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Physics 211: Lecture 18, Pg 1 Physics 211: Lecture 18 Physics 211: Lecture 18 Today’s Agenda Today’s Agenda Angular Motion Angular Motion Rotational kinematics Direction of rotation and the right hand rule Vector cross product A=BXC (not the dot product) Rotational dynamics and torque Work and energy with example

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Physics 211: Lecture 18, Pg 1

Physics 211: Lecture 18Physics 211: Lecture 18

Today’s AgendaToday’s Agenda

Angular MotionAngular Motion

Rotational kinematics

Direction of rotation and the right hand rule Vector cross product A=BXC

(not the dot product)

Rotational dynamics and torque

Work and energy with example

Physics 211: Lecture 18, Pg 2

Comparison of SimpleComparison of SimpleRotational vs. Linear Kinematics Rotational vs. Linear Kinematics

Angular Linear

t0 +=

200 t

21

t ++=

constant= ttanconsa

atvv 0 +=

200 at

21

tvxx ++=

And for a point at a distance R from the rotation axis linear and angular coordinates are related:

dx = dRv = Ra = R

Physics 211: Lecture 18, Pg 3

Example:Example:

A flywheel spins with an initial angular velocity 0 = 500 rad/s. At t = 0 it starts to slow down at a rate of 0.5 rad/s2. How long does it take to stop?

Realize that = - 0.5 rad/s2 (which way does point?)

0 t Use to find when = 0 :

0

0 t

min7.161000/5.0 -

Physics 211: Lecture 18, Pg 4

Rigidly rotating mass distrubutionRigidly rotating mass distrubutionmade to rotate around central axis as shownmade to rotate around central axis as shown

So: but vi = ri

rr1

rr2rr3

rr4

m4

m1

m2

m3

vv4

vv1

vv3

vv2

K m vi ii

12

2

K m r m ri ii

i ii

12

12

2 2 2

which we write as:

K 12

2I

I m ri ii

2

Define the moment of inertiamoment of inertiaabout the rotation axis I has units of kg m2.

Physics 211: Lecture 18, Pg 5

Moment of inertia for highly symmetric mass Moment of inertia for highly symmetric mass distributions - distributions - through the CM (then apply PATthrough the CM (then apply PAT))

Sphere Solid Hollow Shell

Square Two orientations rectangles

Cylinder (around axis, perpendicular to axis, solid, hollow) Hockey puck (cylinder that’s short)

Stick (center, end) Football Deck of cards

If we know ICM , it is easy to calculate the moment of inertia about a parallel axis….

Tipler has summary of moments of inertia for “symmetric” objects like these. Table 9.1 p 274

Physics 211: Lecture 18, Pg 6

Parallel Axis TheoremParallel Axis Theorem

Suppose the moment of inertia of a solid object of mass M about an axis through the center of mass, ICM, is known.

The moment of inertia about an axis parallel to this axis but a distance D away is given by:

IPARALLEL = ICM + MD2

So if we know ICM , it is easy to calculate the moment of inertia about a parallel axis.

And 21

2KE I

Physics 211: Lecture 18, Pg 7

Connection with CM motion...Connection with CM motion...

So for a solid object which rotates about its center or mass and whose CM is moving:

2CM

2CMNET MV

21

21K I

VCM

We will use this formula more in coming lectures.

Physics 211: Lecture 18, Pg 8

Rolling MotionRolling Motion

In rolling systems, CM motion and rolling motion are connected Consider objects of different I rolling down an inclined plane:

h

v = 0 = 0 K = 0

RK = - U = Mgh

22 Mv21

I21

K +=

v = R

M

Roll objects

down ramp

Physics 211: Lecture 18, Pg 9

Rolling...Rolling...

If there is no slipping:

v 2v

In the lab reference frame

v

In the instantaneous CM reference frame all the KE is rotationalIn lab frame add trans KE

v

Where v = R

Physics 211: Lecture 18, Pg 10

Rolling...Rolling...

22 Mv21

I21

K += Use v = R and I = cMR2 .

The rolling speed is always lower than in the case of simple slidingsince the kinetic energy is shared between CM motion and rotation.

We will study rolling more in the next lecture!(the theory of bowling)

hoop: c = 1

disk: c = 1/2

sphere: c = 2/5

etc...( ) 2222 Mv1

21

Mv21

MR21

K +=+= c c

So: ( ) MghMv121 2 =+ 1

1gh2v

+=c c

Physics 211: Lecture 18, Pg 11

Direction of Rotation:Direction of Rotation: In general, the rotation variables are vectors (have direction and magnitude) If the plane of rotation is in the x-y plane, then the convention is

CCW rotation is in the + z direction

CW rotation is in the - z direction

x

y

z

x

y

z

This angular velocity vector doesn’t point to a particular point in space

It describes the magnitude and direction of the angular velocity an object has

Physics 211: Lecture 18, Pg 12

Direction of Rotation:Direction of Rotation:The Right Hand RuleThe Right Hand Rule

To figure out in which direction the rotation vector points, curl the fingers of your right hand the same way the object turns, and your thumb will point in the direction of the rotation vector!

We often pick the z-axis to be the rotation axis as shown. = z

= z

= z

For simplicity we omit the subscripts unless explicitly needed.

x

y

z

x

y

z

Physics 211: Lecture 18, Pg 13

Example:Example:

A flywheel spins with an initial angular velocity 0 = 500 rad/s. At t = 0 it starts to slow down at a rate of 0.5 rad/s2. How long does it take to stop?

Realize that = - 0.5 rad/s2 (which way does point?)

0 t Use to find when = 0 :

t

0

min7.161000/5.0/500

Physics 211: Lecture 18, Pg 14

Lecture 18, Lecture 18, Act 1Act 1RotationsRotations

A ball rolls across the floor, and then starts up a ramp as shown below. In what direction does the angular acceleration vector point when the ball is rolling up the ramp?

(a)(a) down the ramp (b)(b) into the page

(c)(c) out of the page

Physics 211: Lecture 18, Pg 15

Lecture 18, Lecture 18, Act 1Act 1SolutionSolution

When the ball is on the ramp, the linear acceleration a is always down the ramp (gravity).

a

The angular acceleration is therefore counter-clockwise.

Using your right hand rule, is out of the page!

Physics 211: Lecture 18, Pg 16

Rotational Dynamics:Rotational Dynamics:What makes it spin?What makes it spin?

Suppose a force acts on a mass constrained to move in a circle. Consider its acceleration in the direction at some instant: a = r

Now use Newton’s 2nd Law in the direction: F = ma = mr

rF = mr2

^

r

aa

FF

m

rr

F

Multiply by r :

Physics 211: Lecture 18, Pg 17

Rotational Dynamics:Rotational Dynamics:What makes it spin?What makes it spin?

rF = mr2use

Define torque: = rF. is the tangential force F

times the lever arm r.

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

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

spin CW.

I=

I=

2mr=I

r

aa

FF

m

rr

F

Physics 211: Lecture 18, Pg 18

Rotational Dynamics:Rotational Dynamics:What makes it spin?What makes it spin?

So for a collection of many particles arranged in a rigid configuration:

rr1

rr2rr3

rr4

m4

m1

m2

m3

FF4

FF1

FF3

FF2

ii

2ii

iii rmFr

,

i I

ii I

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

INET

The componentof Fi

Physics 211: Lecture 18, Pg 19

Rotational Dynamics:Rotational Dynamics:What makes something spin?What makes something spin?

NET = I

This is the rotational analogue of FNET = ma

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

Moment of inertiaMoment of inertia II is the rotational analogue of mass.is the rotational analogue of mass. 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 = Nm.

torque does

Physics 211: Lecture 18, Pg 20

TorqueTorque

= r F sin = r sin F

= rpF

Equivalent definitions

rp = “distance of closest approach” if the force is constant

= rF

Recall the definition of torque:

rr

rp

FF

F

Fr

Physics 211: Lecture 18, Pg 21

TorqueTorquehow does this depend on how F points?how does this depend on how F points?

= r Fsin

So if = 0o, then = 0

And if = 90o, then = maximum

rr

FF

rrFF

Physics 211: Lecture 18, Pg 22

Lecture 18, Lecture 18, Act 2Act 2TorqueTorque

In which of the cases shown below is the torque provided by the applied force about the rotation axis biggest? In both cases the magnitude and direction of the applied force is the same.

(a)(a) case 1 (b)(b) case 2

(c)(c) same L

L

F F

axis

case 1 case 2

Physics 211: Lecture 18, Pg 23

Lecture 18, Lecture 18, Act 2Act 2SolutionSolution

Torque = F x (distance of closest approach)

LF F

case 1 case 2

L

Torque is the same!

The applied force is the same. The distance of closest approach is the same.

Physics 211: Lecture 18, Pg 24

Torque and the Torque and the Right Hand Rule:Right 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.

r r

FF

x

y

z

Physics 211: Lecture 18, Pg 25

The Cross ProductThe Cross Product

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:

AA X BB = CC

The length of CC is given by: C = AB sin

The direction of CC is perpendicular to the plane defined by AA and BB, and inthe direction defined by the right handrule.

AA

BB

CC

Physics 211: Lecture 18, Pg 26

The Cross ProductThe Cross Product

Cartesian components of the cross product:

C C = AA X BB

CX = AY BZ - BY AZ

CY = AZ BX - BZ AX

CZ = AX BY - BX AY

AA

BB

CC

Note: B X A = - A X B

Physics 211: Lecture 18, Pg 27

Torque & the Cross Product:Torque & the Cross Product:

rr

FF

x

y

z

So we can express the torque as the vector product:

= rr X FF = rF sin

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

look at this in particular…

Physics 211: Lecture 18, Pg 28

Comment onComment on == II

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

z = Izz

We usually omit the z subscript for simplicity.

z

z

z

Iz

Physics 211: Lecture 18, Pg 29

ExampleExample

To loosen a stuck nut, a (stupid) man pulls at an angle of 45o on the end of a 50 cm wrench with a force of 200 N. What is the magnitude of the torque on the nut? If the nut suddenly turns freely, what is the angular

acceleration of the wrench? (The wrenchhas a mass of 3 kg, and its shape is that of a thin rod).

L = 0.5 m

F = 200 N

45o

Physics 211: Lecture 18, Pg 30

ExampleExample

L = 0.5m

F = 200 N45o

Torque = LFsin = (0.5 m)(200 N)(sin 45) = 70.7 Nm

Wrench w/ bolts

If the nut turns freely, = I We know and we want , so we need to figure out I.

222 kgm250m50kg331ML

31 ..I

So = / I = (70.7 Nm) / (0.25 kgm2)

Physics 211: Lecture 18, Pg 31

WorkWork

Consider the work done by a force FF acting on an object constrained to move around a fixed axis. For an infinitesimal angular displacement d:

dW = FF.drdr = FR dcos() = FR dcos(90-) = FR dsin()

= FR sin() ddW = d

If is small we can integrate and find: W = Analogue of W = F •r W will be negative if and have opposite signs!

R

FF

dr = R ddaxis

Physics 211: Lecture 18, Pg 32

Work & Kinetic Energy:Work & Kinetic Energy:

Recall the Work/Kinetic Energy Theorem: K = WNET

This is true in general, and hence applies to rotational motion as well as linear motion.

So for an object that rotates about a fixed axis:

NETif WK 22I21

Physics 211: Lecture 18, Pg 33

Example: Disk & StringExample: Disk & String

A massless string is wrapped 10 times around a disk of mass M = 40 g and radius R = 10 cm. The disk is constrained to rotate without friction about a fixed axis though its center. The string is pulled with a force F = 10 N until it has unwound. (Assume the string does not slip, and that the disk is initially not spinning).

How fast is the disk spinning after the string has unwound?

F

R M

Physics 211: Lecture 18, Pg 34

Disk & String...Disk & String...

The work done is W = The torque is = RF (since = 90o) The angular displacement is

F

R M

So W = (.1 m)(10 N)(20rad) = 62.8 J

Physics 211: Lecture 18, Pg 35

Disk & String...Disk & String...

WNET = W = 62.8 J = K 12

2I

Recall thatIfor a disk aboutits central axis is given by:

I 12

2MR

K MR W

12

12

2 2So

4 4 62 804 12 2

WMR

Jkg

.. .

R M

Flywheel, pulley,

& mass

Physics 211: Lecture 18, Pg 36

Lecture 18, Lecture 18, Act 3Act 3Work & EnergyWork & Energy

Strings are wrapped around the circumference of two solid disks and pulled with identical forces for the same distance. Disk 1 has a bigger radius, but both have the same moment of inertia. Both disks rotate freely around axes though their centers, and start at rest.

Which disk has the biggest angular velocity after the pull ?

(a)(a) disk 1 (b)(b) disk 2

(c)(c) same FF

1 2

Physics 211: Lecture 18, Pg 37

Lecture 18, Lecture 18, Act 3Act 3SolutionSolution

FF

1 2

d

The work done on both disks is the same!W = Fd

The change in kinetic energy of each will therefore also be the same since W = K.

But we know K 12

2I

So since I1 = I2

1 = 2

Physics 211: Lecture 18, Pg 38

Spinning Disk Demo:Spinning Disk Demo:

We can test this with our big flywheel.

22 mv21

21KW I

I

m

negligiblein this case

2WI

In this case, I = 1 kg - m2

W = mgh = (2 kg)(9.81 m/s2)(1 m) = 19.6 J

= 6.26 rad/s ~ 1 rev/s

Physics 211: Lecture 18, Pg 39

Recap of today’s lectureRecap of today’s lecture