Problem finding the total kinetic energy of bicycle system

• tranvannhancu
In summary: Now, rotate the plane around the spin axis by an angle -phi. This will put the vector theta-dot*k in a new position. The component that was aligned with the spin axis will now be theta-dot*cos(phi) and the component that was perpendicular to the spin axis will now be theta-dot*sin(phi). In the now deleted post from CWatters, there was a quotation from a text (apparently) regarding the meaning if Ip and Ir. I would interpret these in this way:Ir = Mass moment of inertia with respect to a radial line (the same for all radial lines by symmetry)Ip = Mass moment of inertia with respect to the axis of rotationIn summary, the conversation discusses the derivation of a
tranvannhancu
I've been working on this problem for a while and here comes the problem: I'm building a balancing system for a bicycle using Gyroscopic effect (take a look at Figure 2 for more details). My system consists of 2 parts:
• the bicycle itself
• a flywheel on top of it
All parameters are given. The question is I have to derive formula for calculating total kinetic energy of the system. Figure 1 is the formula given in the article but I'm a bit confused about how to get it. I can understand the first 3 terms in the formula like this:

(1) translational kinetic energy of the bike's center of mass

(2) translational kinetic energy of the flywheel's center of mass

(3) rotational kinetic energy of the bike about its center of mass

Fig. 1

Since the last 3 terms in the given expression involve moment of inertia, I guess these are rotational kinetic energy of the flywheel about some reference which I don't know yet.

Please tell me how to get these last 3 terms. I would be very appreciated for your help.

Fig. 2

Reference: see attached PDF file

Attachments

• Balancing Control of Bicyrobo - Asia Institute of Technology.pdf
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I'm thinking about rotating frame of reference. Any ideas everyone ?

Sorry I deleted my original post because I was totally wrong. I miss read as φ as ω in a key place.

.

tranvannhancu
In the now deleted post from CWatters, there was a quotation from a text (apparently) regarding the meaning if Ip and Ir. I would interpret these in this way:

Ir = Mass moment of inertia with respect to a radial line (the same for all radial lines by symmetry)
Ip = Mass moment of inertia with respect to the axis of rotation

Does that help at all?

tranvannhancu
Dr.D said:
In the now deleted post from CWatters, there was a quotation from a text (apparently) regarding the meaning if Ip and Ir. I would interpret these in this way:

Ir = Mass moment of inertia with respect to a radial line (the same for all radial lines by symmetry)
Ip = Mass moment of inertia with respect to the axis of rotation

Does that help at all?

Yes and thank you for your reply. Then expression (4) seems like the rotational kinetic energy of the flywheel about gimbal axis which relates to the derivative of φ. But I'm still stuck at finding the meaning of expressions (5) and (6). Any ideas?

In the XYZ coordinate system, the roll angular velocity of the entire assembly is simply theta-dot*k, where k is a unit vector in the Z direction. Now, resolve that velocity component on the body coordinate system of the flywheel. There is an axial component theta-dot*sin(phi) and a component about a radial line in amount theta-dot*cos(phi). These are the two velocity components that go into making up terms 5 and 6.

tranvannhancu
The expression for T in post #1 seems to be lacking a term representing the spin velocity of the flywheel. Is it not spinning?

Assuming that the flywheel is spinning with angular velocity w relative to the frame, there should be a term of the form
(1/2)*Ip*(w +/- theta-dot*sin(phi))^2
where the +/- choice depends on the direction chosen positive for w. This is because w and theta-dot*sin(phi) are colinear angular velocties which should be added before squaring. At least, that is how it looks to me.

tranvannhancu
Dr.D said:
In the XYZ coordinate system, the roll angular velocity of the entire assembly is simply theta-dot*k, where k is a unit vector in the Z direction. Now, resolve that velocity component on the body coordinate system of the flywheel. There is an axial component theta-dot*sin(phi) and a component about a radial line in amount theta-dot*cos(phi). These are the two velocity components that go into making up terms 5 and 6.

First of all, can you show me the way to resolve this roll angular velocity into 2 components above: axial component and radial component

Dr.D said:
The expression for T in post #1 seems to be lacking a term representing the spin velocity of the flywheel. Is it not spinning?

Assuming that the flywheel is spinning with angular velocity w relative to the frame, there should be a term of the form
(1/2)*Ip*(w +/- theta-dot*sin(phi))^2
where the +/- choice depends on the direction chosen positive for w. This is because w and theta-dot*sin(phi) are colinear angular velocties which should be added before squaring. At least, that is how it looks to me.

Second of all, the flywheel is spinning with angular velocity w, indeed. But I think this missing term should be (1/2)*Ip*(w^2), which is the rotational kinetic energy of the flywheel about its center of mass.

In the global XYZ fram, the bicycle roll is described by the vector theta-dot*k where k is a unit vector in the Z-direction.

Now, in the vertical plane, resolve this vector into components along the spin axis and perpendicular to the spin axis to get, respectively
theta-dot*sin(phi) and theta-dot*cos(phi)

They are both in the vertical plane, but one is aligned with the spin axis will the other is normal to it and parallel to a radial line on the flywheel.

The components along the spin axis add together to get the net rotational velocity of the flywheel = (w +/- theta-dot*sin(phi)). They must be added before expressing the kinetic energy which depends on the net rotational velocity (the sum/difference).

If this is hard to follow, I suggest that you draw the appropriate vectors on your original diagram. Then I think it will be evident.

tranvannhancu

1. How is the total kinetic energy of a bicycle system calculated?

The total kinetic energy of a bicycle system is calculated by adding together the kinetic energies of all the individual components of the system. This includes the kinetic energy of the wheels, pedals, chain, and any other moving parts.

2. What is the formula for calculating the kinetic energy of a moving object?

The formula for calculating kinetic energy is KE = 1/2 * m * v^2, where KE is kinetic energy, m is the mass of the object, and v is the velocity of the object.

3. Does the weight of the bicycle affect its kinetic energy?

Yes, the weight of the bicycle does affect its kinetic energy. The greater the mass of the bicycle, the more kinetic energy it will have at a given speed.

4. How does the speed of the bicycle impact its kinetic energy?

The speed of the bicycle has a direct impact on its kinetic energy. As the speed increases, the kinetic energy also increases. This is because the formula for kinetic energy includes the velocity of the object squared.

5. Can the kinetic energy of a bicycle system be changed?

Yes, the kinetic energy of a bicycle system can be changed by altering the speed or mass of the bicycle. It can also be changed by adding or removing components of the system, such as adding a basket or removing a wheel.

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