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Rotational Motion

We consider only rigid (non-squishy) bodies in

this section.

We have already studied translational motion in

considerable detail.

We have already studied translational motion in

considerable detail. We now know that the

translational motion we have been describing is

the motion of the center of mass.

We begin by considering purely rotational motion

(the center of mass does not change its xyz

coordinates)

We begin by considering purely rotational motion

(the center of mass does not change its xyz

coordinates) but soon we will consider objects

which both rotate and translate.

Angular Quantities

For rotational motion, we specify angles in

radians instead of degrees.

Look at a point P somewhere on a rotating

circular disc.

P is a distance r from the center of the disc.

P

r

l

Choose the x-axis to be horizontal. Then the

line from the center of the disc to P makes an

angle with the x-axis.

O

x

l/r, where l is the length of the arc from the

x-axis to P.

If the angle is 360 degrees, the arc length is

2r, so 2 radians are equal to 360 degrees.

This is how I remember the conversion factor

between degrees and radians.

Strictly speaking, radians are not a unit (angles

are unitless).

When you studied kinematics, you defined an

objects motion by specifying its position,

velocity, and acceleration.

Now we are about to study angular kinematics. We

will define an objects angular motion by

specifying its angle (rotational position

relative to some axis), angular velocity, and

angular acceleration.

If we call the z axis the axis of rotation,

angular velocity is defined by avg/t, where

is the angular displacement of a rotating

object in the time t. Instantaneous angular

velocity is d/dt.

Wait a minute! Velocity is a vector! What is

the direction of

Consider our rotating wheel.

Heres a point and its instantaneous linear

velocity.

Another.

Another. And another.

How do we define a unique direction for the

angular velocity

If we put the x and y axes in the plane of the

wheel, then the wheel is rotating about an axis

perpendicular to this plane.

z

We usually call this the z-axis.

To remind you that angular velocity has a

direction, Ill write z,avg/t and zd/dt.

Our rules for vectors apply. You get to choose

the direction of the z-axis. Whether z is

positive or negative depends on the direction of

rotation.

this symbol indicates an arrow coming out of the

screen

Right hand rule curl the fingers of your right

hand in the direction of rotation. Your thumb

points in the direction of the angular velocity

vector.

Youll learn to love the right hand rule next

week!

Note that all points on a rigid body rotate with

the same angular velocity (they go around once in

the same time).

What about the tangential velocities of different

points

Well get to that question in a minute.

Angular velocity z,avg/t and zd/dt.

You can probably figure out how angular

acceleration is defined.

z,avgz/t, where z is the change in angular

velocity of a rotating object in the time t.

Instantaneous angular acceleration is zdz/dt.

The subscript z on and emphasizes the

rotation is relative to some axis, which we

typically label z.

I should really put a subscript z on because it

is also measured relative to the z axis. To be

consistent with Physics 23, I will leave it off.

Because points on a rotating object also have an

instantaneous linear motion, linear and angular

motion must be connected.

vtang

r

z

.

As your book shows, vtangentialrz and

atangentialrz. Note that z and z are the

same for all points on a rotating rigid body, but

vtang and atang are not.

Notation vtangentialvtangv.

, , v and a are all magnitudes of vectors!

z and z are vector components!

We often use frequency and period of rotation

f/2 and T1/f.

Example What are the linear speed and

acceleration of a child seated 1.2 m from the

center of a steadily rotating merry-go-round that

makes one complete revolution in 4.0 s

z 2/4 s-1

vtang rz (1.2 m) (/2 s-1)

vtang 0.6 m/s

z 0 (the angular velocity is not changing)

atang rz 0

aradial ar (0.6 m/s)2 / (1.2 m)

ar 2.96 m/s2

The total acceleration is the vector sum of the

radial and tangential accelerations

New OSEs introduced in this section

Kinematic Equations for Uniformly Accelerated

Rotational Motion

Remember our equations of linear kinematics

Analogous equations hold for rotational motion.

linear angular vxv0xaxt z0z

zt xx0v0xt½axt2 00zt½zt2 vx2v0x2

2ax(x-x0) z2z22z(-0)

new OSEs

Example A centrifuge motor goes from rest to

20,000 rpm in 5 minutes. Through how many

revolutions did the centrifuge motor of example

8.5 turn during its acceleration period Assume

constant angular acceleration.

First we need to calculate

z,avgz/t

z,avg(f-i) / (tf-ti)

f(20000rev/min)(min/60s)(2 radians/rev)

tf(5 min)(60 s/min)

z,avg7.0 rad/s2

Next calculate the angle through which the

centrifuge motor turns.

00zt½zt2

½ (7.0 rad/s2) (300 s)2

3.15x105 radians

There are 2 radians in each revolution so the

number of revolutions, N, is

N(3.15x105 radians)(revolution/2 radians)

N5x104 revolutions).

I chose to do the calculations numerically rather

than symbolically for practice with angular

conversions.

Rolling Motion

Many rotational motion situations involve rolling

objects.

Rolling without slipping involves both rotation

and translation.

Friction between the rolling object and the

surface it rolls on is static, because the

rolling objects contact point with the surface

is always instantaneously at rest.

this point on the wheel is instantaneously at

rest if the wheel does not slip ()

the illustration of in this diagram is

misleading the direction of would actually be

into the screen

Here are some things you need to know about

rolling without slipping.

The point on the rolling object in contact with

the ground is instantaneously at rest.

The center of the wheel moves with the speed of

the center of mass.

The point at the top of the rolling wheel moves

with a speed twice the center of mass.

The two side points level of the center of mass

move vertically with the speed of the center of

mass.

What is important for us is that if an object

rolls without slipping, we can use our OSEs for

z and z, using the translational speed of the

object.

vCMrz

aCMrz

These relationships will be valuable when we

study the kinetic energy of rotating objects.

Torque

We began this course with a section on

kinematics, the description of motion without

asking about its causes.

We then found that forces cause motion, and used

Newtons laws to study dynamics, the study of

forces and motion.

We have had a deceptively brief section on

angular kinematics. It is brief because we

already learned how to solve such problems on the

first day of class.

Whats next kinematics dynamics

Whats next

Whats next kinematics dynamics rotational

kinematics rotational dynamics

Whats next kinematics dynamics rotational

kinematics rotational dynamics The rotational

analog of force is torque.

Consider two equal and opposite forces acting at

the center of mass of a stationary meter stick.

F

F

Does the meter stick move

Fext macm 0 so acm 0.

Consider two equal and opposite forces acting on

a stationary meter stick.

F

F

Does the meter stick move

Fext macm 0 so acm 0.

The center of mass of the meter stick does not

accelerate, so it does not undergo translational

motion.

However, the meter stick would begin to rotate

about its center of mass.

A torque is produced by a force acting on an

extended (not pointlike) object.

F

The torque depends on how strong the force is,

and where it acts on the object.

O

You must always specify your reference axis for

calculation of torque. By convention, we

indicate that axis with the letter O and a dot.

Torques cause changes in rotational motion.

Torque is a vector. It is not a force, but is

related to force.

So never set a force equal to a torque!

Lets apply a force to a rod and see how we get a

torque.

First apply the force.

You need to choose an axis of rotation. Usually

there will be a smart choice. Label it with a

point (or line) and an O.

Choose the direction of rotation that you want to

correspond to positive torque.

Label the positive direction with a curved arrow

and a sign! Do this around the point labeled

O.

Traditionally, the counterclockwise direction is

chosen to be positive. You are free to choose

otherwise.

Draw a vector from the origin to the tail of the

force vector. Give it a name (typically R).

Label the angle between R and F (you may have to

the vectors around to see this angle).

O

The symbol for torque is the Greek letter tau

(). The magnitude of the torque due to F is RF

sin.

F

Look at your diagram and determine if the torque

would cause a or a rotation (according to

your choice of ). In this case, the rotation

would be -, so z-RF sin.

Important is the angle between R and F.

R and F around until their tails touch.

is the angle between them.

O

F

Between means go from R to F.

Dont be fooled by a problem which gives an angle

not between the vectors! (Example coming

soon.)

In this diagram, which is the angle between R and

F

F

R

NO!

There are two choices for the angle between R and

F.

Because sin()-sin(-), either choice will give

you the correct answer (switch direction of

rotation and switch sign on sine gives no net

switch in sign).

There are other ways to find the torque.

Often it is easy to visualize F, the component

of F which is perpendicular to R.

O

The magnitude of the torque due to F is RF, and

in this case z-RF. (Note FF sin.)

R

Sometimes it is easier to visualize R, the

component of R which is perpendicular to F.

F

The magnitude of the torque due to F is RF, and

in this case z-RF.

The magnitude of the torque due to F is RF, and

in this case z-RF. (Note RR sin.)

Summarizing

R is called the lever arm or moment arm.

R is called the lever arm or moment arm. The

line along which F is directed is its line of

action.

The z axis passes through the point O and is

perpendicular to the plane of the paper.

To find the direction of the torque, curl your

fingers around the direction of rotation from R

into F. The thumb of your right hand points in

the direction of the torque.

Important reminder label the point O about which

your torques are calculated and draw a curved

arrow around it with a sign to show what you

have chosen for a positive sense of direction.

Draw the curved arrow around the point O, not

somewhere else!

A torque producing a rotation is . A torque

producing a - rotation is -.

Example The biceps muscle exerts a vertical

force of 700 N on the lower arm, as shown in the

figure. Calculate the torque about the axis of

rotation through the elbow joint.

There is no new litany for torques. You should

adapt the litany for force problems.

When you work with torques, the first thing you

need to do is draw an extended free-body diagram.

Before that, we need to have a diagram of the

thing we are investigating.

We are not interested in the upper arm!

We have our diagram. Now we must do a free-body

diagram. For rotational motion, we must do an

extended free-body diagram, which shows where the

forces are applied.

Label the rotation axis.

Choose a direction for rotation.

O

How about this for an OSE z RF sin

No! No! No!

From the extended free-body diagram, I see that

the angle between r and F is 90, so zRF

sin(90) would work.

I think it is better to look for r or F. In

this case, r is easy to see.

From the diagram rr cos.

OSE z RF RF cos. Done! (Except for

plugging in numbers.)

That was a lot of work for something that takes

2 lines in the textbook!

No. I showed you a general approach to torque

problems. The text just solved one simple

problem.

If more than one torque acts on an object, the

net torque is the algebraic sum of the two

torques (algebraic means there may be signs

involved).

Example Calculate the net torque on the compound

wheel shown in the drawing.

The diagram will serve as an extended free-body

diagram. No need for a separate one.

z,net z z,F1 z,F2

z,net r1F1 r2(-F2cos)

z,net r1F1 - r2F2cos

Rotational Dynamics Torque and Rotational Inertia

We saw in our study of dynamics that forces cause

acceleration F ma.

Torques produce angular acceleration, and the

rotational equivalent of mass is the moment of

inertia, I OSE z Iz.

This is really a vector equation, but our

problems will all have a unique axis of rotation,

which is like a one-dimensional problem, so

that the only vestiges of the vector nature of z

will be the sign.

What is this moment of inertia, I

It is the rotational analog of mass.

I depends on the mass of the object. It also

depends on how the mass is distributed relative

to the axis of rotation.

The figure on the nextgives I for various

objects of uniform composition. You should use

this figure (or its equivalent, or appropriate

portions of it) in homework solutions.

Solid cylinder, mass M, radius R I½MR2 It

doesnt matter how thick the cylinder is!

This means a single object can have different

Is for different axes of rotation!

(No Transcript)

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