Two-Dimensional Rotational Dynamics 8.01 W10D2 Fall, 2006.

Post on 17-Jan-2016

216 views 0 download

Transcript of Two-Dimensional Rotational Dynamics 8.01 W10D2 Fall, 2006.

Two-Dimensional Rotational Dynamics

8.01

W10D2 Fall, 2006

Main Idea: Fixed Axis Rotation of Rigid Body

Torque produces angular acceleration about center of mass

is the moment of inertial about the center of mass

is the angular acceleration about center of mass

I

cm

cmtotal I

cm

cm

cm

Torque as a Vector Force exerted at a point P on a rigid body.

Vector from a point S to the point P.

S S ,P P r F

PF

S ,Pr

P

PF

,rS P

S

Torque about point S due to the force exerted at point P:

Summary: Cross ProductMagnitude: equal to the area of the parallelogram defined by the two vectors

Direction: determined by the Right-Hand-Rule

sin sin sin (0 ) A B A B A B A B

Properties of Cross Products

( )

( )

c c c

A B B A

A B A B A B

A B C A C B C

Cross Product of Unit Vectors

• Unit vectors in Cartesian coordinates ˆ ˆ ˆ ˆ| || | sin 2 1

ˆ ˆ ˆ ˆ| || | sin(0) 0

i j i j

i i i j

ˆ ˆ ˆ ˆˆ

ˆ ˆ ˆ ˆˆ

ˆ ˆˆ ˆ ˆ

i j k i i 0

j k i j j 0

k i j k k 0

Components of Cross Product

x y z x y zˆ ˆ ˆ ˆˆ ˆA A A , B B B A i j k B i j k

ˆ ˆ ˆ( ) ( ) ( )

ˆ ˆ ˆ

y z z y z x x z x y y x

x y z

x y z

A B A B A B A B A B A B

A A A

B B B

A B i j k

i j k

Concept Question: Cross Product

Consider two vectors with x > 0 and

with Fx > 0 and Fz > 0 . The cross product

points in the

1) + x-direction2) -x-direction3) +y-direction4) -y-direction5) +z-direction6) -z-direction7) None of the above directions

x̂r i

r F x z

ˆ ˆF F F i k

Recall from Last Class: Rotational Kinematics

• Individual element of mass

• Radius of orbit

• Tangential velocity

• Tangential acceleration

• Radial Acceleration

mi

v

tan,ir,i

a

tan,ir,i

arad,i

v

tan,i2

r,i

r,i 2

r,i

Dynamics: Newton’s Second Law and Torque about S

Tangential force on mass element produces torque

Newton’s Second Law

Torque about S

z-component of torque about S

tan, tan, tan,ˆ ˆ

i i i iF m a F

F

tan,im

ir,i

(

z ,S)

ir,i

Ftan,i

mi(r,i

)2

, , tan,S i i i r F

Moment of Inertia and Torque

Component of the total torque about an axis passing through S is the sum over all elements

Recall: Moment of Inertia about and axis passing through S :

Summary:

z ,Stotal (

z ,S)

1 (

z ,S)

2 (

z ,S)

ii1

iN

mi(r,i

)2i1

iN

I

S m

i(r,i

)2

i1

iN

z ,Stotal I

S

Concept Question: Chrome Inertial Wheel

A fixed torque rotates is applied to the shaft of the chrome inertial wheel. If the four weights on the arms of the are slid out, the component of the angular acceleration along the shaft direction will

1) increase.

2) decrease.

3) remain the same.

Demo: Chrome Inertial Wheel

For a fixed torque: vary the moment of inertia will vary the angular acceleration.

z ,Stotal I

S

Demo: Moment of Inertia Wheel

Measuring the moment of inertia.

Analysis: Measuring Moment of Inertia

Free body force diagrams and force equations:

Rotational equation:

Constraint:

Solve for moment of inertia:

Time to travel distance s:

F T mpg 0

mg T maRdT Icm

a Rd

Icm mRd2 (

g

a 1)

a

2s

t2 Icm mRd

2 (gt 2

2s 1)

Properties of Wheel

Radius of disc:

Mass of disc:

Mass of weight holder:

Theoretical result:

md 5.223 kg

m 0.150 kg

Rd 0.50 m

I

cm

1

2m

dR

d2

Group Problem: Atwood’s Machine

A pulley of mass mp, radius R, and moment of inertia Icm about the center of mass, is suspended from a ceiling. An inextensible string of negligible mass is wrapped around the pulley and attached on one end to an object of mass m1 and on the other end to an object of mass m2 , with

m1 > m2 . At time t = 0, the objects are released from rest. Find the magnitude of the acceleration of the objects.

Analysis: Atwood’s Machine

Free body force diagrams and force equations:

T3 T1 T2 mpg 0 m1g T1 m1a T2 m2g m2a

Analysis: Atwood’s Machine

Rotational motion equation:

Force equations:

Constraint condition:

Acceleration:

Time to travel distance d:

m1g T1 m1a

T2 m2g m2a

cm, ztotal I

cm

z

R T1 T2 Icm z

a Rz

a m1 m2 g

(Icm / R2 ) m1 m2

2cm 1 2

1 2

( / )2 d I R m mdt

a m m g

Experiment 05: Moment of Inertia

Apparatus Connect output of

phototransistor to channel A of 750.

Connect output of tachometer generator to channel B of 750.

Connect power supply.

Red button is pressed: Power is applied to motor.

Red button is released: Rotor coasts: Read output voltage using LabVIEW program.

Use black sticker or tape on white plastic rotor for generator calibration.

22

Calibrate Tachometer-generator

Spin motor up to full speed, let it coast. Measure and plot voltages for 0.25 s period. Sample Rate: 5000 Hz.

Count rotation periods to measure ω.

Program calculates average output voltage, angular velocity, and the calibration factor angular velocity per volt periods.

Rotor Moment of Inertia

Plot only the generator voltage for rest of experiment.

Use a 55 gm weight to accelerate the rotor.

Analysis: moment of inertia

Force and rotational equations while weight is descending:

Constraint:

Rotational equation while slowing down

Solve for moment of inertia:

mg T ma1

rT f IR1

a1r

1

Speeding up Slowing down f IR2

rm(g r1) IR2 IR1

IR rm(g r1)

(1 2 )

Understand graph output to measure IR

The angular frequency along line A-B is increasing because

1. the weight has hit the floor and is tension in the string is no longer applying a torque.

2. the weight is descending and the tension in the string is applying a torque.

3. for some other reason.

Understand graph output to measure IR

The slope of the line B-C is equal to

1. the angular acceleration after the weight has hit the floor.

2. angular acceleration before the weight has hit the floor.

3. Neither of the above.

Measure IR: results

Measure and record α1 and α2.

For your report, calculate IR:

IR rm(g r1)

(1 2 )