I finished reading the theory a
nd doing the exercises of HRK Chapter 10 today (Angular Momentum). I also started on the problems.
The coolest problem was a one about rolling without slipping.
Problem:
A cylinder rolls down an inclined plane of angle $\theta$. Show that the acceleration of the center of mass is $\frac{2}{3}g \sin \theta$.
Solution:
One obvious solution is to do this without using angular momentum, by making one $F=MA$ and one $\tau = I\alpha$ equation, and then solving for $a$.
However, since this is the angular momentum chapter, we will do that approach. It turns out that angular momentum makes this much simpler!
We will use $\tau_{\text{ext}}=\frac{dL}{dt}$ about the axis that goes through the bottom point. Note that since the cylinder is rolling without slipping, this axis is instantaneously at rest, which means that we can indeed use this equation in this case. The only torque acting with respect to this axis is the gravitational torque, since the friction and normal force torque have a moment arm of $0$. This torque is $$\tau_{\text{ext}}=(Mg \sin \theta)R$$ Now, it's well-known property that the angular velocity around the bottom point is the same as the angular velocity around the CM. Hence, $$L_{\text{bottom point}} = (I_{\text{bottom point}})(\omega)=\left(\frac{1}{2}MR^2 + MR^2 \right)(\omega)=\frac{3}{2}MR^2 \omega$$
$$\implies \frac{dL_{\text{bottom point}}}{dt}=\frac{3}{2}MR^2 \frac{d\omega}{dt}=\frac{3}{2}MR^2 \alpha$$
So, $$(Mg \sin \theta)(R) = \frac{3}{2}MR^2 \alpha$$
$$\implies g \sin \theta = \frac{3}{2}R \alpha$$
$$\implies \alpha = \frac{2}{3R} g \sin \theta$$
$$\implies a = \alpha R = \frac{2}{3} g \sin \theta,$$ as desired. $\square$
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