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Honors Physics 1Class 14 Fall 2013

The rotating skew rod

Rotational Inertia Tensor

Stability of spinning objects

The spinning top in gravity

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The angular momentum vector and angular velocity do not necessarily point in the same direction

Consider a rigid body consisting of two particles of equal mass on the ends of a massless rigid rod of length 2l . The midpoint of the rod is attached to a vertical axis which rotates at angular speed . The rod is skewed at an angle from the axis. Find the angular momentum of the system.

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Each mass moves in a circle of radius cos with angular speed .

cos

Taking the midpoint of the rod as origin, .

2 cos .

is perpendicular to the rod and lies in the plan

i ii

i

L r p

l

p m l

r l

L m l

L

e of rod and the z axis.

turns with the rod and traces a circle about the z axis.L

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In the previous example, we point out that L is not parallel to .

This is generally the case for non-symmetric bodies.

The fact that rotates means that there must be a torque on it.

The component o

L

f , , parallel to the z axis is constant.

The horizontal component L = sin rotates with the rod.

Choosing a starting phase,

L cos sin cos

sin sin

ˆˆ ˆsin cos sin cos

and the

z

h

x h

y

L L

L

L t L t

L L t

L L ti tj L k

ˆ ˆ torque is = sin sin cos

so sin

The larger the angular momentum, the more torque to rotate.

dLL ti tj

dtL

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Geometric interpretation of precessionof the skewed rod

( )hL t

( )hL t t

hL

so sin and points along the tangent in the xy plane.

Remember that , so F points at right angles to the

change in L.

h h

hh h

L L

dL dL L

dt dtL

r F

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Moment of Inertia Tensor

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Because we have found a case where L is not parallel to

should be modified.

The proper expression is where is a 3x3 matrix.

( , , )

xx xy xz

yx yy yz

zx zy zz

xx

L I

L I I

I I I

I I I I

I I I

I x x y z d

��

�

( , , ) product of inertia

Note: It is always possible to find a set of three orthogonal axes

about which the products of inertia are zero.

body

xy yxbody

x xx x xy y xz z

xdydz

I xy x y z dxdydz I

L I I I

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The spinning bicycle wheel(Gyroscope)

The angular momentum is along the axis of the spinning wheel.

(horizontal if we did it right)

The torque about the support point is due to the weight

and is at right angles to L and g.

Since

r mg

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is at right angles to L, the magnitude of L does not change.

Assuming that all of the mass of the wheel is at a distance R

from the axis and a distance D from the support point.

L=M R ; ; DW

DW L

2M R

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Moment of inertia of various objects

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4 22 3

The moment of a bicycle wheel is easy. The mass is

all at a distance of the radius:

A disk requires a little more work:

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4 2A rectangular plate is solved using the para

I MR

R MRI rr drd r dr

2

2 2

llel axis theorem

1for one of the dimensional integrals and for a rod.

121

12

cm x

plate x y

I ML

I M L L

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Stability of spinning objectsApplications: Rolling hula hoops,

flying saucers, footballs, rifle bullets...

Consider a cylinder moving parallel to its axis and what happens if we

exert a small perturbing force at right angles to the cylinder axis for a

short time t.

In the case where the cylinder is not initi

ally spinning

so L so = .

If the cylinder is rapidly spinning with angular momentum .

Torque causes precession while the torque is applied

= so the axis rotates by = t=

AA

s

s

Fl tFl t Fl t

I

L

Fl Fl

L

.

Instead of tumbling, the cylinder changes orientation slightly and

then stops precessing.

Note that spin has no effect on center of mass motion.

s

t

L

F