PES 1110 Fall 2013, Spendier Lecture 27/Page 1

Today: - The Cross Product (3.8 Vector product) - Relating Linear and Angular variables continued (10.5) - Angular velocity and acceleration vectors (not in book) - Office hours this week Friday and next week Monday cancelled - Friday lecture will be held by Prof. Karen Livesey Last time: Related the Linear and Angular Variable (10.5) v rw We know how to relate linear to angular speed. Now we need to relate direction to talk about linear and angular velocities. To get the correct direction, we use the cross product (vector product) of vectors. This is the other way to multiply two vectors. We have already learned bout the first way – the scalar product. The Cross Product (3.8 Vector product) - A way to multiply two vectors. The result of which is a new vector. (Note: This is different from the scalar product where the result was a scalar.)

Magnitude of Cross Product Here the vectors A

and B

are separated by an angle θ. The magnitude of their cross

product is

sinA B AB q

So the cross product is at its maximum when θ = 90º ( A

and B

are perpendicular) and

zero when θ = 0º ( A

and B

are parallel). Direction of Cross Product The cross product has to be perpendicular to both vectors A

and B

A B

is perpendicular to both A

and B

So, which vector can we draw on the page that is perpendicular to both A

and B

? None.

But there is a vector coming out of the page or into the page that is perpendicular to both A

and B

. We need to expand our 2D problem into 3D for it to work. When we do a cross product on the page the resulting vector is on the z-axis. But which way on the z-axis? To decide this we need another Right-Hand-Rule (RHR) to give direction.

PES 1110 Fall 2013, Spendier Lecture 27/Page 2

Right-Hand-Rule (RHR) Take the fingers of the right hand and “sweep" A

into B

, i.e, curl fingers from A

toward B

(using smaller angle, i.e. 0 180q ). Extended thumb points in direction of cross product. A B

(out of the page, dot inside the circle) B A

(into the page, cross inside a circle)

In magnitude A B

and B A

are the same, but they are into opposite directions! Hence

the cross product is a vector manipulation that depends on order. It is a type of multiplication where order is important. A B B A

Calculating the cross product in unit vector notation:

ˆ ˆˆ ˆ ˆ ˆx y z x y zA B A i A j A k B i B j B k

which can be expanded according to the distributive law; that is, each component of the first vector is to be crossed with each component of the second vector. The cross products of unit vectors are given in Appendix E (see “Products of Vectors”). For example, in the expansion of the above equation

ˆ ˆ ˆ ˆ 0x x x xA i B i A B i i

because the two unit vectors i and i are parallel and thus have a zero cross product. Similarly, we have

ˆˆ ˆ ˆ ˆx y x y x yA i B j A B i j A B k

Continuing to expand one can show that ˆˆ ˆ

y z y z z x z x x y x yA B A B B A i A B B A j A B B A k

Example 1: Using the cross product, determine the vector perpendicular to r1 = (2, −3, 1) and r2 = (−2, 1, 1)

PES 1110 Fall 2013, Spendier Lecture 27/Page 3

How do we use this in Physics? Linear and Angular Velocities (Important this is not in the book) Imagine a wheel rotating counterclockwise. Using the right hand rule, we know that the direction of angular velocity is out of the page. In this case we drew a point which is not on the edge.

r

- The vector from the center (axis of rotation) to the point of interest. This is our position vector. At an instant of time, the direction of linear velocity v is tangent to the circular path pointing in the direction of rotation. Hence, the angel between the position vector and the linear velocity vector is 90º. The angle between v , w , and r

is 90º. So the cross product is made to give you the

direction. Before, in terms of magnitude v rw To get the linear velocity v , is it rw or r w ? These two options do not result in the same vector. We just learned that the order when taking the cross product is important. r w : point your fingers along r

and curl your fingers out of the page – towards w . In

this case your thump point into the opposite direction of r

. It is the wrong way. Therefore v rw This takes a lot of practice!

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Linear Accelerations This is quite complicated since every point on a rotating object has two acceleration components.

We already know about one. We talked about this one when we did uniform circular motion. When the dot goes around the circle its linear velocity is constantly changing direction. We know this acceleration is called centripetal acceleration, lets call it

rada - changes in direction In what direction is rada ? It is towards the center. How big is rada ?

222

rad

rva rr r

ww (r is distance from the center to the point of interest)

rada is always opposite to the position vector , hence

2

rada rw

This is the first component. The other component, we did not have to worry about before because we were doing uniform circular motion. Something is going around a circle with a constant speed. Now, we maybe want to speed or slow down the rotating object. For this we need the tangential component of linear acceleration:

tana - changes in speed

The direction of the tangential component of linear acceleration is 90º to the radial component and points into the direction of the linear velocity v .

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We learned this previously: - Acceleration at 90º (perpendicular) to velocity causes only a change in direction. - Acceleration at 0º (parallel) to velocity causes only a change in speed.

What quantity deals with changes in angular speed - the angular acceleration α. What is the unit of α? It is rad/s2. What do I need to multiply this quantity with to turn it into m/ s2? Just by meters since rad has no units!

tana ra When you do the vectors:

tana ra If it is speeding up - a would be out of the page (same direction as w in Figure on previous page). In this case if I do ra , then I would get tana as shown in the picture above. If it is slowing down - a would be into the page, and tana would be in the opposite direction as shown in the picture above. The angel between rada and tana is always 90º. So the linear acceleration is:

tanrada a a (there are two components to the net linear acceleration!)

Therefore the magnitude of the net linear acceleration vector is:

2 2tanrada a a

Non-Circular Objects Everything we have done so far is applicable to non-circular objects as well. For non-circular objects, we need to put the origin of the coordinate system at the axis of rotation. This allows us to use all of the equations for circular objects we have learned so far. r = distance from axis of rotation

The black dot in the center at the axis of rotation will not move, while the red dot will move in a circular path while the non-circular object rotates.

PES 1110 Fall 2013, Spendier Lecture 27/Page 6

Example 2: The l = 2.5 m long stick has, at the instant shown, angular speed ω = 3 rad/s and angular velocity α = 5 rad/s2 both in the clockwise sense. What is the magnitude and direction of a) the angular velocity, b) the angular acceleration, c) the linear velocity, and d) the linear acceleration of the point at a distance 2 m from O?

PES 1110 Fall 2013, Spendier Lecture 27/Page 7

Example 3: An early method of measuring the speed of light makes use of a rotating slotted wheel. A beam of light passes through one of the slots at the outside edge of the wheel travels to a distant mirror, and returns to the wheel just in time to pass through the next slot in the wheel. One such slotted wheel has a radius of 5.0 cm and 500 slots around its edge. Measurements taken when the mirror is L = 500 m from the wheel indicate a speed of light of 3.0 x 105 km/s. a) What is the (constant) angular speed of the wheel? b) What is the linear speed of a point on the edge of the wheel?