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I have a question behind the conceptual understanding of the following equation: $$\frac{\text{d}}{\text{d}t}\mathbf{L}_G = \sum_i \ \mathbf{r}_i\times \mathbf{f}_i$$ where $\mathbf{L}_G$ is the total angular momentum of a rigid body about the center of mass $G$, $\mathbf{r}_i$ is the position vector of a point part of the rigid body taken from $G$ and $\mathbf{f}_i$ is the external force at that point.

Suppose we have a cylinder rolling down that hill as shown on pages 86-87 here. Why is it the case that $C\ddot{\theta} = aF$ as shown on page 87? How does one leverage the formula above to arrive at this equation? Conceptually speaking, one sums the moment of the forces about the center of mass for each point over the entire body, so what exactly are the forces (i.e. what are $\mathbf{f}_i$?) at various points on the cylinder (gravity, reaction force, etc.)?

Thank you for any help.

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  • $\begingroup$ In the frame where $G$ is at rest, only one force ($F$) acts on the cylinder with a non-zero radius. Gravity and the normal force from the ramp act through the center of mass. $\endgroup$ – BowlOfRed Jun 2 '16 at 23:47
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The $\vec f_i$ are the external forces acting on the cylinder, i.e. gravity, reaction force and friction force.

The reaction and the friction act at a single point, the contact with the hill. The torque about the center of the disc due to the reaction force is zero since the force is anti-parallel to the lever arm. The torque due to the friction $F$ is simply $aF$ since it is perpendicular to the lever arm $a$.

For the weight you can imagine dividing the body into many small pieces of mass $\Delta m_i$, with positions $\vec r_i$ with respect to the center of mass. The total torque due to these small weights is $$\sum_i \vec r_i\times \Delta m_i\vec g=\left(\sum_i\Delta m_i\vec r_i\right)\times\vec g=M\vec R_{cm}\times\vec g,$$ where $\vec R_{cm}$ is the center of mass coordinate. Whenever the gravitational field is constant, we can consider all the mass of the body in its center of mass when calculating the torque due to weight. This torque vanishes since we are calculating it about the center of mass.

Therefore we are left with $$\frac{d\vec L}{dt}=aF\hat k.$$ Since $\vec L=C\dot\theta\hat k$ you get the result.

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  • $\begingroup$ Thanks for your reply, made good sense. However, I have a question about splitting the body into small pieces. For each of these pieces, is the only external force on them gravity? What about the reaction force on each piece from the pieces around it to keep it "rigid"? Is that considered an internal force and ignored? $\endgroup$ – user334263 Jun 3 '16 at 13:27
  • $\begingroup$ @user334263 There is a force acting on the piece $\Delta m_i$ due to the neighbor $\Delta m_j$ resulting in an inner torque $\vec\tau_{ij}$. But then when you go to the piece $\Delta m_j$ there is another inner torque $\vec\tau_{ji}$ due to the contact force coming from $\Delta m_i$. Using the (strong form of the) third Newton's law you show these torques cancel each other out. The bottom line is that resultant of the inner forces and torques vanish. Otherwise we would be lost having to calculate forces and torque due to every single molecule. $\endgroup$ – Diracology Jun 3 '16 at 13:34
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Cylinder on an incline.

Why is it the case that $C\ddot{θ}=aF$ as shown on page 87? How does one leverage the formula above to arrive at this equation? Conceptually speaking, one sums the moment of the forces about the center of mass for each point over the entire body, so what exactly are the forces (i.e. what are $f_i$?) at various points on the cylinder (gravity, reaction force, etc.)?

There's really no need to invoke:

$$\frac{\text{d}}{\text{d}t}\mathbf{L}_G = \sum_i \ \mathbf{r}_i\times \mathbf{f}_i$$

... for this problem.

We know there's only one force that can produce torque about $\text{G}$ and that is $F$. The torque generated is $\tau=F \times a$. Using Newton's second law for acceleration, applied here to rotation, then gives us:

$$C\ddot{\theta}=Fa$$

where $C$ is the inertial moment of the cylinder about the axis $\text{G}$ and $\ddot{\theta}$ the angular acceleration.

But if there was a set of torques caused by couples $(r_1,f_i)$ acting on the cylinder then we would write:

$$C\ddot{\theta}=\sum_i \ \mathbf{r}_i\times \mathbf{f}_i$$

The various forces $f_i$ can be any forces that apply to your specific real world problem.

Remember that:

$$\frac{\text{d}}{\text{d}t}\mathbf{L}_G=\frac{\text{d}}{\text{d}t}\mathbf{C\dot{\theta}}=C\ddot{\theta}$$

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