# Homework Help: Trebuchet physics problem

1. May 13, 2017

### vishnu 73

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

The trebuchet is a siege engine that was employed in the Middle Ages to smash castle walls or to lob projectiles over them. A simpliﬁed version of a trebuchet is shown in the following ﬁgure. A heavy weight of mass M falls under gravity, and thereby lifts a lighter weight of mass m. The motion of the mass M is blocked as shown in the ﬁgure, which launches the lighter mass m; the blockade forms an angle θ with the vertical. The mass of the blockade is much larger than all other masses. The shorter arm of the trebuchet is of length H , whereas the longer arm is of length l; the whole beam (both arms) are of mass µ.
2. Relevant equations
conservation of energy

3. The attempt at a solution
ok i did the solution but i got a super complicated expression for omega i will is it supposed to be that way? and is it possible to get a numerical value for omega from the given information? i have just described my solution in words if you want my actual equations i would gladly post thanks

x y a b are heights above ground
I moment of inertia
v translational velocity of masses

Ei = Ef
Mgx + μ(L+H)gy = Mga + μ(L+H)gb + 1/2 I ω2 + 1/2 Mv2 + 1/2 m v2

y = (L+H)/2 sin tan-1(H/L)
x = (L+H) sin tan-1(H/L)

a = H(1 - cosθ)
b = H + (L-H)/2 cosθ

I = 1/12 μ(L+H)3 + μ(L+H)(L-H)2/4

v of M : ω* H
v of m : ω * L

is this correct?

2. May 13, 2017

### Staff: Mentor

μ is given as mass of the beam, you shouldn't multiply it by (L+H) to get the mass.

The expressions can get a bit lengthy, yes.

3. May 13, 2017

### kuruman

I haven't worked out the problem (yet), but this item in your attempt attracted my attention. Expressions such as $\sin \left [ \tan^{-1}(H/L) \right ]$ are ugly and can be simplified. You are looking for $\sin \theta$ where $\tan \theta = H/L = opp/adj.$ Well,
$$\sin \theta = \frac{opp}{hyp}=\frac{H}{\sqrt{H^2+L^2}}.$$
Also,
$$\cos \theta = \frac{adj}{hyp}=\frac{L}{\sqrt{H^2+L^2}}.$$

On edit: Your expression for Ef does not include the potential energy of mass m.
On second edit: Angle $\theta$ mentioned in this post is not the same as angle $\theta$ given by the problem. It is the angle formed w.r.t. the horizontal by the arm in its initial position.

Last edited: May 13, 2017
4. May 13, 2017

### vishnu 73

@mfb
oh yah got confused with mass and linear mass density sorry about that

@kuruman
oh ya i forgot about potential energy of small mass
and simplification too

i dont understand what you mean by they are different θ they are the same in my working

because for a it is the height above the ground which H(1-cosθ) like a pendulum)

and for b it can be calculated similarly
and finally for small mass m the final height above ground is H + L cosθ
i dont get what you mean by they are different angles

5. May 14, 2017

### ehild

The heights of what and when are x, y, a, b?
I is moment of inertia about what axis?
v is linear velocity, but it is different for m and M. Use different symbol for them.

6. May 14, 2017

### vishnu 73

@ehild
sorry for the ambiguity was in a rush that day let me explain myself

x is the initial height of mass M above the ground
y is the initial height of centre of mass of rod above the ground
a is the final height of mass M above the ground
b is the final height of centre of mass of rod above the ground
new variable i forgot to include
c is the final height of mass m above the ground

I is the moment of inertia about where the rod is pivotted i had used the parallel axis theorem

vm = ωL
vM = ωH

am i right now?

7. May 14, 2017

### ehild

H/L=sin(φ), x=(L+H)sin(φ) and y=sin(φ)(L+H)/2. What are the initial and final potential energies?
Initial state:

Final state

Last edited: May 14, 2017
8. May 15, 2017

### vishnu 73

initial potential energy :
μ sinφ (L+H)/2 g + M g (L+H) sinφ

final potential energy
μ (H + (L-H)/2 )cosθ g + M g ( H - H cosθ) + mg(H + Lcosθ)

final kinetic energy
I ω2 + 1/2 M (ω H)2 + 1/2 m (ωL)2

is this correct ?

edit :
what did you use to draw that thanks next time it would be helpful for me to post diagrams thanks.

9. May 15, 2017

### kuruman

What is your zero of potential energy? If it is the ground, then the initial PE is correct. The final PE is not. You need to add distance a (see the diagram posted by @ehild) to all the final vertical distances.

10. May 15, 2017

### ehild

It is "Paint" drawing program of Windows.

11. May 16, 2017

### vishnu 73

@kuruman
yes my zero potential energy is ground
but i don't see what height i need to add as far as i know i have accounted for all the heights
please point out to me

@ehild
thanks

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12. May 16, 2017

### ehild

See figure, and check your heights.

13. May 16, 2017

### kuruman

I did in post #9.

14. May 17, 2017

### vishnu 73

a = H - Hcosθ

M is a above ground

centre of rod is
H + (L-H)/2 above the ground

the small m is Lcosθ + H above the ground
i still don't see why you need to add another a to the heights

15. May 17, 2017

### ehild

a and c are correct, but b is not. The angle is missing.

16. May 17, 2017

### vishnu 73

oh ya forgot the cosθ

but isn't this the heights i have i been saying since the start so is it correct? thanks!

17. May 17, 2017

### vishnu 73

just for clarification i drew this please don't mind my ugly drawing skills thanks!
is this corrects?

18. May 17, 2017

### ehild

The drawing is all right. So what is b? Where is cos(theta)?

19. May 17, 2017

### vishnu 73

cos theta is the vertical height from edge of blockade to b
that is b is
H + (L-H)/2 cosθ

20. May 17, 2017

### ehild

correct at last.

21. May 18, 2017

### vishnu 73

thanks i will do the problem now and post result for ω soon

22. May 21, 2017

### vishnu 73

this is the final expression i am getting for omega is it correct please don't mind because i am not good with latex don't how to type such a long expression in latex

23. May 21, 2017

### ehild

The kinetic energy of the beam is 1/2 I ω2. It seems to me that you forgot the factor 1/2.

24. May 21, 2017

### vishnu 73

oh ya other than is the rest correct i just multiply the numerator by 2

25. May 21, 2017

### ehild

No, the kinetic energy of m and M were correct.
And express sinΦ with the given quantities H and L.