phy1012f rotation ii

CONSERVATION LAWS

PHY1012FROTATION

II

Gregor Leighgregor.leigh@uct.ac.za

CONSERVATION LAWS ROTATIONAL ENERGYPHY1012F

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ROTATIONAL ENERGYLearning outcomes:

At the end of this chapter you should be able to…Use the laws of conservation of mechanical energy and angular momentum to solve rotational problems, including those involving rolling motion.Use vector mathematics to describe and solve problems involving rotational problems.

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ROTATIONAL ENERGYEach particle in a rigid rotating body has kinetic energy.

axle

The sum of all the individual kinetic energies of each of the particles is the rotational kinetic energy of the body:

Krot = ½ m1v12 + ½ m2v2

2 + …

Krot = ½ m1r122 + ½ m2r2

22 + …

Krot = ½ (m1r12)2

Krot = ½ I2

m1

m2 m3r3r2

r1

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K U Emech = K + U

Emech = 0

CONSERVATION OF ENERGYAs usual, energy is conserved (in frictionless systems).If, however, a horizontal axis of rotation does not coincide with the centre of mass, the object’s potential energy will vary.So we write: Emech = Krot + Ug

= ½ I2 + MgyCM

axle

NEWTON’S LAWS ROTATION OF A RIGID BODYPHY1012F

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f2 = i

2 + 2

f2 = 0 + 2(–15)(–0.5 )

f 15 6.8 rad/s

vt = r

vt = –6.8 1 = 6.8 m/s

ROTATION ABOUT A FIXED AXISA 70 g metre stick pivoted freely at one end is released from a horizontal position. At what speed does the far end swing through its lowest position?

7. Use rotational kinematics to find angular positions and velocities.

pivot

= –15 rad/s2

i = 0 rad i = 0 rad/s

f = –0.5 rad f = ?

You canNOT use rotational

kinematics to solve this problem!

Why not?

(Not this time!)

(Not constant)

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A 70 g metre stick pivoted freely at one end is released from a horizontal position. At what speed does the far end swing through its lowest position?

M = 0.07 kgyCM = 0 m0 = 0 rad/s

pivotx

y

Before:

L = 1 m

yCM = –0.5 m1 = ?vtip = ?

After:½ I0

2 + MgyCM 0 = ½ I12 + MgyCM 1

21 1 12 23

210 ML Mg L

13gL

1tip 3 5.42 m/sL gv L

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ROLLING MOTIONRolling is a combination of rotation and translation. (We shall consider only objects which roll without slipping.)As a wheel (or sphere) rolls along a flat surface…

each point on the rim describes a cycloid;the axle (the centre of mass) moves in a straight line, covering a distance of 2R each revolution;the speed of the wheel is given by

CM2 Rv RT

2R

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FUNNY THING ABOUT THE CYCLOID…If a farmer’s road surface were rutted into a cycloid form, the smoothest way to get his sheep to market would be to use a truck with…

SQUARE wheels!

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ROLLING MOTIONThe velocity of a particle on a wheel consists of two parts: CM , reli iv v v

So the point, P, at the bottom of an object which rolls (without slipping) is instantaneously at rest…

+ =

TRANSLATION

P

vCM

vCM

vCM

R

0

–R

v = R

v = 2vCM = 2R

v = 0

+ ROTATION = ROLLING

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KINETIC ENERGY OF A ROLLING OBJECTIf we regard P as an instantaneous axis of rotation, the object’s motion simplifies to one of pure rotation, and thus its kinetic energy is given by:

K = Krot about P = ½ IP2 v = R

v = 2R

v = 0

P

Using the parallel axis theorem,IP = (ICM + MR2)

K = ½ ICM2 + ½ M(R)2

K = ½ ICM2 + ½ M(vCM)2

I.e. K = Krot + KCM

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CMvR

21CM2 1M c v Mgh

THE GREAT DOWNHILL RACE

½ ICM2 + ½ M(vCM)2 = Mgh

Kf = Ui

ICM = cMR2 and

2

22 CM1 1CM2 2

vcMR M v MghR

CM21

ghvc

I.e. The actual values of M and R do not feature, but where the mass is situated is of critical importance.

h

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THE GREAT DOWNHILL RACE

vCM2 = 0 + 2ax

where

2 2

CM 1CM 2

sin2

ghc

hv

a x

I.e. The acceleration of a rolling body is less than that of a particle by a factor which depends on the body’s moment of inertia.

h x

sinhx

CM21

ghvc

particleCM 1

aa c

sin1

gc

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VECTOR DESCRIPTION OF ROTATIONAL MOTION

Using only “clockwise” and “counterclockwise” is the rotational analogue of using “backwards” and “forwards” in rectilinear kinematics. A more general handling of rotational motion requires vector quantities.The vector associated with a rotational quantity…

has magnitude equal to the magnitude of that quantity;has direction as given by the right-hand rule.

E.g. The angular velocity vector, , of this anticlockwise-turning disc points…

in the positive z-direction.

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ˆ ˆ ˆi j k

THE CROSS PRODUCT

Once again, the quantity rF sin is the product of two vectors, and , at an angle to each other. This time, however, we use the orthogonal components to determine the cross product of the vectors: .

The magnitude of the torque exerted by force

applied at displacement from the turning point is: rFsin

F

r

F

r

r F

y

x

1

1

ij

ˆ ˆ ˆ ˆ ˆ ˆi i = j j = k k = 0

ˆ ˆ ˆˆ ˆj k i k j

ˆ ˆ ˆˆ ˆk i j i k 1z

k

In RH system: ˆ ˆj i

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THE CROSS PRODUCT

The more orthogonal the vectors, the greater the cross product; the more parallel, the smaller…

Notes:

sinr F rF

F

r

Since it is a vector quantity, the cross product is also known as the vector product. .Derivative of a cross product:

dpd drr p p rdt dt dt

A B C A B A C

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ANGULAR MOMENTUMWe have shown that in circular motion (where vt and r are perpendicular) a particle has angular momentum L = mrvt.mvt = p

L = rp

z

p mv

r

pMore generally (allowing for

and to be at any angle )…

pr

L r p

= (mrv sin, direction from RH rule) dpdL d drr p p rdt dt dt dt

netv p r F

I.e. netdLdt (Cf. in linear motion:

)net

dpF dt

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Linear Rotational

ROTATIONAL MOMENTUM & ENERGYSummary of corresponding quantities and relationships:

(around an axis of symmetry)

Linear momentum, , is con-served if there is no net force

KCM = ½ MvCM2 Krot = ½ I2

CMP Mv

L I

netdLdt

netdPF dt

Angular momentum, , is con-served if there is no net torque

P

L

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ROTATIONAL ENERGYLearning outcomes:

At the end of this chapter you should be able to…Use the laws of conservation of mechanical energy and angular momentum to solve rotational problems, including those involving rolling motion.Use vector mathematics to describe and solve problems involving rotational problems.