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

@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.