What is the Velocity of a 2-Wheeled Vehicle with Weight on Top?

In summary: Once you have that, you can use the basic velocity equation (v = u + at) to calculate the linear kinetic energy of the system (E). The rotational kinetic energy will be equal to the sum of the individual rotational inertias of the individual parts (KE = I(R1+R2+...+Rn)). Hope that makes sense.
  • #1
henry1117
8
0
Basically, I need to find the mathematical modelling of a 2 wheeled vehicle if you like, that has a weight positioned at the top of the vehicle but inside the wheels if that makes sense. Essentially, when you let go of the wheel, the vehicle starts spinning until around half a revolution where the weight drops off and the vehicle continues to go.

This is actually an engineering project but since I thought that it's more of a physics thing, I decided to post it here. So if anyone could help me with this and help me find the acceleration of the vehicle before the weight drops off and then the velocity of the vehicle at the moment which the weight drops off, that would be really good thanks.
 
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  • #2
Thread moved to the Homework Help forums.

Welcome to the PF, henry. What do you think the relevant equations are for this project? What are the equations for angular acceleration? What would a free body diagram look like for the weight, as it starts at the top and descends to the bottom of the wheel as it accelerates the vehicle?

You need to show us some of your work before we can offer any hints or tutorial help.
 
  • #3
i actually have no idea. In class we've only learned really basic physics type stuff. I know all the velocity equations like v = u +at and the 2 other ones like that and i know the basic inertia equation but i haven't really learned about this so i was wondering if you could give me a starting point at least.

For the free body diagram, I suppose that it would just be a semicircle in the shape of the letter 'C'
 
  • #4
I would approach this problem using conservation of energy: the change in gravitational potential energy of the block is equal to the change in kinetic energy of the (block+vehicle) system. But it's a bit more complicated than it looks because you need to include both the rotational kinetic energy and the linear kinetic energy of the system. This problem will take some time to work out.

First, try to relate the vehicle's linear velocity to its angular velocity (in radians/second). Do you know how to do that? Get a formula for the linear velocity in terms of angular velocity and the radius of the wheels.

Then it's pretty easy to get a formula for linear kinetic energy. Put this in terms of the angular velocity (it'll keep the problem simpler).

Then for the rotational kinetic energy. From your description it sounds like the system is made up of four components, each of which will have to be treated separately: two wheels (you can treat these as a single cylinder), one axle (a smaller cylinder), one weight (treat this as a point mass), and one whatever-it-is you use to connect the weight to the axle (probably a rod with a pivot point at one end). You'll want to look up the formulas for rotational inertia for each of these types of objects, and use those to obtain a formula for the rotational kinetic energy of each.

The sum of all those kinetic energies is going to be equal to the change in gravitational potential energy of the weight (by conservation of energy). So you're going to need to look up the formula for gravitational potential energy. Set up your equation and solve for angular velocity (it should be the only unknown in the whole big equation). Using the equation you derived first, relate that angular velocity to the linear velocity you're trying to find.

The acceleration involves some more computations. Work on velocity first.

(By the way, if you were given any simplifying assumptions for the problem, you might want to post those. There could be an easier way to attack it if you're allowed to ignore some things.)
 
  • #5
Sorry if i didn't make it clear but the vehicle is at a standing start and essentially uses the weight to make the vehicle move. So I'm not entirely sure that I can get a velocity.

Also, this isn't really a question it's more of a research type thing for a project before I make the vehicle which is why I'm rather clueless about it.

And i applogise for my lack of knowledge in the area
 
  • #6
Lack of knowledge is okay :) It just looks like you're trying to do something that's a little beyond your physics level. But you can do it.

re: velocity, it will have one when the weight drops off, won't it?

Your first step will have to be finding a general formula for the vehicle's linear velocity (v) (how fast it moves across the table) in terms of its angular velocity (how fast the wheels are turning at the time). The symbol for angular velocity is omega, and it's expressed in units of radians per second. Think about how far the vehicle moves across the table when its wheels turn 1 revolution (= 2*pi radians).
 
  • #7
ok thanks.

wouldnt linear velocity = angular velocity x (Pi x D)

where D is the diamater of the wheel.
 
  • #8
That's right if you're measuring angular velocity in revolutions/second, but for most physics we use radians/second, so the factor of 2 pi is canceled out and we have:
linear velocity = angular velocity x r
v = r * omega

Now, you need to find the linear kinetic energy of your vehicle when it's rolling at a certain angular velocity. The equation for linear kinetic energy is KE = 1/2 m*v^2 where m is mass and v is linear velocity. Using the equation relating v to r and omega (above), put the kinetic energy equation in terms of m, r, and omega. Note that the mass here refers to the mass of the entire system, including the weight.

Your next step will be to look up the formulas for rotational inertia (I) of the following shapes:
- a cylinder or disc rotating around its central axis
- a point mass
- a rod pivoting around one end

You'll also want to figure out a formula for the change in gravitational potential energy of your weight as it moves from the "up" position to the "down" position (note that this change in energy is completely independent of the path it takes, so you can just do your calculations as if the weight were falling straight down the same vertical distance).
 
  • #9
ok thanks for all that.

So substituting the variables for linear kinetic energy i get KE = 1/2 x m x (r x Omega)^2

And the inertia formulas are:

Inertia of a point = mr^2
Inertia of a disk = ½ x mr^2
Inertia of the end of a rod = 1/3 x mL^2

Ok, now for the change in gravitational potential energy, I wasn't too sure about what was required but at the top, isn't the potential energy MGH and at the bottom the potential energy, because height is 0, total PE is also 0?
 
  • #10
That's right. And yes, the change in gravitational potential energy is mgh, where h is the vertical distance from the "up" position to the "down" position.

Now, the formula for rotational kinetic energy of an object is 1/2 I omega^2, where I is the rotational inertia. So you can get expressions for the rotational kinetic energy of each element of your system by just substituting in the formulas you found for I.

So you need to set up an equation that should be in this form:
(linear kinetic energy of the whole system) + (rotational kinetic energy of the wheels) + (rotational kinetic energy of the axle) + (rotational kinetic energy of the rod holding the weight) + (rotational kinetic energy of the weight) = (change in gravitational potential energy of the weight).

In the above equation, you'll have probably four different masses (the mass of the whole system is the sum of all its parts) and three different radii and a length and a height. Be sure to use meaningful labels for them so you can keep them straight.

Once you have that big equation set up, you're going to have to solve it for omega, which shouldn't be terribly hard (you should be able to factor out an omega^2 from all the terms on the left side). Once you've solved it for omega, you'll have a formula for the angular velocity of the vehicle at the moment the weight drops off in terms of all the physical characteristics of your vehicle. You can take the angular velocity and plug it back into the first equation (v = r*omega) to get the linear velocity, which is what you're interested in.

On the acceleration: If you think about the physical situation a bit, you should be able to come up with the acceleration right before the weight drops off without doing any calculations. Now, if you want the acceleration as e.g. a function of time, that's more complicated (but might be doable - I have to think about it).
 
  • #11
ok that was great. i should be fine from there on.

with the acceleration, i don't think that i'll actually need that.

just another question, is there anyway other measure which i could use to compare this vehicle with another vehicle that would be useful? like i can compare the velocity at which the weight falls of each of the vehicles but is there anything else i could do?

also, does the final velocity answer take into account that the weight falls off after it reaches the bottom of the vehicle because that would effect the mass of the system.

also, is it possible to work out the coefficient of kinetic friction mathematically?
 
  • #12
No problem :)

re: comparing it with other vehicles, I'm not sure what you're thinking of comparing. Top speed? Time to get to that speed? Max power output? Total power output? Energy efficiency? (this would have to be experimentally-tested)

Re: velocity, the final velocity right before the weight drops off is, for all intents and purposes, the same as the velocity right after it drops (there might be a negligible difference due to effects of friction, but in the idealized world where you do your computations, the velocities are equal). Basically, when the weight drops off, it takes its component of the energy with it, and the rest of the system keeps on doing what it was doing unaffected.

re: coefficient of friction, no, that's an experimentally determined quantity that depends on the characteristics of the surfaces involved. However, for an idealized universe, we model wheels as rolling without slipping, so it's actually static (not kinetic) friction that's involved - and when the wheels are rolling without slipping, they don't lose energy. You can leave friction out of your preliminary calculations.

Just make sure you design your wheels so that they have enough static friction that they don't slide when you're vehicle starts rolling - one possibility is that you could make "tires" for them out of wide rubber bands. (In the real world there's always some slippage and some sticking, so energy is lost, but the extent to which that happens is something you have to figure out experimentally. If you do things right it will be pretty negligible.)
 
  • #13
ok a new problem...

basically now I'm looking at a normal 4 wheeled vehicle with some string attached to one axel which then goes up to the top of the vehicle on a pully where a weight sits at the top.

Same as the question before, could you give me some pointers on where to start off the find the velocity? I'm guessing i'ld need to look at angular velocity again
 

Related to What is the Velocity of a 2-Wheeled Vehicle with Weight on Top?

1. What is circular acceleration?

Circular acceleration is the rate of change of the tangential velocity of an object moving in a circular path. It is measured in meters per second squared (m/s^2).

2. How is circular acceleration different from linear acceleration?

Circular acceleration is specific to objects moving in a circular path, while linear acceleration applies to objects moving in a straight line. Circular acceleration also has two components - tangential and centripetal - while linear acceleration only has one direction.

3. What is the formula for calculating circular acceleration?

The formula for circular acceleration is a = v^2/r, where a is the circular acceleration, v is the tangential velocity, and r is the radius of the circular path.

4. How does circular acceleration affect objects in motion?

Circular acceleration causes objects to constantly change their direction of motion, even if their speed remains constant. This is because the direction of the tangential velocity is constantly changing due to the circular acceleration.

5. What is the real-life application of circular acceleration?

Circular acceleration is important in many real-world situations, such as in the motion of objects in a circular orbit, such as planets around the sun. It is also used in amusement park rides, such as roller coasters, to create thrilling and dynamic experiences for riders.

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