Simplified Trebuchet Physics: Solving for Omega | Homework Equations

[timetraveller123]

Homework Statement

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.

Homework Equations

conservation of energy

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

E_i = E_f
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?

 

[mfb]

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.

 

[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 E_f 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.

 

[timetraveller123]

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

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

i don't 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 don't get what you mean by they are different angles

 

[ehild]

Homework Statement

View attachment 203463
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.

Homework Equations

conservation of energy

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

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.

 

[timetraveller123]

@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

v_m = ωL
v_M = ωH

am i right now?

 

[ehild]

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

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

 

[timetraveller123]

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.

 

[kuruman]

is this correct ?

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.

 

[ehild]

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

It is "Paint" drawing program of Windows.

 

[timetraveller123]

@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

 

[ehild]

@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

See figure, and check your heights.

 

[kuruman]

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

I did in post #9.

You need to add distance a (see the diagram posted by @ehild) to all the final vertical distances.

 

[timetraveller123]

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

 

[ehild]

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

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

 

[timetraveller123]

oh you forgot the cosθ

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

 

[timetraveller123]

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

 

[ehild]

oh you forgot the cosθ

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

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

 

[timetraveller123]

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

 

[ehild]

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

correct at last.

 

[timetraveller123]

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

 

[timetraveller123]

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

 

[ehild]

View attachment 203926

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

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

 

[timetraveller123]

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

 

[ehild]

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

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

 

[timetraveller123]

omg yes so careless i will just put 2 in the denominator of the first two terms in the denominator
and
sin φ as you said is H/L

 

[ehild]

Correct!