the kinematic equations will
almost certainly be of the translational variety. However, it’s worth noting just how deep the
parallels between translational and rotational kinematics run.
Vector Notation of Rotational Variables
Angular velocity and angular acceleration are vector quantities; the equations above define their
magnitudes but not their directions. Given that objects with angular velocity or acceleration are
moving in a circle, how do we determine the direction of the vector? It may seem strange, but the
direction of the vector for angular velocity or acceleration is actually perpendicular to the plane in
which the object is rotating.
We determine the direction of the angular velocity vector using the right-hand rule. Take your
right hand and curl your fingers along the path of the rotating particle or body. Your thumb then
points in the direction of the angular velocity of the body. Note that the angular velocity is along
the body’s axis of rotation.
The figure below illustrates a top spinning counterclockwise on a table. The right-hand rule
shows that its angular velocity is in the upward direction. Note that if the top were rotating
clockwise, then its angular velocity would be in the downward direction.

To find the direction of a rigid body’s angular acceleration, you must first find the direction of the
body’s angular velocity. Then, if the magnitude of the angular velocity is increasing, the angular
acceleration is in the same direction as the angular velocity vector. On the other hand, if the
magnitude of the angular velocity is decreasing, then the angular acceleration points in the
direction opposite the angular velocity vector.
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