What is the angular velocity of the disk

[vworange]

1. A disk of radius 2.90 cm and mass 1 kg is pulled by a string wrapped around its circumference with a constant force of 0.46 Newtons. What is the angular velocity of the disk to three decimal places after it has been turned through 0.90 of a revolution?

So I've been using:
\alpha=\tau/I

More specifically:
\alpha=(r*F)/(.5mr^2)

I know I'm getting the acceleration just fine. I've then been using:
\omega^2 = 2\alpha\Delta\Theta

I know the problem is coming in here. Probably in conversion of units (rad/s^2) to (rev/s^2) or something along those lines.

2. A large uniform "butcher block" rests on two supports and has a weight hanging from its end. The block has a mass of 100 kg and a length of 2 meters. If L = 1.50 meters and the hanging weight is 157 Newtons, what is the force on the left support to the nearest Newton?

The ball hangs on the right end of the block, the first support is on the left end of the block and the second support is 1.5 meters away from the left end of the block.

I've been trying all sorts of things, including:
L = 2.0 m
d = 1.5 m
F2 = mg(L/d)
F1 = mg - F2

I'm pretty sure the answer should be negative.. but i don't have it right yet.

 

[dalitwil]

you must totally be in the same physics class as me, homework quiz 11? ... heh, okay so i figure out how to do the last one:

(mass of the ball)(2-L)=981*(L-1) + x*L
Solve for x and that's your answer.

 

[vworange]

Yeah.. that equation isn't working for me. Why (L-1), btw? And mass of the ball.. are you dividing the N value by 9.81 or did you mean the force of the ball (in N)?

 

[dalitwil]

My bad, maybe i made that sound confusing...

mass of the ball is given in Newtons (157N)... using the numbers you have, i get:

(157)*(2-1.5)=981*(1.5-1)+ (1.5x)... solving for x you get -274.667. The answer they want is +275. (Just ignore the negative.)

 

[dalitwil]

AAAAAAAND I just figured out how to do the other problem:

So they give you the .05 (or whatever) of a revolution it turns. Convert that to radians by multiplying with 2pi.

Then you need to find your delta x, which you do by multiplying the radians you found in the above by your radians (don't forget to convert cm to m!)

By having x, you can now find what W equals, using W=F*delta x

When you found your W, set this equal to .5*(.5MR^2)ω^2 and solve for ω.

(it's going to be the positive of the two answers)

 

[vworange]

Here's a site I found with the same problem. You can view the correct answers and try to work to it, but I still haven't figured out the proper equation yet.

http://wps.prenhall.com/wps/media/objects/1088/1114211/ch8/Chapter_8.html

 

[dalitwil]

AAAAAAAND I just figured out how to do the other problem:

So they give you the .05 (or whatever) of a revolution it turns. Convert that to radians by multiplying with 2pi.

By having x, you can now find what W equals, using W=F*delta x

When you found your W, set this equal to .5*(.5MR^2)ω^2 and solve for ω.

(it's going to be the positive of the two answers)

MY MISTAKE: just noticed "Then you need to find your delta x, which you do by multiplying the radians you found in the above by your radians (don't forget to convert cm to m!)"

WHEN I SAID MULTIPLYING THE RADIANS YOU FOUND ABOVE BY YOUR RADIANS, I MEANT YOUR RADIUS.

sorry.