lecture 7:rotational motion

Lecture 7:Rotational motion

Physics 1, NTC

Angular Motion, General Notes•When a rigid object rotates about a fixed axis in a given time interval, every portion on the object rotates through the same angle in a given time interval and has the same angular speed and the same angular acceleration.

• So , , q w a all characterize the motion of the entire rigid object as well as the individual particles in the object.

Section 10.1

Angular Position

•As the particle moves, the only coordinate that changes is q•As the particle moves through q, it moves though an arc length s.•The arc length and r are related:

• s = q r

Section 10.1

Conversions

•Comparing degrees and radians

•Converting from degrees to radians

3601 57.3

2rad

180

rad degrees

Section 10.1

Directions

•Strictly speaking, the speed and acceleration (w, a) are the magnitudes of the velocity and acceleration vectors.•The directions are actually given by the right-hand rule.

Section 10.1

Relationship Between Angular and Linear Quantities

•Every point on the rotating object has the same angular motion.•Every point on the rotating object does not have the same linear motion.•Displacements

• s = θ r

•Speeds• v = ω r

•Accelerations• a = α r

Section 10.3

ds dv r r

dt dt

Acceleration Comparison •The tangential acceleration is the derivative of the tangential velocity.

•Centripetal acceleration

t

dv da r r

dt dt

Section 10.3

22

C

va r

r

Rotational Kinetic Energy

•The total rotational kinetic energy of the rigid object is the sum of the energies of all its particles.

•I is called the moment of inertia.

•Rotational kinetic energy is not a new type of energy, the form is different because it is applied to a rotating object.

•The units of rotational kinetic energy are Joules (J).

2 2

2 2 2

1

2

1 1

2 2

R i i ii i

R i ii

K K m r

K m r I

Section 10.4

Moment of Inertia•Defined by

dimensions = ML2 or unit of kg.m2.

•Mass = inherent property, but the moment of inertia depends on the choice of rotational axis.•Moment of inertia is a measure of the resistance of an object to changes in its rotational motion, similar to mass being a measure of an object’s resistance to changes in its translational motion.

• The moment of inertia depends on the mass and how the mass is distributed around the rotational axis.

2i i

i

I r m

Section 10.5

Moment of Inertia, cont•For a continuous rigid object, imagine the object to be divided into many small elements, each having a mass of Δmi.

•If r is constant, the integral can be evaluated with known geometry, otherwise its variation with position must be known.

lim 2 20im i ii

I r m r dm 2I r dV

Section 10.5

Moments of Inertia of Various Rigid Objects

Section 10.5

Moment of Inertia of a Uniform Rigid Rod•The shaded area has a mass

• dm = l dx•Then the moment of inertia is

/ 22 2

/ 2

21

12

L

y L

MI r dm x dx

L

I ML

Section 10.5

Moment of Inertia of a Uniform Solid Cylinder•Divide the cylinder into concentric shells with radius r, thickness dr and length L.

•dm = r dV = 2 (p rLr) dr

•Then for I

z

z

I r dm r L r dr

I MR

2 2

2

2

1

2

Section 10.5

9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis

Example 9 The Moment of Inertia Depends on Wherethe Axis Is.

Two particles each have mass and are fixed at theends of a thin rigid rod. The length of the rod is L.Find the moment of inertia when this object rotates relative to an axis that is perpendicular to the rod at(a) one end and (b) the center.

(a) 22222

211

2 0 LmmrmrmmrI

Lrr 21 0mmm 21

2mLI

9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis

(b) 22222

211

2 22 LmLmrmrmmrI

22 21 LrLr mmm 21

221 mLI

Parallel-Axis Theorem•In the previous examples, the axis of rotation coincided with the axis of symmetry of the object.•For an arbitrary axis, the parallel-axis theorem often simplifies calculations.•The theorem states I = ICM + MD 2

• I is about any axis parallel to the axis through the center of mass of the object.

• ICM is about the axis through the center of mass.• D is the distance from the center of mass axis to the arbitrary axis.

Section 10.5

Moment of Inertia for a Rod Rotating Around One End – Parallel Axis Theorem Example

•The moment of inertia of the rod about its center is

•D is ½ L•Therefore,

21

12CMI ML

2CM

22 21 1

12 2 3

I I MD

LI ML M ML

Section 10.5

Torque•Torque, t, is the tendency of a force to rotate an object about some axis.

• Torque is a vector, but we will deal with its magnitude here:

•t = r F sin f = F d • F is the force• f is the angle the force makes with the horizontal • d is the moment arm (or lever arm) of the force

• There is no unique value of the torque on an object.• Its value depends on the choice of a rotational axis.

Section 10.6

Torque is a vector!•The moment arm, d, is the perpendicular distance from the axis of rotation to a line drawn along the direction of the force.

• d = r sin Φ•The horizontal component of the force (F cos f) has no tendency to produce a rotation.•Torque will have direction.

• If the turning tendency of the force is counterclockwise, the torque will be positive.

• If the turning tendency is clockwise, the torque will be negative.

Section 10.6

Net Torque

•The force will tend to cause a counterclockwise rotation about O.•The force will tend to cause a clockwise rotation about O.•St = t1 + t2 = F1d1 – F2d2

2F

1F

Section 10.6

Torque and Angular Acceleration, Extended•Consider the object consists of an infinite number of mass elements dm of infinitesimal size.

•Each mass element rotates in a circle about the origin, O.

•Each mass element has a tangential acceleration.

•From Newton’s Second Law

• dFt = (dm) at•The torque associated with the force and using the angular acceleration gives

• dt ext = r dFt = atr dm = ar 2 dm

• This becomes S t = IaSection 10.7

Torque and Angular Acceleration, Extended cont.

•rigid body under a net torque.•The result also applies when the forces have radial components.

• The line of action of the radial component must pass through the axis of rotation.

• These components will produce zero torque about the axis.

Section 10.7

Falling Smokestack Example

•When a tall smokestack falls over, it often breaks somewhere along its length before it hits the ground.•Each higher portion of the smokestack has a larger tangential acceleration than the points below it.•The shear force due to the tangential acceleration is greater than the smokestack can withstand.•The smokestack breaks.

Section 10.7

9.3 Center of Gravity

Conceptual Example 7 Overloading a Cargo Plane

This accident occurred because the plane was overloaded towardthe rear. How did a shift in the center of gravity of the plane cause the accident?

Summary of Useful Equations

Section 10.8

9.2 Rigid Objects in Equilibrium

Example 3 A Diving Board

A woman whose weight is 530 N is poised at the right end of a diving boardwith length 3.90 m. The board hasnegligible weight and is supported bya fulcrum 1.40 m away from the leftend.

Find the forces that the bolt and the fulcrum exert on the board.

9.2 Rigid Objects in Equilibrium

022 WWF

N 1480

m 1.40

m 90.3N 5302 F

22

WWF

9.2 Rigid Objects in Equilibrium

021 WFFFy

0N 530N 14801 F

N 9501 F

9.2 Rigid Objects in Equilibrium

Example 5 Bodybuilding

The arm is horizontal and weighs 31.0 N. The deltoid muscle can supply1840 N of force. What is the weight of the heaviest dumbell he can hold?

9.2 Rigid Objects in Equilibrium

0 Mddaa MWW

0.13sinm 150.0M

9.2 Rigid Objects in Equilibrium

N 1.86

m 620.0

0.13sinm 150.0N 1840m 280.0N 0.31

d

Maad

MWW

9.3 Center of Gravity

DEFINITION OF CENTER OF GRAVITY

The center of gravity of a rigid body is the point at whichits weight can be considered to act when the torque dueto the weight is being calculated.

9.3 Center of Gravity

When an object has a symmetrical shape and its weight is distributed uniformly, the center of gravity lies at its geometrical center.

9.3 Center of Gravity

21

2211

WW

xWxWxcg

9.3 Center of Gravity

Example 6 The Center of Gravity of an Arm

The horizontal arm is composedof three parts: the upper arm (17 N),the lower arm (11 N), and the hand (4.2 N).

Find the center of gravity of thearm relative to the shoulder joint.

9.3 Center of Gravity

21

2211

WW

xWxWxcg

m 28.0

N 2.4N 11N 17

m 61.0N 2.4m 38.0N 11m 13.0N 17

cgx

Energy in an Atwood Machine, Example•The system containing the two blocks, the pulley, and the Earth is an isolated system in terms of energy with no non-conservative forces acting.

•The mechanical energy of the system is conserved.

•The blocks undergo changes in translational kinetic energy and gravitational potential energy.

•The pulley undergoes a change in rotational kinetic energy.

•Find ?

Section 10.8

Rolling Object

•The red curve shows the path moved by a point on the rim of the object.• This path is called a cycloid.

•The green line shows the path of the center of mass of the object.•In pure rolling motion, an object rolls without slipping.•In such a case, there is a simple relationship between its rotational and translational motions.

Section 10.9

rv

The tangential speed of apoint on the outer edge ofthe tire is equal to the speedof the car over the ground.

ra

Pure Rolling Motion, Object’s Center of Mass•The translational speed of the center of mass is

•The linear acceleration of the center of mass is

CM

ds dv R R

dt dt

CMCMdv d

a R Rdt dt

Section 10.9

Rolling Motion Cont.•Rolling motion can be modeled as a combination of pure translational motion and pure rotational motion.•The contact point between the surface and the cylinder has a translational speed of zero (c).

Section 10.9

Total Kinetic Energy of a Rolling Object•The total kinetic energy of a rolling object is the sum of the translational energy of its center of mass and the rotational kinetic energy about its center of mass.

K = ½ ICM w2 + ½ MvCM2

• The ½ ICMw2 represents the rotational kinetic energy of the cylinder about its center of mass.

• The ½ Mv2 represents the translational kinetic energy of the cylinder about its center of mass.

Section 10.9

Total Kinetic Energy, Example•Accelerated rolling motion is possible only if friction is present between the sphere and the incline.

• The friction produces the net torque required for rotation.

• No loss of mechanical energy occurs because the contact point is at rest relative to the surface at any instant.

• In reality, rolling friction causes mechanical energy to transform to internal energy.

• Rolling friction is due to deformations of the surface and the rolling object.

Section 10.9

Find and at the bottom of the incline?

9.5 Rotational Work and Energy

Example 13 Rolling Cylinders

A thin-walled hollow cylinder (mass = mh, radius = rh) anda solid cylinder (mass = ms, radius = rs) start from rest atthe top of an incline.

Determine which cylinder has the greatest translationalspeed upon reaching the bottom.

9.5 Rotational Work and Energy

mghImvE 2212

21

iiifff mghImvmghImv 2212

212

212

21

iff mghImv 2212

21

ENERGY CONSERVATION

rv ff

9.5 Rotational Work and Energy

iff mghrvImv 22212

21

2

2

rIm

mghv of

The cylinder with the smaller momentof inertia will have a greater final translationalspeed.

Lecture 8: Angular Momentum

Vector Product (cross product)

•The quantity AB sin q is equal to the area of the parallelogram formed by •The direction of is perpendicular to the plane formed by .

•The best way to determine this direction is to use the right-hand rule.

Section 11.1

Properties of the Vector Product•The vector product is not commutative. The order in which the vectors are multiplied is important.

• If A // B (parallel q = 0o or 180o), then• Therefore

•If A is perpendicular to B, then •The vector product obeys the distributive law.

0 A A

A B B A

0 A B

Section 11.1

AB A B

x ( + ) = x + x A B C A B A C

Final Properties of the Vector Product•The derivative of the cross product with respect to some variable such as t is

where it is important to preserve the multiplicative order of the vectors.

d d d

dt dt dt

A BA B B A

Section 11.1

Vector Products of Unit Vectors

ˆ ˆ ˆ ˆ ˆ ˆ 0

ˆ ˆ ˆ ˆ ˆ

ˆ ˆ ˆ ˆ ˆ

ˆ ˆ ˆ ˆ ˆ

i i j j k k

i j j i k

j k k j i

k i i k j

Section 11.1

Signs in Cross Products •Signs are interchangeable in cross products

•

• and

jiji ˆˆˆˆ

- A B A B

Section 11.1

Using Determinants•The cross product can be expressed as

•Expanding the determinants gives

y z x yx zx y z

y z x yx zx y z

A A A AA AA A A

B B B BB BB B B

ˆ ˆ ˆ

ˆ ˆ ˆ i j k

A B i j k

y z z y x z z x x y y xA B A B A B A B A B A Bˆ ˆ ˆ A B i j k

Section 11.1

Vector Product Example•Given•Find•Result

ˆ ˆ ˆ ˆ2 3 ; 2 A i j B i j

A B

ˆ ˆ ˆ ˆ(2 3 ) ( 2 )

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ2 ( ) 2 2 3 ( ) 3 2

ˆ ˆ ˆ0 4 3 0 7

A B i j i j

i i i j j i j j

k k k

Section 11.1

Torque Vector Example•Given the force and location

•Find the torque produced

ˆ ˆ(2.00 3.00 )

ˆ ˆ(4.00 5.00 )

N

m

F i j

r i j

ˆ ˆ ˆ ˆ [(4.00 5.00 )N] [(2.00 3.00 )m]

ˆ ˆ ˆ ˆ[(4.00)(2.00) (4.00)(3.00)

ˆ ˆ ˆ ˆ(5.00)(2.00) (5.00)(3.00)

ˆ2.0 N m

r F i j i j

i i i j

j i i j

k

Section 11.1

Angular Momentum•Consider a particle of mass m located at the vector position r and moving with linear momentum p . •Find the net torque.

•This looks very similar to the equation for the net force in terms of the linear momentum since the torque plays the same role in rotational motion that force plays in translational motion.

Add the term sinceit 0

( )

d

dtd

dtd

dt

pr F r

rp

r p

Section 11.2

Angular Momentum, cont

•The instantaneous angular momentum of a particle relative to the origin O is defined as the cross product of the particle’s instantaneous position vector and its instantaneous linear momentum.

L r p

Section 11.2

Torque and Angular Momentum•The torque is related to the angular momentum.

• Similar to the way force is related to linear momentum.

•The torque acting on a particle is equal to the time rate of change of the particle’s angular momentum.•This is the rotational analog of Newton’s Second Law .

• must be measured about the same origin.• This is valid for any origin fixed in an inertial frame.

d

dt L

and L

Section 11.2

Angular Momentum•The SI units of angular momentum are (kg.m2)/ s.•Both the magnitude and direction of the angular momentum depend on the choice of origin.•The magnitude is L = mvr sin f

• f is the angle between and .

•The direction of L is perpendicular to the plane formed by r and p .p

r

Section 11.2

Angular Momentum of a Particle, Example•The vector is pointed out of the diagram.

•The magnitude is L = mvr sin 90o = mvr• sin 90o is used since v is perpendicular to r.

•A particle in uniform circular motion has a constant angular momentum about an axis through the center of its path.

= L r p

Section 11.2

Angular Momentum of a Rotating Rigid Object

•The rigid object is a non-deformable system.

•Each particle of the object rotates in the xy plane about the z axis with an angular speed of w

•The angular momentum of an individual particle is

Li = mi ri2 w

•L and are directed along the z axis.

Section 11.3

9.6 Angular Momentum

DEFINITION OF ANGULAR MOMENTUM

The angular momentum L of a body rotating about a fixed axis is the product of the body’s moment of inertia and its angular velocity with respect to thataxis:

IL

Requirement: The angular speed mustbe expressed in rad/s.

SI Unit of Angular Momentum: kg·m2/s

9.6 Angular Momentum

PRINCIPLE OF CONSERVATION OFANGULAR MOMENTUM

The angular momentum of a system remains constant (is conserved) if the net external torque acting on the system is zero.

9.6 Angular Momentum

Conceptual Example 14 A Spinning Skater

An ice skater is spinning with botharms and a leg outstretched. Shepulls her arms and leg inward andher spinning motion changesdramatically.

Use the principle of conservationof angular momentum to explainhow and why her spinning motionchanges.

9.6 Angular Momentum

Example 15 A Satellite in an Elliptical Orbit

An artificial satellite is placed in an elliptical orbit about the earth. Its pointof closest approach is 8.37x106mfrom the center of the earth, andits point of greatest distance is 25.1x106m from the center ofthe earth.

The speed of the satellite at the perigee is 8450 m/s. Find the speedat the apogee.

9.6 Angular Momentum

IL

angular momentum conservation

PPAA II

rvmrI 2

P

PP

A

AA r

vmr

r

vmr 22

9.6 Angular Momentum

PPAA vrvr

sm2820

m 1025.1

sm8450m 1037.86

6

A

PPA r

vrv

P

PP

A

AA r

vmr

r

vmr 22