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Impule & Momentum

  • Thread starter joemama69
  • Start date
  • #1
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Homework Statement



Note Diagram.

If they are released from rest, determine the time required for block A to attain a speed of 2 m/s. Neglect the mass of the ropes


Homework Equations





The Attempt at a Solution



I = mk2 = (30)(.252) = 1.875 kg m2

X Direction

mv + [tex]\int[/tex] Fdt = mv2

0 + Oxt = 0

Y Direction

mv + [tex]\int[/tex] Fdt = mv2

0 + Oyt - 30(9.81)t = 0

Oy = 294.3

Moments

Iw1 + [tex]\int[/tex]MOdt = Iw2

0 + 25(.3) - 10(.18) = 1.875w2

w = [tex]\sqrt{3.04}[/tex]

How do i relate this to velocity and time. I can use v = wr, but how can i solve for t when v = 2 m/s

2 = .3w, therefore i need to find t when w = 2/.3
 

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Answers and Replies

  • #2
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I'm having trouble understanding the image. Can you get a better one or explain the image further?
Anyway, it seems you are making things much more complicated than they have to be.
Draw a FBD, and apply Newton's second law in linear and angular forms to each mass. The rest should just be kinematics.
 
  • #3
399
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The spool has two radiuses. one is .18m and one is .3m. they start from rest. when the blocks are released, block A weighs more so it will fall, thus lifting block B. When will block a reach a speed of 2 m/s.
 
  • #4
399
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But ya i dont think i needed most of what i posted first. Heres what im at

I = mk2 = 30(.25)2 = 1.875

25(.3) - 10(.18) = 1.875w2

w = [tex]\sqrt{3.04}[/tex]

w = v/r

[tex]\sqrt{3.04}[/tex](.3) = v

how do i get a t in there
 
  • #5
3,003
2
I think that I would take a different approach. You know that

[tex]\sum M =I\alpha[/tex]

and there is a relationship between alpha and the acceleration of the blocks.

So you can write the sum of the moments in terms of a

Since [itex]a=\frac{dv}{dt}[/itex] you should be able to find t using the initial conditions.

Edit: Was there a reason you thought that it was an Impulse & Momentum problem? I am only wondering in case I overlooked something.
 
  • #6
399
0
okok i think i got it

I = 1.875

Iw2 + + [tex]\sum\int[/tex]Mdt = Iw2

0 + [tex]\int[/tex](25(.3) - 10(.18)dt = 1.875w2

5.7t = 1.875w2 therefore w = [tex]\sqrt{3.04t}[/tex]

w = v/r

[tex]\sqrt{3.04t}[/tex](.3) = v = 2

3.04t = 4

t = 1.32 seconds

Is this right
 
  • #7
399
0
I thought it was impulse and momentum because the chapter is called

Planar Kinetics of a Rigid Body: Impulse and Momentum
Section 19.2 Principle of Impulse and Momentum

Im assuming there is more than one way to solve it, im just suppose to use impulse and momentum
 
  • #8
3,003
2
Edit: I can multiply :rofl: I think you're good!

You can always do it the other way as a check.

You have all the numbers necessary.
 
Last edited:
  • #9
399
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but u do agree with my answer based on my method
 

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