Solving Impulse & Momentum for Time Needed for Block A to Reach 2 m/s

[joemama69]

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 + $$\int$$ Fdt = mv₂

0 + O_xt = 0

Y Direction

mv + $$\int$$ Fdt = mv₂

0 + O_yt - 30(9.81)t = 0

O_y = 294.3

Moments

Iw₁ + $$\int$$M_Odt = Iw₂

0 + 25(.3) - 10(.18) = 1.875w²

w = $$\sqrt{3.04}$$

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

 

[tjkubo]

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.

 

[joemama69]

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.

 

[joemama69]

But you i don't think i needed most of what i posted first. Heres what I am at

I = mk² = 30(.25)² = 1.875

25(.3) - 10(.18) = 1.875w²

w = $$\sqrt{3.04}$$

w = v/r

$$\sqrt{3.04}$$(.3) = v

how do i get a t in there

 

[Saladsamurai]

I think that I would take a different approach. You know that

$$\sum M =I\alpha$$

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 $a=\frac{dv}{dt}$ 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.

 

[joemama69]

okok i think i got it

I = 1.875

Iw^{2 +} + $$\sum\int$$Mdt = Iw²

0 + $$\int$$(25(.3) - 10(.18)dt = 1.875w²

5.7t = 1.875w² therefore w = $$\sqrt{3.04t}$$

w = v/r

$$\sqrt{3.04t}$$(.3) = v = 2

3.04t = 4

t = 1.32 seconds

Is this right

 

[joemama69]

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, I am just suppose to use impulse and momentum

 

[Saladsamurai]

Edit: I can multiply I think you're good!

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

You have all the numbers necessary.

 

[joemama69]

but u do agree with my answer based on my method