PHY1012F ROTATION II

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PHY1012F ROTATION II. Gregor Leigh gregor.leigh@uct.ac.za. ROTATIONAL ENERGY. Use the laws of conservation of mechanical energy and angular momentum to solve rotational problems, including those involving rolling motion.

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### PHY1012FROTATION II

Gregor Leighgregor.leigh@uct.ac.za

ROTATIONAL ENERGY

### ROTATIONAL ENERGY

• 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.

Learning outcomes:At the end of this chapter you should be able to…

ROTATIONAL ENERGYROTATIONAL ENERGY

Each particle in a rigid rotating body has kinetic energy.

m2

r3

m3

r2

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

r1

axle

m1

Krot = ½m1v12 + ½m2v22 + …

 Krot = ½m1r122 + ½m2r222 + …

 Krot = ½(m1r12)2

 Krot = ½I2

CONSERVATION OF ENERGY

As usual, energy is conserved (in frictionless systems).

PHY1012F

ROTATIONAL ENERGY

If, however, a horizontal axis of rotation does not coincide with the centre of mass, the object’s potential energy will vary.

axle

So we write:

Emech = Krot + Ug= ½I2 + MgyCM

K U

Emech=K+U

Emech=0

4

ROTATION OF A RIGID BODYROTATION ABOUT A FIXED AXIS

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?

pivot

f2= i2 + 2

• You canNOT use rotational kinematics to solve this problem! Why not?

 f2= 0 + 2(–15)(–0.5 )

• (Not constant)

vt = r

 vt = –6.8 1 = 6.8 m/s

• Use rotational kinematics to find angular positions and velocities.
• (Not this time!)

y

• L = 1 m
• x

ROTATIONAL ENERGY

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?

pivot

• M = 0.07 kg
• yCM = 0 m

Before:

½I02 + MgyCM 0 = ½I12 + MgyCM 1

• yCM = –0.5 m
• 1 = ?
• vtip = ?

After:

2R

ROTATIONAL ENERGY

ROLLING MOTION

Rolling 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 2R each revolution;
• the speed of the wheel is given by
ROTATIONAL ENERGYFUNNY 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!

ROTATIONAL ENERGYROLLING MOTION

The velocity of a particle on a wheel consists of two parts:

TRANSLATION

+ ROTATION

= ROLLING

v=2vCM =2R

vCM

R

vCM

v = R

+

=

0

v = 0

–R

vCM

P

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

ROTATIONAL ENERGYKINETIC ENERGY OF A ROLLING OBJECT

If 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:

v=2R

K =Krot about P = ½IP2

v = R

Using the parallel axis theorem,

IP = (ICM + MR2)

v = 0

K =½ICM2 + ½M(R)2

P

 K =½ICM2 + ½M(vCM)2

K =Krot+ KCM

I.e.

h

ROTATIONAL ENERGY

THE GREAT DOWNHILL RACE

Kf = Ui

 ½ICM2 + ½M(vCM)2 = Mgh

ICM = cMR2

and

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

h

ROTATIONAL ENERGY

THE GREAT DOWNHILL RACE

vCM2 = 0 + 2ax

x

where

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.

ROTATIONAL ENERGYVECTOR 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.

ROTATIONAL ENERGYTHE CROSS PRODUCT

The magnitude of the torque exerted by force applied at displacement from the turning point is:  rFsin

Once again, the quantity rFsin 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: .

• y

1

In RH system:

• x

1

1

• z
ROTATIONAL ENERGYTHE CROSS PRODUCT

Notes:

• The more orthogonal the vectors, the greater the cross product; the more parallel, the smaller…
• Since it is a vector quantity, the cross product is also known as the vector product.
• .
• Derivative of a cross product:
ROTATIONAL ENERGYANGULAR MOMENTUM

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

mvt = p

• z

 L = rp

More generally (allowing for and to be at any angle )…

= (mrvsin, direction from RH rule)

I.e.

(Cf. in linear motion: )

ROTATIONAL ENERGYROTATIONAL MOMENTUM & ENERGY

Summary of corresponding quantities and relationships:

KCM = ½MvCM2

Krot = ½I2

(around an axis of symmetry)

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

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

ROTATIONAL ENERGY

### ROTATIONAL ENERGY

• 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.

Learning outcomes:At the end of this chapter you should be able to…