# Trebuchet, max speed

1. Aug 16, 2008

### Zukie91

1. The problem statement, all variables and given/known data

1. The problem statement, all variables and given/known data
A war-wolf, or trebuchet, is a device used during the Middle Ages to throw rocks at castles and now sometimes used to fling pumpkins and pianos. A simple trebuchet is shown in Figure P8.77. Model it as a stiff rod of negligible mass 3.00 m long and joining particles of mass 60.0 kg and 0.120 kg at its ends. It can turn on a frictionless horizontal axle perpendicular to the rod and 14.0 cm from the particle of larger mass. The rod is released from rest in a horizontal orientation. Find the maximum speed that the object of smaller mass attains.
2. The attempt at a solution

don't even know where to start, any help is much appreciated
(the answer is 24m/s but i don't know how to get that)
Thanks

2. Aug 16, 2008

### tiny-tim

Welcome to PF!

Hi Zukie91! Welcome to PF!

Hint: use conservation of energy (KE + PE = constant).

3. Aug 16, 2008

### Zukie91

i noticed that there is another thread asking the same question which nobody replied to. On that thread, under relevant equations, KvnBushi listed this
$$K = \frac{1}{2} I_{cm} w^2 + \frac{1}{2} M v_{cm}^2$$
is that actually relevant?
ok i can't get that latex code or whatever it is to work properly, i'll just link the other thread

4. Aug 16, 2008

### tiny-tim

(LaTeX isn't working at the moment. )

The Iw2 part isn't relevant.

The mv2 part is.

5. Aug 16, 2008

### Zukie91

i appreciate you help btw,
do i need to use torque?
i'm sorry i'm just not getting this problem
and how do i use pe if i don't have the height?
oh wait, would the 14 cm be the height?
and i made a typo on the answer, it should be 24.5 m/s not 24 m/s

Last edited: Aug 16, 2008
6. Aug 16, 2008

### tiny-tim

No … just KE + PE = constant.
sort-of … but 14cm is the length of the whole rod, and it isn't the end of the rod that's fixed, is it?

7. Aug 16, 2008

### Zukie91

it says the whole rod is 3.00 m long, and that the axel is 14.0 cm from the larger mass, which means its 2.86 m from the projectile right?

8. Aug 16, 2008

### tiny-tim

… oops!

oops … I misread the question!

Yes, you're right … the PE for the two masses will be based on .14 and 2.86 (at maximum height difference).

Now how will you work out the KEs?

9. Aug 16, 2008

### Zukie91

so am i going to be using
1/2mv^2i + mghi = 1/2mv^2f + mghf
i'm not sure when i should use which mass, or if i should add/subtract them

10. Aug 16, 2008

### tiny-tim

ah … KE + PE = constant, so if you increase the KE from 0, you must decrease the PE … so it's minus.

11. Aug 16, 2008

### Zukie91

what is minus?
and how does this only have to do w/ pe and ke, and nothing to do with rotation, if its rotating around an axle?

12. Aug 16, 2008

### tiny-tim

The gh has a minus in front of it: m(v2/2 - gh).

And the question tells you to model the masses as particles, so all you need to know is their speeds.

13. Aug 16, 2008

### Zukie91

were you able to get the correct answer with this method, because i think i get what you are saying, and cannot get the problem right using any combination of m and h

14. Aug 16, 2008

### Zukie91

ok i don't think i'm really getting anywhere, would you mind setting up the eq, and i'll try to figure out what i was doing wrong from there?

15. Aug 17, 2008

### tiny-tim

Hi Zukie91!
Been asleep … :zzz:

I've checked the answer, and 24.5 is roughly correct.

Hint: if the small mass has speed 24.5, then the large mass has speed 24.5 x 0.14/2.86 = 1.2

Does that help?

(if not, show us your calculations)

16. Aug 17, 2008

### Zukie91

still not getting it :( i'm closer but not right yet
i'm doing
deltaPE=deltaKE
mgh1-mgh2=1/2Mv12+1/2mv22
(60*9.8*.14) - (.12*9.8*2.86) + 30v12 + .06v22
78.96=30v12 +.06v22
then using the proportion v1/v2 = .14/2.87 i get that v2 = 20.5 v1
substituting that in for v2
i get 78.96=31.23v2
which simplifies to v=1.59 which when divided by .14/2.86 is 32.48 which is still wrong :(

17. Aug 17, 2008

### tiny-tim

Hi Zukie91!

Your 31.23 is wrong … you forgot to square.

18. Aug 17, 2008

### Zukie91

wait, what did i forget to square?

19. Aug 17, 2008

### tiny-tim

20.5

20. Aug 17, 2008

### Zukie91

oh, duh, thank you so much. really appreciate all your help