physics 2211: lecture 36

Lecture 36, Physics 2211 Spring 2005© 2005 Dr. Bill Holm

Physics 2211: Lecture 36Physics 2211: Lecture 36

Rotational Dynamics and Torque


Summary Summary (with comparison to 1-D kinematics)(with comparison to 1-D kinematics)

And for a point at a distance R from the rotation axis:

s = Rv = Ra = R

2 20 02

210 0 2s s v t at

2 20 02v v a s s

constantconstanta

AngularLinear

0v v at

210 0 2t t

0 t


Rotation & Kinetic EnergyRotation & Kinetic Energy

The kinetic energy of a rotating system looks similar to that of a point particle:

Point ParticlePoint Particle Rotating System Rotating System

v is “linear” velocity

m is the mass.

is angular velocity

I is the moment of inertia

about the rotation axis.

21

2K I21

2K mv

2i i

i

I m r


Suppose a force acts on a mass constrained to move in a

circle. Consider its acceleration in the direction at some

instant.

Multiply by r :

F ma

Fr mra

Newton’s 2nd Law in the direction:

m

F

F

a

rr

x

y

Rotational DynamicsRotational Dynamics


Torque has a direction:+ z if it tries to make the system

turn CCW.- z if it tries to make the system

turn CW.

Define torque: is the tangential force F

times the distance r.

Fr

m

F

F

a

rr

x

y



rp = “distance of closest approach” or “lever arm”

FFr

r

F

sinr F sinr F Fr

Fr

rp



For a collection of many particles arranged in a rigid configuration:

2

i

i i i i ii i

r F m r

Since the particles are connected rigidly,they all have the same .

NET I

m4

m1

m2

m3

1r

4F

3F

2F

4r

3r

2r

1F


2

i i ii i

I

m r


This is the rotational analog of FNET = ma

Torque is the rotational analog of force:Torque is the rotational analog of force: The amount of “twist” provided by a force.

Moment of inertiaMoment of inertia I I is the rotational analog of mass, i.e.,is the rotational analog of mass, i.e.,

““rotational inertial.”rotational inertial.” If I is big, more torque is required to achieve a given angular acceleration.

Torque has units of kg m2/s2 = (kg m/s2) m = N-m.

NET I



When we write = I we are really talking about the z component of a more general vector equation. (More on this later.) We normally choose the z-axis to be the rotation axis.)

z = Izz

We usually omit the

z subscript for simplicity.

Comment onComment on == II

z

z

z

Iz


Torque and the Torque and the Right Hand RuleRight Hand Rule

The right hand rule can tell you the direction of torque:Point your hand along the direction from the axis to the

point where the force is applied.Curl your fingers in the direction of the force.Your thumb will point in the direction

of the torque.

x

y

z

r

F


We can describe the vectorial nature of torque in a compact form by introducing the “cross product”.The cross product of two vectors is a third vector:

The Cross (or Vector) ProductThe Cross (or Vector) Product

BA C

The direction of is perpendicular to the plane defined by and and the “sense” of the direction is defined by the right hand rule.

C

A

B

B

A

C

The length of is given by:

C = AB sin C


Cross product of unit vectors:

The Cross ProductThe Cross Product

ˆ ˆ ˆ ˆ ˆ 0i i j j k k

ˆ ˆ ˆi j k ˆ ˆj k i

ˆ ˆk i j

ˆ ˆ ˆj i k ˆ ˆk j i

ˆ ˆ ˆi k j

i

jk

++

+


The Cross ProductThe Cross Product

Cartesian components of the cross product:

ˆ ˆ ˆ ˆxx y yz zA A A AB B B BC i j k i j k

B

A

C

Note:

AB A B

ˆ ˆ ˆ

x y z

x y z

A A A

i j k

B B B

C or

zx zyyC B ABA

xy xzzC B ABA

yz yxxC B ABA


Torque & the Cross ProductTorque & the Cross Product

So we can define torque as:

x

y

z

r

F

sinFr

r F

X = rY FZ - FY rZ = y FZ - FY z

Y = rZ FX - FZ rX = z FX - FZ x

Z = rX FY - FX rY = x FY - FX y


Center of Mass RevisitedCenter of Mass Revisited

Define the Center of Mass (“average” position):For a collection of N individual pointlike particles whose

masses and positions we know:

(In this case, N = 4)

y

x

m1

m4

m2

m3

1r

4r

2r

3r

CMR1

N

i ii

CM

m rR

M

1

N

ii

M m

(total mass)


System of Particles: Center of MassSystem of Particles: Center of Mass

The center of mass is where the system is balanced!Building a mobile is an exercise in finding centers of mass.Therefore, the “center of mass” is the “center of gravity” of an object.

m1

m2

+m1 m2

+



For a continuous solid, we have to do an integral.

y

x

dm

where dm is an infinitesimal

element of mass.

r

CM

rdm rdmR

Mdm



The location of the center

of mass is an intrinsic

property of the object!!

(it does not depend on where

you choose the origin or

coordinates when

calculating it).

y

x

We find that the Center of Mass is at the “mass-weighted” center of the object.

CMR



The center of mass (CM) of an object is where we can freely pivot that object.

Force of gravity acts on the object as though all the mass were located at the CM of the object. (Proof coming up!)

If we pivot the objectsomewhere else, it willorient itself so that theCM is directly below the pivot.

This fact can be used to findthe CM of odd-shaped objects.

+ CM

pivot

+

CM

pivot

+

pivot

CM

mg



Hang the object from several pivots and see where the vertical lines through each pivot intersect!

pivot

pivotpivot

+

CM

The intersection point must be at the CM.


Torque on a body in a uniform Torque on a body in a uniform gravitation fieldgravitation field

What is the torque exerted by the force of gravity on a body of total mass M about the origin?

d r dF r dm g rdm g

x

y

z

dm

dF dm g

M

origin r

1CMrdm g rdm g rdm Mg r Mg

M

CMr dF r Mg


Torque on a body in a uniform Torque on a body in a uniform gravitation fieldgravitation field

x

y

z

dm

dF

M

origin r

r dF

CMr

x

y

zMg

Morigin

CMr

CMr Mg

Equivalent Torques

physics 2211: lecture 36

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