What Is the Maximum Speed of a Trebuchet's Payload?

[Zukie91]

Homework Statement

1. Homework Statement
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.
Image: http://www.grabup.com/uploads/e7490c945283118e547daa00676d2d28.png
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

 

[tiny-tim]

Welcome to PF!

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

Hi Zukie91! Welcome to PF!

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

 

[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
https://www.physicsforums.com/showthread.php?t=200958

 

[tiny-tim]

… 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 …

(LaTeX isn't working at the moment. )

The Iw² part isn't relevant.

The mv² part is.

 

[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

 

[tiny-tim]

do i need to use torque?

No … just KE + PE = constant.

and how do i use pe if i don't have the height?
oh wait, would the 14 cm be the height?

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?

 

[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?

 

[tiny-tim]

… oops!

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?

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?

 

[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

 

[tiny-tim]

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

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

 

[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?

 

[tiny-tim]

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?

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.

 

[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

 

[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?

 

[tiny-tim]

Hi 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

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?

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)

 

[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 :(

 

[tiny-tim]

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

Hi Zukie91!

Your 31.23 is wrong … you forgot to square.

 

[Zukie91]

wait, what did i forget to square?

 

[tiny-tim]

20.5

 

[Zukie91]

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