lecture 9: vector cross product - zhejiang...

General Physics IGeneral Physics I

Lecture 9: Vector Cross Lecture 9: Vector Cross ProductProduct

Prof. WAN, Xin （万歆）

[email protected]://zimp.zju.edu.cn/~xinwan/

OutlineOutline

● Examples of the rotation of a rigid object about a fixed axis

− Force/torque point of view− Energy point of view

● Vector cross product− Example: Coriolis effect− Torque

Example: Rotating RodExample: Rotating Rod

● A uniform rod of length L and mass M is attached at one end to a frictionless pivot and is free to rotate about the pivot in the vertical plane. The rod is released from rest in the horizontal position. What is the initial angular acceleration of the rod and the initial linear acceleration of its right end?


● The torque due to this force about an axis through the pivot is

● The moment of inertia for this axis of rotation

Can you show?


● The linear acceleration of the right end of the rod

● This result—that at > g for the free end of the rod—is rather interesting. It means that if we place a coin at the tip of the rod, hold the rod in the horizontal position, and then release the rod, the tip of the rod falls faster than the coin does!

Example: Atwood's MachineExample: Atwood's Machine

● Two blocks having masses m1 and m2 are connected to each other by a light cord that passes over two identical, frictionless pulleys, each having a moment of inertia I and radius R. Find the acceleration of each block and the tensions T1, T2, and T3 in the cord. (Assume no slipping between cord and pulleys.)


● Discussion:− Equal mass: At equilibrium.− Unequal mass: Normally we assume a

direction for the acceleration. If the result is negative, the real acceleration is in the opposite direction.

− I = 0: Goes back to the same old Newton's laws.

Example: Connected CylindersExample: Connected Cylinders

Consider two cylinders having masses m1 and m2, where m1 ¹ m2, connected by a string passing over a pulley. The pulley has a radius R and moment of inertia I about its axis of rotation. The string does not slip on the pulley, and the systemis released from rest. Find the linear speeds of the cylinders after cylinder 2 descends through a distance h, and the angular speed of the pulley at this time.

Detailed AnalysisDetailed Analysis

● We are now able to account for the effect of a massive pulley. Because the string does not slip, the pulley rotates.

● We neglect friction in the axle about which the pulley rotates for the following reason:

− Because the axle’s radius is small relative to that of the pulley, the frictional torque is much smaller than the torque applied by

the two cylinders, provided that their masses are quite different. ● Mechanical energy is constant; hence, the increase in the

system’s kinetic energy (the system being the two cylinders, the pulley, and the Earth) equals the decrease in its potential energy.

Example: Connected CylindersExample: Connected Cylinders

Torque AgainTorque Again

● The tendency of a force to rotate an object about some axis is measured by a vector quantity called torque

− r: the distance between the pivot point and the point of application of F.

− d: the perpendicular distance from the pivot point to the line of action of F.

− This quantity d is called the moment arm of F.

Vector RepresentationVector Representation

● The torque t involves the two vectors r and F, and its direction is perpendicular to the plane of r and F. We can establish a mathematical relationship between t, r, and F, using a new mathematical operation called the vector product, or cross product

Will discuss more next week.

The Vector ProductThe Vector Product

The direction of C is perpendicular to the plane formed by A and B , and the best way to determine this direction is to use the right-hand rule.

Properties of Vector ProductProperties of Vector Product

● Unlike the scalar product, the vector product is not commutative. Instead, the order in which the two vectors are multiplied in a cross product is important:

● The vector product obeys the distributive law:

More PropertiesMore Properties

● If A is parallel or antiparallel to B, then A ´ B = 0; therefore, it follows that A ´ A = 0.

● If A is perpendicular to B, then |A ´ B| = AB● 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 A and B

Elementary ResultsElementary Results

Generic ResultsGeneric Results

A=Ax i+Ay j+Az k

B=Bx i+By j+Bz k

Ax i×By j=AxBy k

Ay j×Bx i=−Ay Bx k

C= A×B=Cx i+Cy j+Cz k

k∣Ax Ay

Bx By∣

= i∣Ay Az

By Bz∣+ j∣Az Ax

Bz Bx∣+ k∣Ax Ay

Bx By∣=∣

i j kAx Ay Az

Bx By Bz∣

In a Rotating SystemIn a Rotating System

Newton's Second Law:

T=mv2

rT−m

v2

r=0

Fictitious centrifugal force

Typhoon (Low Pressure)Typhoon (Low Pressure)

Rotation: Clockwise in the southern hemisphere and counterclockwise in the northern hemisphere.

Coriolis EffectCoriolis Effect

● The Coriolis effect is a deflection of moving objects when the motion is described relative to a rotating reference frame.

Vector form

for point object at rest in the rotational frame

Graphic IllustrationGraphic Illustration

● In the inertia frame of reference

drdt ∣I =

drdt ∣R + ω × r

r

r+dr∣I ω×r

d r∣R

change of the displacement within the rotational frame

change of the displacement due to the rotation of the reference frame

Mathematical OriginMathematical Origin

This is generically true for any vector u.

dudt ∣I =

dudt ∣R + ω × u

d(ux i)

dt=

dux

dti + ux

d idt

=dux

dti + ω×(ux i)

That is, we need to take into account the change of directions of the axes.

x:

d(uy j)

dt=

duy

dtj + uy

d jdt

=duy

dtj + ω×(uy j)y:

d(uz k)

dt=

duz

dtk + uz

d kdt

=duz

dtk + ω×(uz k)z:

Acceleration as the 2nd DerivativeAcceleration as the 2nd Derivative

● In the inertia frame of reference

v∣I =drdt ∣I =

drdt ∣R + ω × r

d vdt ∣I =

dvdt ∣R + ω × v

d2 rdt2∣I = [ ddt∣R + ω × ]

2

r

=d2 rdt2∣R + 2 ω ×

drdt ∣R + ω × (ω × r )

r

r+Δ r∣I

ω×r

Δ r∣R

ω×r

Δ r∣R

““Force” in the Rotating FrameForce” in the Rotating Frame

● In the rotating frame of reference

Fnet = md2 rdt2∣R

= md2 r

dt2∣I − 2mω×drdt ∣R − mω×(ω×r )

= F − 2mω×vR − m ω×(ω×r )

= F − 2mω×vR + mω2 r

a×(b×c) = b(a⋅c) − c(a⋅b)Identity:

centrifugal forceCoriolis force

for r⊥ω

WarningWarning

● The treatment of the Coriolis force is to study the motion of an object in the rotating frame of reference (typically on another rigid object, like the earth, which rotates with a constant angular velocity).

● This is different from studying the rotation of a rigid object in the inertia frame of reference, which is the main subject of our interest.

● Once again, stay in the inertial frame of reference if you can.

lecture 9: vector cross product - zhejiang...

Documents