10.1.11
9.1.11
15.11.10
Webassign: 118 point beast
This weeks webassign happens to have more than 100 submissions. It is a monstrosity looming over you. However, it should in theory weigh much more than other smaller webassigns, so if you get it done it'll be great. Yes, there are multiple choice. Just don't wait until Thursday night. I wouldn't even wait until wednesday night, the amount of clicking you would have to do might result in carpal tunnel if you tackle it all at once. I've got the monitor I plan on smashing up, and an old analog tv set that needs to go, so expect to see my "experiment" sometime soon.
UPDATE: I have almost reached 20%. That may be all I do tonight, and reading the chapter is a must.
UPDATE: I have almost reached 20%. That may be all I do tonight, and reading the chapter is a must.
Rotational Dynamics Equation Derivation
I=∑miri2= ∫r2dm
Ip= ∫r2dm
= ∫(h+r)2dm
= ∫(h2+2hr=r2)dm
= ∫h2dm + ∫r2dm + ∫2hr dm
= h2 ∫dM + r2 ∫dm
Ip = Mh2 + Icm
τ = r x F
= rFsinθ
F=ma
τ=rF=rma
= rm(rα)
=mr2α
=(∑miri2)α
= Iα
p=mv
l=m(r x v)
dl/dt = m[dr/dt x v + r x dv/dt]
= m [0 + r x a]
=mαr2
dl/dt=Iα
Ip= ∫r2dm
= ∫(h+r)2dm
= ∫(h2+2hr=r2)dm
= ∫h2dm + ∫r2dm + ∫2hr dm
= h2 ∫dM + r2 ∫dm
Ip = Mh2 + Icm
τ = r x F
= rFsinθ
F=ma
τ=rF=rma
= rm(rα)
=mr2α
=(∑miri2)α
= Iα
p=mv
l=m(r x v)
dl/dt = m[dr/dt x v + r x dv/dt]
= m [0 + r x a]
=mαr2
dl/dt=Iα
7.10.10
Webassign Assignment #5
Problem 8 in text, 5 on webassign.
Two horizontal forces act on a 1.3kg disk
Force one is in the negative x direction. -1î
Force 2 has a magnitude of 9N
3m/s•s
maî + maj
F1=-1i
F2 = 9cos∂i+ 9sin∂j The key here is that you’re looking for the x component only.
mai = -1 + 9cos∂
1.3•3 = -1+9cos∂
=57.013º
NOTE: In a case where you have a negative F1, you'll have to do 180 minus your angle.
Forces on monkey: Gravity (mg), Monkey pulling on rope, force of rope on monkey F=T
Up is positive
Forces on package: Gravity (mg), Tension up, normal force
Package is 14kg
acceleration is up
T - 9g = 9•a
T - 14g + F(normal) = m(package)a
a) F(normal) = 0, a(package) = 0
T -14g = 0
T=137.34
Monkey must have acceleration of 5.45
b) acceleration after? F(normal) is still zero. a(package) is a number
a(package) = -a(monkey)
a(monkey)= 5/23•g should be 2.1326
EDIT: Awesome, I can upload my notes directly to the blog.
Two horizontal forces act on a 1.3kg disk
Force one is in the negative x direction. -1î
Force 2 has a magnitude of 9N
3m/s•s
maî + maj
F1=-1i
F2 = 9cos∂i+ 9sin∂j The key here is that you’re looking for the x component only.
mai = -1 + 9cos∂
1.3•3 = -1+9cos∂
=57.013º
NOTE: In a case where you have a negative F1, you'll have to do 180 minus your angle.
- 9kg monkey goes up rope, frictionless tree limb
Forces on monkey: Gravity (mg), Monkey pulling on rope, force of rope on monkey F=T
Up is positive
Forces on package: Gravity (mg), Tension up, normal force
Package is 14kg
acceleration is up
T - 9g = 9•a
T - 14g + F(normal) = m(package)a
a) F(normal) = 0, a(package) = 0
T -14g = 0
T=137.34
Monkey must have acceleration of 5.45
b) acceleration after? F(normal) is still zero. a(package) is a number
a(package) = -a(monkey)
a(monkey)= 5/23•g should be 2.1326
EDIT: Awesome, I can upload my notes directly to the blog.
Doing Physics: Impact resistant glass
Well, I think we all can agree that physics is a useful tool. But for what, you might ask? Everything. So, I think we need to use it to figure out how we can break impact resistant glass. I heard computers have cool looking parts in them anyways, so we'll go with an old computer moniter as soon as I find a good one.
28.9.10
#4 problem 6
We did the setup for problem 6 in class today.
The way you need to set it up is by giving each particle an equation with two components: X and Y. So, whatever height your first particle was at will be its y equation, I.E. Y=25. The x equation will be X = vt, I.E. x=2.5t. This is the particle moving at constant velocity. For the second particle, the one starting at the origin, your equations will need to have cos(∂) or sin(∂). These will use the acceleration equation for distance. So 1/2at^2 cos/sin (∂). In my case, I ended up with x=.195t^2sin(∂) and y=.195t^2cos(∂). I won't go any farther than this, other than to say that you need to have all the t's be the same, x=x, and y=y. If you need more help, ask me in class or ask Doc Lane.
The way you need to set it up is by giving each particle an equation with two components: X and Y. So, whatever height your first particle was at will be its y equation, I.E. Y=25. The x equation will be X = vt, I.E. x=2.5t. This is the particle moving at constant velocity. For the second particle, the one starting at the origin, your equations will need to have cos(∂) or sin(∂). These will use the acceleration equation for distance. So 1/2at^2 cos/sin (∂). In my case, I ended up with x=.195t^2sin(∂) and y=.195t^2cos(∂). I won't go any farther than this, other than to say that you need to have all the t's be the same, x=x, and y=y. If you need more help, ask me in class or ask Doc Lane.
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