Question

The length of the flat 2.4 kg rod is 1.2 m. The rod is attached the other end with a joint so that it can rotate freely. In the initial situation, the rod is held stationary horizontally. a) what is the angular acceleration of the rod at α release? b) what are the rods tangential accelerations of tip and center of gravity at? c) if the rod tip would there be a coin free, so what would happen to it at the time of release?

Answer 1

a) Take total initial mechanical energy relative to the lowest possible position of the rod with length $L$. Obviously, it is the potential energy only:

$$ME_\text{i}=mg\frac{L}{2}.$$

Total final mechanical energy is the kinetic energy of the rotating rod with moment of inertia $J$:

$$ME_\text{f}=\frac{1}{2}J\omega^2=\frac{1}{2}\cdot \frac{1}{3}mL^2\cdot \omega^2.$$

Law of conservation of energy states:

$$ME_\text{i}=ME_\text{f} \iff mgL=\frac{1}{3}mL^2\omega^2,$$ $$\omega=\sqrt{\frac{3g}{L}}.$$

Meanwhile the rod passed an angle $\phi=\pi/2$, and the angular acceleration is

$$\alpha=\frac{\omega^2}{2\phi}=\frac{3g}{L\pi}=\frac{3\cdot 9.8}{1.2\cdot 3.14}=7.8\text{ s}^{-2}.$$

b) The tangential acceleration of the tip:

$$a=\omega^2L=3g=3\cdot9.8=29.4\text{ m/s}^2,$$

of the center of mass:

$$a_\text{cm}=3g/2=1.5\cdot 9.8=14.7\text{ m/s}^2.$$

c) The rod would just fall down parallel to the ground if no other forces influence it.