Question

Test Review....

Answer 1

Wanna start from chap 10 on some easier stuff?

Answer 2

Awesome.

Answer 3

hey, we can start anywhere

Answer 4

I wanted to lok over the basics from the biginning real quick.....unless you want to keep going with chap 36.

Answer 5

I was looking at 100 real quick.

Answer 6

chap 10...

Answer 7

cool

Answer 8

Want to pick a problem?

Answer 9

Maybe problem 100...I know its from HW but maybe a good one to do again.

Answer 10

works for me

Answer 11

So other than common sense why can we say it moves to the right? Because the sum of the forces are in the right direction?

Answer 12

since it starts from rest, the CM moves in the direction of the net force. If it had an initial velocity that wouldn't necessarily be the case

Answer 13

\[V _{f}^2 = V _{i}^2 +2a(Deltax)\]

Answer 14

this one isn't on the sheet but is important to remember

Answer 15

and Im assuming since the force is making a tangential velocity that this will be the veloicty at center of mass.....so we can use that formula.

Answer 16

\[v = \sqrt{2*(35/.85)*5.5}\]

Answer 17

swap .85 with 21

Answer 18

Cool. So the next part.

Answer 19

\[\tau = I \times \alpha\]

Answer 20

but we want omega.
\[\alpha = tao/I \]

Answer 21

So how does alpha become omega?

Answer 22

\[\alpha = \Delta \omega/\Delta t \]

Answer 23

\[aplha = \omega/deltaT\]

Answer 24

since it starts at rest there is only one value each for \[\alpha and t\]

Answer 25

ok.

Answer 26

so then \[\tau = Fr\] can be used since those values are also given

Answer 27

Looks good. Do you want to move on?

Answer 28

sure

Answer 29

stay in 10?

Answer 30

Maybe do 1 from each chapter.......but whatever you want to do.

Answer 31

I'm game for anything

Answer 32

How about you pick one from 11.

Answer 33

We will be doing a cross product, would hate to screw that up. But I wanna do an angular momentum one at some point.

Answer 34

I'm looking now

Answer 35

36?

Answer 36

ok.

Answer 37

so L = rxp for this one?

Answer 38

yep

Answer 39

so would the momentum be 75 grams times the velocity vector treating mass as a scalar?

Answer 40

yes, since mass is a scalar, you can do m * (r X v)

Answer 41

but I can also dist right into v correct? You way will be cleaner...

Answer 42

that should work too

Answer 43

We have to convert to kg dont we....

Answer 44

yes

Answer 45

Thats a Dennis kind of problem......

Answer 46

haha

Answer 47

Did ulises send you the pictures?

Answer 48

yes

Answer 49

want to run through those problems?

Answer 50

Yeah starting from the beginning...Im gonna grab some tea brb. I hate the way dennis poses questions.....

Answer 51

brb

Answer 52

back

Answer 53

ok

Answer 54

\[F=ma _{t}\]

Answer 55

\[a _{t}= r\]

Answer 56

=\[\alpha * r\]

Answer 57

what up ladies, this is Ulises, roger that?

Answer 58

\[F = \alpha timesm timesr\]

Answer 59

lol

Answer 60

Greg whats what does the t stand for on accel.

Answer 61

tangential

Answer 62

I'm not sure how theta plays into it

Answer 63

me neither..

Answer 64

\[\theta = l/r\]?

Answer 65

what are we given to work with......we have theta, d and mass.

Answer 66

and isnt d actually the radius in this case?

Answer 67

no, still d here

Answer 68

are you guys still on part (a)?

Answer 69

yes.

Answer 70

If we dont treat d as a radius can we still use theta = L/r ?

Answer 71

= l/r
r=d/2

Answer 72

Wanna do the next problem and come back to this?

Answer 73

I'm not understanding what relationship theta has in it at all

Answer 74

me either.

Answer 75

what if we use conservation of angular momentum?

Answer 76

if we use the magnitude of the definition of angular momentum rosin(theta) is what I'm thinking of to make use of theta

Answer 77

sorry I meant rosin(theta)

Answer 78

what the heck I'm trying to put a p not an o

Answer 79

jason - what does that mean?

Answer 80

\[Atan = R \alpha\]
\[Atan =(d/2) \alpha = (d/2)(\Delta \omega/ \Delta t)\]

Answer 81

try this one out...

Answer 82

torque = Frsin(theta) = Ia
=mgsin(theta) = mr^2*a

Answer 83

a=t/I
a=mgsin(theta)/mr^2
a= gsin(theta)/r

Answer 84

whoa....where did that come from?

Answer 85

but g isn't right because g doesn't = a here

Answer 86

but if it's rotating that means there's also a centripetal acceleration, which is different from the tangential acceleration

Answer 87

I'm off the tangential acceleration kick

Answer 88

Lets come back to it...

Answer 89

How about 3.