Rotational Motion

We consider only rigid (non-squishy) bodies in this section.

We have already studied translational motion in considerable detail. 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)… 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.

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 r  P ℓ 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 2  r, 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).

O  P ℓ When you studied kinematics, you defined an object’s motion by specifying its position, velocity, and acceleration.

x Now we are about to study angular kinematics. We will define an object’s 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.

Here’s a point and its instantaneous linear velocity.

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.

We usually call this the z-axis.

 z The direction of  is along the z axis, perpendicular to the wheel, and is given by the right hand rule.

To remind you that angular velocity has a direction, I’ll write  z,avg =  /  t and  z =d  /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.

You’ll 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?

We’ll get to that question in a minute.

Angular velocity  z,avg =  /  t and  z =d  /dt.

You can probably figure out how angular acceleration is defined.  P ℓ  z,avg =  z /  t, where  z is the change in angular velocity of a rotating object in the time  t.

Instantaneous angular acceleration is  z =d  z /dt.

The subscript z on  and  emphasizes the rotation is relative to some axis, which we typically label “z.” x I should really put a subscript z on measured relative to the z axis. To be consistent with Physics 23, I will leave it off.

 because it is also

Because points on a rotating object also have an instantaneous linear motion, linear and angular motion must be connected.

As your book shows, v tangential =r  z and a  z are the same for all points on a rotating rigid body, but v a tang tangential are not.

=r  z . Note that tang  z and and  z Notation: v tangential =v tang =v  .

 ,  , v and a are all magnitudes of vectors!

 z and  z are vector components!

v tang .

r

Acceleration has both tangential and radial components. We know from before that a r =v 2 /r. Total acceleration is a=a tangential +a r .

We often use frequency and period of rotation: f=  /2  and T=1/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 v = 2  /4 s -1 tang = r  z = (1.2 m) (  /2 s -1 )  z  r v tang v tang = 0.6

 m/s  z = 0 (the angular velocity is not changing)

a a tang radial = r  = a r z = 0 = (0.6  m/s) 2 / (1.2 m)  z  r v tang a r = 2.96 m/s 2 The total acceleration is the vector sum of the radial and tangential accelerations: a = 2.96 m/s 2 , towards center of merry-go-round New OSE’s introduced in this section:  z, =d  /dt  z =d  z /dt v tang =r  z a tang =r  z You can also use a radial = a R =R  z 2

Kinematic Equations for Uniformly Accelerated Rotational Motion

Remember our equations of linear kinematics? Analogous equations hold for rotational motion.

linear

v x =v 0x +a x t x=x 0 +v 0x t+½a x t 2 v x 2 =v 0x 2 +2a x (x-x 0 )

angular

 z =  0z +  z t θ=θ 0 +  0z t+½  z t 2  z 2 =  z 2 +2  z (θ-θ 0 ) new OSE’s

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,avg =  z /  t z,avg =(  f  i ) /(t f -t i ) 0  f =(20000rev/min)(min/60s)(2  radians/rev) t f =(5 min)(60 s/min)  z,avg =7.0 rad/s 2

Next calculate the angle through which the centrifuge motor turns.

θ=θ 0 +  0z 0  z t 2 θ=½ (7.0 rad/s 2 ) (300 s) 2 θ=3.15x10

5 radians There are 2  radians in each revolution so the number of revolutions, N, is N=(3.15x10

5 radians)(revolution/2  radians) N=5x10 4 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 object’s 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 (slide) the illustration of the direction of

 

in this diagram is misleading; 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.

v cm v top =2v cm v cm v cm The center of the wheel moves with the speed of the center of mass.

v bot =0 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 OSE’s for object.

 z and  z , using the translational speed of the v CM =r  z a CM =r  z v cm v bot v v top cm =0 =2v v cm cm 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 Newton’s 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.

What’s 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?



F

ext = ma cm = 0 so a cm = 0.

Consider two equal and opposite forces acting on a stationary meter stick.

F F Does the meter stick move?



F

ext = ma cm = 0 so a cm = 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.

The torque depends on how strong the force is, and where it acts on the object.

O

F

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,

*

force.

but is related to

*So never set a force equal to a torque!

Let’s 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.

* + O Label the positive direction with a curved arrow and a “+” sign! Do this around the point labeled “O.”

F *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

slide the vectors around to see this angle)

.

+ O

R

The symbol for torque is the Greek letter tau ( to F is RF sinθ.

 ). The magnitude of the torque due θ

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 sinθ.

Important:

between

R

θ is the angle and

F

.

“Slide”

R

and

F

around until their tails touch. θ is the angle between them.

+ O θ

F R R

θ “Between” means go from

R

to

F

.

F

Don’t be fooled by a problem which gives an angle θ not “between” the vectors! (Example coming soon.)

This θ is the correct “angle between.” Watch out for diagrams containing some other angle!

In this diagram, which is the angle between

R

and

F

?

F R

θ?

NO!

There are two choices for the angle between

R

and

F

.

Because sin(θ)=-sin(-θ), 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 the component of

F

perpendicular to

R

.

F

 , which is The magnitude of the torque due to F is RF  , and in this case  z =-RF  . (Note F  =F sinθ.) Sometimes it is easier to visualize

R

 , the component of

R

which is perpendicular to

F

.

R

 + O

R

 θ

F



F

The magnitude of the torque due to F is R case  z =-R  F.

 =R sinθ.)  F, and in this

Summarizing: OSE:  z = RF  = R  F = RF sinθ

R

 is called the lever arm or moment arm.

which action.

F

is directed is its line of

R

 + O

R

The z axis passes through the point O and is perpendicular to the plane of the paper.

θ

F



F

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!

R F

A torque producing a + rotation is +. A torque producing a rotation is -.

O +

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.

r r=5 cm

F F

 We are not interested in the upper arm!

30°

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.

r

F

 O +

r

θ

F

Label the rotation axis.

Choose a + direction for rotation.

How about this for an OSE?

 z = RF sinθ No! No! No!

From the extended free-body diagram, I see that the angle between

r

and

F

is 90+θ, so  z =RF sin(90+θ) would work.

I think it is better to look for

F

 . In this case,

r



r

 or is easy to see.

O +

r r

 θ

F

From the diagram r  =r cosθ.

OSE:  z = R  F = RF cosθ. 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:

the compound wheel shown in the drawing.

Calculate the net torque on The diagram will serve as an extended free-body diagram. No need for a separate one.

 z,net =  z =  z,F1 +  z,F2

r 1 r 2 F 2

θ  z,net = +r 1 F 1 + r 2 (-F 2 cosθ)

F 1

 z,net = r 1 F 1 - r 2 F 2 cosθ

Rotational Dynamics; Torque and Rotational Inertia

We saw in our study of dynamics that forces cause acceleration: 

F

= m

a

.

Torques produce angular acceleration, and the rotational equivalent of mass is the moment of inertia, I: OSE:  z = I  z .

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 next slide gives 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 I=½MR 2 It doesn’t matter how thick the cylinder is!

*This means a single object can have different I’s for different axes of rotation!