# Homework Help: Principle of Work and Energy

1. Nov 2, 2012

### zack7

1. The problem statement, all variables and given/known data
I have to solve this question in two ways, the first is from kinetics and kinematics and the second from the principle of work and energy. For the the first problem I obtained 3.160643122 for the acceleration at b(pulley with force t) and proceeded to find to find the velocity at a which I found out to be v = 1.256980509 by using v2=2ad

I then proceed to use the principle of work and energy but can never match the answers. I am not sure where I am going wrong.

The goal is to find the velocity at velocity at B(second pulley/pulley with force T)
2. Relevant equations
Note the question changed a little with the force T moving upward 6 inches

3. The attempt at a solution

Last edited: Nov 3, 2012
2. Nov 3, 2012

### zack7

Basically by using kinematics and Newton's Equations I came to the answer that the velocity of B the (second pulley with T force ) is 1.256980509 by using v2=2ad

Then I would have to use the principle of work and energy to find out the same velocity which should be the same as question 1

F ds +Ʃ M dθ=$\frac{1}{2}$mv2+$\frac{1}{2}$IW2

Kinematics
V=wr

FA+($\frac{1}{2}$)(ma+mc)gA+(mbgA)=$\frac{1}{2}$(ma+mc)V2f+$\frac{1}{2}$mbv2f+$\frac{1}{2}$IbW2+$\frac{1}{2}$IaW2

a=pulley with weight
b=pulley with T force
c=weight
A=distance traveled by the pulley

Last edited: Nov 3, 2012
3. Nov 3, 2012

### tiny-tim

hi zack7!

thanks for the pm

4. Nov 3, 2012

### xur54

Glancing at this, my initial reaction is that you should not be getting velocity numbers or distances. Why is the goal to find the velocity at B? That's not what the question is asking for in the book.
I'm going to work the problem right now and let you know what I find.

5. Nov 3, 2012

### zack7

My teacher changed the question up a little bit and we have to find the velocity of B with the force T when it moves up 6 inches.

6. Nov 3, 2012

### xur54

Gotcha gotcha. I thought maybe that was what it was. I'm assuming it starts from rest (safe to assume). I'm also assuming gravity is in this problem (although it doesn't have to be).

7. Nov 3, 2012

### PhanthomJay

Your kinematic approach looks excellent, although I haven't cranked out the numbers or checked all your subscripts carefully. You've put in a lot of good work.

For the energy approach, looks like you slipped a minus sign, regarding the work done by gravity. And what happened to the M dθ term?

8. Nov 3, 2012

### zack7

My teacher said that since no moment was applied it would be zero

9. Nov 3, 2012

### xur54

The good news: our answers are somewhat close which in a problem like this is probably a good thing.
I got 1.77776 ft/s for vB (which means vA is .88888 ft/s=.5*vB)
Using work: The work from the tension adds energy which becomes PE, KE, and rotational KE.
Simple equation (I converted the 3 masses to slugs and used 32.2 ft/s for gravity and converted all lengths to feet. Note 6 inches=.5 feet. Sorry, I had to leave units out of calculations for speed. I also use parentheses for multiplication. I=.022 r=.3333):
work: 15*.5=Wtension=
PE: +10(.5).5 +10(.25).5 +20(.25).5
KE: +.5(mpulley)vB^2+.5(mpulley)(.5*vB)^2+.5*mass*(.5*vB)^2
RotKE: +.5(I)(vB/r)^2 +.5(I)(.5*vB/r)^2+0
which the calculator or computer program can solve quickly
I need to look at your work more tomorrow and finish my work as well. When is your homework due? Hope I'm not too late.
(edit: fixed wrong units. no change in the numbers.)

Last edited: Nov 4, 2012
10. Nov 4, 2012

### xur54

Ugh. I forgot the M dθ term. I agree with PhantomJay. I need to fix that tomorrow. Your teacher says there is none, hmmmm. I'm not so sure.

11. Nov 4, 2012

### zack7

That is great and your are not late, I will compute it, and thank you for the help

12. Nov 4, 2012

### PhanthomJay

It seems to me that the applied force F produces moment about the center of the top pulley.

13. Nov 4, 2012

### xur54

Actually I'm going with his teacher (again) now. The ropes cause moments on both pulleys but that gets accounted for in the $\frac{1}{2}$IbW2+$\frac{1}{2}$IaW2 terms. The rotational positions of the pulleys don't have different energy levels so no Mdθ terms are needed in this case I feel. Quick analogy: if you push a block on a frictionless horizontal surface, all the work done goes into KE. Similar for the pulleys.
What do you think PhantomJay? Mdθ terms still needed?
Edit: I see. You can use the moment to calculate how much work is done on the pulleys. The problem is the moment is unknown because the ropes that go around the pulleys change tension some. Simpler to go from the work put into the whole system becomes PE, KE, and rotationalKE in this linked system. Mdθ will come up in different methods.

Last edited: Nov 4, 2012
14. Nov 4, 2012

### xur54

Phantom Jay correctly pointed out missing negative signs because gravity force is down. Found something else too though. In your attached work and in this equation, the distances traveled are messed up some. Notice that the rope end with the force travels the farthest. It does not travel the same distance as pulley B which is what you have in your equation. FA should be 2*FA (if I understand correctly that F is the tension).
Agree zack7?

Last edited: Nov 4, 2012
15. Nov 4, 2012

### zack7

Yup that makes sense and after incorporating the negative and the distance traveled, I am getting close to the answer but still am no able to find one. It seems it I am missing a term or I am not doing the work done part properly.

16. Nov 4, 2012

### xur54

I looked at your formula on the computer screen and plugged it into the my TI-89 and used the solve function and got 1.77776. Worked for me. I would rewrite you equation with corrections. I think your subscript on a moment of inertia might be incorrect. And the way the linear velocity terms are written could lead to error. (mc)V2f+$\frac{1}{2}$mbv2f. Note that Vf does not equal vf here.
Similar issue on W and W. Those two are different but with same variable. Caution caution caution on that.

Last edited: Nov 4, 2012
17. Nov 4, 2012

### zack7

The W for the pulley from the bottom would be $\frac{v}{2R}$ and the other $\frac{v}{R}$

2FA-($\frac{1}{2}$)(ma+mc)gA-(mbgA)=$\frac{1}{2}$(ma+mc)2V2f+$\frac{1}{2}$mbv2f+$\frac{1}{2}$IbW2+$\frac{1}{2}$IaW2

Is that how you did it ? I did it this way and got 1.265

At first even for my kinematics approach I got 1.777 when I used d=d but I thought it would move half of the rope therefore d= d/2 , which is how I got the 1.25 and I am getting 1.26 using the approach above.

Last edited: Nov 4, 2012
18. Nov 4, 2012

### xur54

Here is my other method:
The letters in the picture mean tensions.
k=mass pulley. l=mass of the hanging mass. AA=acceleration of A and hanging mass. AB=acceleration of B. Parentheses often used as multiplication.
I drew a free body diagram (of both pulleys) with linear inertial force and rotational inertia force. I used the picture to do the sum of y forces which equals 0. This gives two equations.
0=(-H-T)+U+(10+k*AB)
0=(-E-U)+(20+l*AA)+(10+k*AA)
Now (T-H)r=a torque=I(alphaB)=I(AB/r) which means H=-(I*AB/r^2-T)
and (U-E)r=I(AA/r) which means E=-(I*AA/r^2-U)
and AB=2*AA
Substitute
0=((I*2AA/r^2-T)-T)+U+(10+k*2*AA)
0=((I*AA/r^2-U)-U)+(20+l*AA)+(10+k*AA)
Using a some sort of solver
AA=3.16045 ft/s^2 and U=16.7852 lb
means AB=6.3209 ft/s^2
accleration=AB
velocity=AB*t+0
position=.5*AB*t^2+0+0
position=.5*.5 which means t=.281252 sec
which means velocity=1.77777 ft/s which agrees as it should

#### Attached Files:

• ###### pulley diagram.jpg
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19. Nov 4, 2012

### xur54

A .01 difference could be rounding but I don't think 1.265 or 1.25 is correct.
Also I tried solving the new equation and I got 1.016 (and 1.295 with a slight change) but never 1.265. Solving mistake??
You put a 2 before Vf which should be .5 and should be (.5*Vf)^2 not 2*Vf^2 and not (2*Vf)^2.

20. Nov 4, 2012

### zack7

Yup the velocity should be 1.77, but I just can't seem to get it with the equation below

The W for the pulley from the bottom would be $\frac{v}{2R}$ and the other $\frac{v}{R}$

2FA-($\frac{1}{2}$)(ma+mc)gA-(mbgA)=$\frac{1}{2}$(ma+mc)(0.5V)2f+$\frac{1}{2}$mbv2f+$\frac{1}{2}$IbW2+$\frac{1}{2}$IaW2