Maximum speed of smaller mass in trebuchet with stiff rod and two masses

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Homework Help Overview

The discussion revolves around a physics problem involving a trebuchet with two point masses, one significantly larger than the other, and the goal is to determine the maximum speed of the smaller mass. The setup includes considerations of torque, moment of inertia, and energy conservation principles.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the use of moment of inertia and torque to find angular acceleration, while others suggest using potential and kinetic energy conservation instead. Questions arise regarding the treatment of the masses as point masses and the implications of torque direction.

Discussion Status

There is an ongoing exploration of different methods to approach the problem, with some participants providing guidance on simplifying the analysis by focusing on energy conservation rather than torques. The discussion reflects a mix of interpretations and attempts to clarify concepts related to the problem.

Contextual Notes

Participants note the lack of information regarding the shape and density distribution of the masses, which influences their treatment as point masses. There is also a discussion about the complexities introduced by calculating torque in this scenario.

semc
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Imagine a trebuchet with a stiff rod of 3m and neligible mass.Two masses 60kg and 0.12kg are at its end with the bigger mass 0.14m away from the pivot point. Find the maximum speed the smaller mass attain.

Alright the question is somewhat like this and i start of by finding the moment of inertia and the torque which gives me the angular acceleration but how do i continue from here?
 
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Hi semc! :smile:

They're point masses, so I wouldn't bother with moment of inertia …

just use (ordinary) PE + KE = constant, to find the speeds when the heavy mass is at the bottom. :wink:
 
Hi tim i don't understand why the two masses can be treated as point mass since the question doesn't specify that the masses have uniform density distribution. So if we are asked to find the velocity of the mass at a given specific time we should calculate the angular acceleration and from there find the velocity or is there a easier method?

I have another question: I learn that force cross d ,where d is the perpendicular distance from the point the force is acting to the pivot point, gives the torque cause by this force correct? So what does the direction of the torque tell us? Does it merely tell us which way the object is rotating or does it gives us information like the direction of the force? Ain't torque the tendency to rotate an object by a force?
 
The forces can be treated as point masses because we have to :smile:

It's indeed true that we don't know what the shape or density distribution of the masses is.
If the bigger mass would be an iron sphere it would have a radius of about 0.12 m so it certainly matters. The smaller mass can be treated as a point mass with very high accuracy.

Even if you assume that \frac{1}{m} \int r^2 dm of the larger mass is 0.14
you still wouldn't know what its centre of mass is, which you need for calculating it's gravitational potential energy.

In any case, it's much easier to only consider potential and kinetic energy and no forces or torques.
 
Hi semc! Hi willem2! :smile:
So what does the direction of the torque tell us? Does it merely tell us which way the object is rotating or does it gives us information like the direction of the force? Ain't torque the tendency to rotate an object by a force?

The direction of the torque is the axis of rotation, nothing more.
Hi tim i don't understand why the two masses can be treated as point mass since the question doesn't specify that the masses have uniform density distribution.

As willem2 :smile: says, there's no choice … we're not told the shape or size, so we have to assume they're point masses!
semc said:
I learn that force cross d ,where d is the perpendicular distance from the point the force is acting to the pivot point, gives the torque cause by this force correct?

Almost correct. :smile: it's the perpendicular distance from the line of the force to the pivot point. :wink:

Since the only force is gravity, and the line of that force is always vertical, that means the perpendicular distance in this case keeps changing, so the torque keeps changing …

so if you use torque, you'll have a very messy unnecessary integral. :frown:

As willem2 says, it's much easier in this case to only consider PE and KE and no forces or torques (or angular anything). :rolleyes:

Try it! :smile:
 
I understand now. Thanks for he help guys!:biggrin:
 

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