Conservation of Energy spool

[joemama69]

Homework Statement

Use conserbvation of Energy

The spool has a mass of 50 kg and a raius of gratino k = .280 m. If the 20 kg block A is released from rest, determine the distance the block must fall inorder for the spool to have an angulr velocity w = 5 rad/s. Also what is the tension in the cord while the block is in motion.

The spool has a radius of .3m, but where the cord is wrapped around it is a radius of .2m

Homework Equations

The Attempt at a Solution

T₁ + v₁ = T₂ + V₂

T & V 1 are both 0 because it starts from rest soooooo...

I = mk² = 50*.28² = 3.92

0 = 1/2*20*v² + 1/2*3.92*5² + Wy + 1/2*.25*s²

0 = 59 + Wy + .14s²

s = $$\sqrt{(-Wy - 59)/.14}$$

y = (.14s + 59)/W How do i find W

 

[joemama69]

ah W = mass*gravity

W = 50(9.81) = 490.5

so how do i get the y and the s

 

[joemama69]

can anyone give me a hint

 

[Redbelly98]

I'm not following everything you did, so here are some questions for you:

Which object is W the weight of?

What does "Wy" represent in the energy equation?

What does "1/2*.25*s²" represent in the energy equation?

 

[joemama69]

0 = .5mv² + .5Iw² + Wy + .5ks²

I = mk² = 50(.28²) = 3.92

0 = .5(20)v² + .5(3.92)w² + 20(9.81)y + .5(.28)s²

oops, my W was wrong, should be 20(9.81)

Wy represents the Potential Energy of Gravity, i would have to solve for y to find the distance the block falls

.5*.25*s² represents the elastic potential energy, i would have to solve for s to find the stretch of the rope

0 = 10v² + 1.96w² + 196.2y + .14s²

is this wrong, how do i finish from here

 

[Redbelly98]

.5*.25*s² represents the elastic potential energy, i would have to solve for s to find the stretch of the rope

Where did you get the 0.25 from in that equation? It wasn't given in the problem statement.

 

[Redbelly98]

Hint: forget about rope stretch in this problem. Assume the rope does not stretch.

However, you will have to figure out how v and ω are related.

 

[joemama69]

the .25 should be the .280 = k, sorry i should have been more carfull

 

[joemama69]

I = 3.92, v = wr = 1, w = 5

0 = 1/2(20)(1²) + 1/2(3.92)(5²) + 20(9.81)y

y = 59/196.2 = .3007 m,

Now i need to get the stretch of the rope, hint please

 

[Redbelly98]

The problem asks for the tension in the cord, not the amount of stretch.

 

[Redbelly98]

the .25 should be the .280 = k, sorry i should have been more carfull

Units!

k=0.28 m. It is the radius of gyration of the spool, not the spring constant of the cord.

Spring constants have units of N/m, so this k=0.28m cannot be a spring constant.

y = 59/196.2 = .3007 m,

Looks good!

 

[joemama69]

On Wikipedia it says Tension (T) is ... T + mg = 0 where velocity is constant, and T + mg $$\neq$$ 0 where velocity is increasig

T + 20(9.81) = 0 when T = -196.2 is that correct

 

[Redbelly98]

Since the velocity is increasing, T + mg ≠ 0.

Again, think in terms of energy: the work done on the brick, and the change in energy of the brick.

 

[joemama69]

Work done on brick

Wy= 20(9.81)(.300713) = 59

The energ goes from potentil to kinetic,

is there an equation for this, i can't find one

 

[Redbelly98]

is there an equation for this, i can't find one

(Work done on brick) = (Change in total energy of brick)

= (Change in kinetic energy of brick) + (Change in potential energy of brick)​

 

[Redbelly98]

(Work done on brick) = (Change in total energy of brick)

= (Change in kinetic energy of brick) + (Change in potential energy of brick)​

FYI, you can treat "Work done on brick" as entirely due to the tension in the cord. So if you know the change in the brick's kinetic and potential energy, as well as the distance the it moved, you can solve for the cord tension.