USAPhO 2000, Четвъртфинал — Задача 3

• a. $\omega=\sqrt{3g/L}$
• b. $\omega_F=\frac14\sqrt{3g/L}$
• c. 0.41 rad

a. The rotational inertia of a uniform thin rod about its end is found using the parallel axis theorem:

$I=I_{cm}+M(L/2)^2=(\frac1{12})ML^2+(\frac14)ML^2=(\frac13)ML^2$ [2]

Using energy conservation: $K_0+U_0=K_F+U_F$ [2]

Let the lowest position of the rod’s center of mass ($L/2$ below the axis) be the zero of gravitational potential energy. Then $U_0=Mg(L/2)$ and $U_F=0$ [1+1]

For kinetic energy: $K_0=0$, $K_F=(1/2)I\omega^2$ [1+1]

Substitution into the energy conservation equation yields

$Mg(L/2)=(\frac12)(\frac13)ML^2\omega^2$. [1]

Dividing by $M,L$, and $1/2$

$g=(\frac13)L\omega^2$

or $\omega=\sqrt{\frac{3g}{L}}$ [1]

b. The collision is inelastic. Using angular momentum conservation

$I_0\omega_0=I_F\omega_F$ [2]

Only the rod is moving initially, so $I_0=(\frac13)ML^2$ and $\omega_0=\sqrt{\frac{3g}{L}}$ [1+1]

The final rotational inertia is the sum of the rod’s and point mass’s

$I_F=(\frac13)ML^2+ML^2=(\frac43)ML^2$ [2]

Therefore:

[2]

c. Once the collision is over, mechanical energy is again conserved. Equating total mechanical energy immediately after the collision and at the highest point

$K_0+U_0=K_F+U_F$

At the highest point $K_F=0$ [1]

[1]

Choosing the lowest point as the zero of gravitational potential energy, $U_0=0$.

In swinging to an angle $\theta$, the center of mass of the rod (marked by an x in the diagram) has moved up a distance $h'=(L/2)(1-\cos\theta)$, [1]

While the point mass at the end has moved up $h=L(1-\cos\theta)$. [1]

Therefore the final potential energy is

[1]

Substituting into the energy conservation equation

$(\frac18)MgL=(\frac32)MgL(1-\cos\theta)$ [1]

$(\frac1{12})=1-\cos\theta$

$\theta=\cos^{-1}(\frac{11}{12})=0.41\ rad$ [1]