Rotational mechanics and Angular Momentum- help needed to check over work

AI Thread Summary
The discussion revolves around a series of physics problems related to rotational mechanics and angular momentum, where the original poster is struggling to arrive at correct answers. In Problem 1, the poster calculates angular acceleration and linear speed but suspects an oversight regarding the wheel's spokes may have led to an incorrect answer. Problem 2 involves calculating torque and angular acceleration, with a query about using degrees instead of radians for cosine calculations. In Problem 3, the poster attempts to apply a formula found online but ends up with an incorrect mass calculation and time result. The poster seeks assistance to identify errors in their approach before a homework deadline.
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Please note, this is a series of problems, with a series of answers. I believe that I understand how to do the problems, but for some reason, my answer is wrong, and I can think of no reason why.

Homework Statement


Problem 1: http://img355.imageshack.us/my.php?image=physics1jw9.png

i) I need to find the speed of the mass as it passes point B in m/s

ii) I need to find the Tension of the string (Which I am sure I can do, but without a correct answer for part i, I'm not going to attempt yet)

Problem 2: http://img186.imageshack.us/my.php?image=physics2pt6.png

Problem 3: http://img355.imageshack.us/my.php?image=physics3el2.png
(Note, the word that is cut off is longitudinal. I had to copy and paste parts of it into one picture, sorry it's a little shoddy <.<)

Homework Equations


All the lovely rotational mechanics equations, and their linear counterparts


The Attempt at a Solution



Problem 1
i) Ok, I did not take into account the spoke on teh wheel whatsoever, since the mass of inertia was given to me, and the spokes of the wheel did not seem to have their own mass. Not sure if this would effect the problem.

T=I(angular a)
Angular acceleration = Torque/I

Torque= 46 KG * 9.8 m/s^2 * 3m = 1352.4
I= 3/4* 23 KG * 3^2 = 155.25
Ang A= 8.7111

a = (Ang A)*r

a= 8.711 * 3 = 26.133

Plugging this acceleration into the classic Xf= Xi + Vi +(at^2)*2 (assuming Vi and Xi are 0)
I get t=2.106

Then, using the Vf= Vi +at
I get Vf= 55.05

This answer is wrong though, not sure where I screwed up




Problem 2
This one seemed simple enough, just a basic torque= I * ang a

Ang a = T/I

I=(1/12)*(2.7)*(3^2)= 2.025

T= rxF
F= M*G= 9.8*2.7= 26.46
r= 3cos(43)=2.045
T= 54.137

Ang a= 26.734

It wants it in Radians/sec^2. Is it bad that I was using degrees for the cos?


Problem 3
I searched on google, and did come across this thread: https://www.physicsforums.com/showthread.php?t=17605

However, upon trying the forumulas given there, I still got it wrong. Using the formula ShawnD had at the end of his post, I got the wrong answer.

m= IW/rV

I= 2950
W=2 rev/min= .2094... rad/sec
r=2.9
V=755
I got M to be .282 kg

I then divided by .012 kg, and got that time = 21.87 s

This answer is also wrong, not sure what happened.

Thank you in advance for looking over this work.
 
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Slight bump, sorry, but homework's due by Thursday Night. >.>
 
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