# I Why don't we include the kinetic moments in the summation of the moments of a rigid body?

Tags:
1. Apr 14, 2017

### zachdr1

Why don't we include the kinetic moments in the summation of the moments of a rigid body? In other words, why are the kinetic moments on the opposite side of the equation as the moments due to external forces? I understand that the math works out but it's not making sense to me conceptually.

The equation I'm talking about is

ΣM = Σm + Iα

2. Apr 16, 2017

### Andrew Mason

I am not sure what you mean by the "kinetic moments". In mechanics, a torque is sometimes referred to as a moment of force. The vector sum of all torques on a body is equal to its moment of inertia x its angular acceleration: $\sum{\vec{M}} = I\vec{\alpha}$ . The use of the term "moment" in this way can be very confusing, because "moment of inertia" is not a torque. Similarly, we speak of magnetic dipole moments and electric dipole moments, which are not torques either. The torque experienced by an electric or magnetic dipole is the cross product of the dipole moment and the field.

AM

Last edited: Apr 16, 2017
3. Apr 16, 2017

### zachdr1

This is only true if the net torque is calculated about the center of mass on a rigid body. If you are calculating the moment of an accelerating rigid body about a point other than the center of mass, you have to include the moment that maG (where G is the center of mass) makes about that point, this moment is what I'm talking about when I say "kinetic moment". The part that I don't understand is why it is not just included in the summation of the moments on the left side of the equation ΣM = ΣMk + Iα where ΣMk is the summation of the kinetic moments.

4. Apr 16, 2017

### Andrew Mason

The equation $\sum{\vec{M}} = I\vec{\alpha}$ assumes that the torques and moment of inertia are about the same axis (ie. the axis of rotation).

Can you provide an example of where the kinetic moment is used this way? If $\sum{\vec{M}}$ is the vector sum of the torques about an axis other than the centre of mass and $I$ is the moment of inertia about a parallel axis through the centre of mass, one simply applies the parallel axis theorem (which effectively adds to the moment of inertia about the c.om. the moment of the c.o.m. about the axis of rotation).

AM

5. Apr 16, 2017

### zachdr1

Take a look at this example.
http://imgur.com/a/Gfxq5

and the solution
http://imgur.com/a/8B7Rq

And are you sure about being able to apply the parallel axis theorem to that equation? You would get ∑Mp = (I + md2)α; that just doesn't seem right to me.

Edit: Oh and I just realized that Iα is included in the kinetic moments, so its actually just ∑M = ∑Mk, not ∑M = ∑Mk + Iα

Last edited: Apr 16, 2017
6. Apr 16, 2017

### Andrew Mason

In this problem, in order for the motorcycle to do a wheelie, the net torque about the centres of mass of the rider and motorcycle must not be in the clockwise direction. At that point, there is no normal force on the front wheel so the torque is provided by two forces: the normal force on the rear wheel (upward) and the horizontal force between the rear wheel and the road. The moments of these two forces about the centres of mass must sum to 0. At that point, there is no angular acceleration. Moment of inertia does not matter here because we are not concerned with finding the rate of angular acceleration.

The thing you refer to as a "kinetic moment" appears to be the moment of applied force about the centres of mass. ∑M = ∑Mk is just another way of saying that the clockwise torque must be equal in magnitude to the counter-clockwise torque.

AM

7. Apr 16, 2017

### zachdr1

In the example they never summed the moments of the applied forces about the centers of mass though? The values they use for the kinetic moments are m1aG1rB and m2aG2rB

8. Apr 16, 2017

### Andrew Mason

Yeah. That appears to be a mistake. The force of gravity is mg where g = 9.8m/sec2 and the horizontal force between the road and the rear wheel is max where ax is the horizontal acceleration. (m is the combined mass of the rider and motorcycle).

AM

9. Apr 16, 2017

### zachdr1

So you're saying the book made a mistake? This whole chapter is about how when a rigid body is accelerating, you have to account for the kinetic moments which are the moments of maG about whatever point you're summing the moments about + Iα

Here is the diagram containing the kinetic forces that cause the kinetic moments needed to solve the problem

http://imgur.com/SIp8cp9

Last edited: Apr 16, 2017
10. Apr 16, 2017

### Andrew Mason

You will have to clarify what aG is. I was asssuming it was g, the freefall acceleration due to gravity.

The concept of "kinetic moment" seems a bit awkward. The force moments are not related to motion (kinetic) but to change in motion. The force of the road on the tire accelerates the motorcycle and rider and, by the third law, the motorcycle and rider exert exerts an equal and opposite force on the road. Those force pairs create a torque.

There is more than one way to analyse the torques here. If one analyses the torques about the c.o.m., the force moments are the normal force on the rear tire about the c.o.m. and the friction force of the road on the rear tire about the c.o m. If one analyses the torques about the point of contact between rear tire and road, the force moments are gravity applied at the centre of mass and the rearward force of the motorcycle and rider about the point of road contact.

AM

11. Apr 16, 2017

### zachdr1

Oh, aG is the acceleration of the center of mass.

And yes I understand how to calculate moments and torque, it's just summing the moments about a point other than the center of mass or the rotational axis when considering an accelerating rigid body that I am confused about.

12. Apr 17, 2017

### Andrew Mason

The term "kinetic moment" is not exactly a standard term used in mechanics. It may be in engineering. Is this an engineering text?

Based on what you have given us so far, "kinetic moments" occur if the centre of mass is accelerating due to a force being applied at a point whose displacement from the centre of mass does not have the same direction as the direction of the force. This causes an opposite torque about the point of application of that force due to inertia - i.e. the third law "reaction" force is always opposite to the applied force that causes the acceleration. So to find the total torque on the accelerating body about a particular point, you must sum the external force moments about that point and add the sum of the kinetic moments about the same point. Since the kinetic moments are going to have the opposite direction to the applied force moments, you could write this as: ∑M - ∑mk = Iα or ∑M = ∑mk + Iα, where ∑M is the sum of all applied force moments, assuming there are no other external force moments. Note: if there are external force moments other than from the applied force that is causing the c.o.m. to accelerate, you will have to add those in separately. Some of those moments may operate in the same direction as the kinetic moments, so it is better to write the equation using vectors, where the directions are all taken into account: $\sum{\vec{M}} + \sum{\vec{m_k}} = I\vec{\alpha}$ where $\sum{\vec{M}}$ is the vector sum of all external force moments excluding the kinetic moments.

AM

Last edited: Apr 17, 2017