8/8/2023

Today I continued HRK Chapter 9 and did problems 9 through 17. Hopefully I can get 18 and 19 in before going to bed today. (edit: lol i didn't) If not, I plan to finish this chapter tomorrow.

The only problem I learned from today was #17.

Problem:

 Figure 9-65 shows two blocks each of mass $m$ suspended from the ends of a rigid weightless rod of length $L_1+L_2$, with $L_1=20\text{cm}$ and $L_2=80\text{cm}$. The rod is held in the horizontal position shown in the figure and then released. Calculate the linear accelerations of the two blocks as they start to move.

My Solution: 

Let the tension in the two strings be $T_1,T_2$, and the accelerations of the two blocks be $a_1,a_2$, respectively. Now, we can make two F=MA equations:

$$T_1-mg=m{a}_1(1)$$

$$T_2-mg=m(-a_2)(2)$$ Also, note that the angular acceleration of the two blocks will be the same, and $\alpha =\frac{a}{R}$, so we have $$\frac{a_1}{L_1}=\frac{a_2}{L_2}(3)$$

We have four variables, and we only have three equations as of now. We need one more constraint.

Note that the problem says that the rod is weightless, which implies that it has $0$ rotational inertia. This means that the net torque on the rod must be $0$, according to $\tau =I\alpha$. This fact gives us our last equation:

$$T_1{L}_1=T_2{L}_2(4)$$

Solving these four equations (although it may seem so, these four equations aren't hard to solve), we get $$a_1=\frac{g(L_2-L_1)(L_1)}{L_1^2+L_2^2}=\boxed{{\displaystyle 1.73\frac{\text{m}}{{\text{s}}^2}}}$$

Plugging this into $(3)$, we get$$a_2=\boxed{{\displaystyle 6.93\frac{\text{m}}{{\text{s}}^2}}}$$

I think the solution I did ALWAYS works on these type of problems. In retrospection, though, there was a simpler solution in this case:

Simpler Solution: 

The net torque on the system is $$\tau =mg{L}_2-mg{L}_1$$ and the rotational inertia of the system is $$I=m{L}_1^2+m{L}_2^2$$ Hence, the angular acceleration of the whole system (this includes both blocks) is $$\alpha =\frac{\tau }{I}=\frac{mg{L}_2-mg{L}_1}{m{L}_1^2+m{L}_2^2}=\frac{g(L_2-L_1)}{L_1^2+L_2^2}$$ Thus, the linear accelerations of the blocks, using $a=\alpha R$, are $$a_1=\frac{g(L_2-L_1)(L_1)}{L_1^2+L_2^2}$$ 

$$a_2=\frac{g(L_2-L_1)(L_2)}{L_1^2+L_2^2}$$ which give us the same answers as before.