USAPhO 1998, Четвъртфинал — Задача 1

• (a, 4 points) $\mathbf v_0=R\omega_0\mathbf i$
• (b, 7) $x=R\omega_0\sqrt{2R/g}$
• (c, 7) $v_B=[(R\omega_0)^2+2gR]^{1/2}$
• (d, 7) $\omega\prime=\omega_0$

(a, 4) $\mathbf v_0=R\omega_0\mathbf i$ (in terms of the coordinate system shown) (4 points)

(b, 7) This is a projectile problem: with $t=0$ at the top of the disc, we have

$x=v_0t=R\omega_0t$ (2)

and $y=-\frac12gt^2$. (2)

At point B, $y=-R$, so that $t=\sqrt{2R/g}$ (2)

and therefore $x=R\omega_0\sqrt{2R/g}$ (1)

(c, 7) One may use either conservation of energy, or projectile kinematics.

Energy conservation approach:

$E_A=E_B$ (recognizing the principle) (3)

$\frac12m(R\omega_0)^2+mgR=\frac12mv_B^2+0$ (applying it) (2)

$v_B=[(R\omega_0)^2+2gR]^{1/2}$ (solving) (2)

Projectile kinematics approach:

$v_B^2=v_x^2+v_y^2$ (2)

where $v_x=R\omega_0$ (2)

and $v_y^2=2gR$ [from $v_y^2=v_{0y}^2+2a_y(y-y_0)$] (2)

and thus, $v_B=[(R\omega_0)^2+2gR]^{1/2}$ (1)

(d, 7) By conservation of angular momentum,

$L=L\prime$ (recognizing the principle) (2)

which becomes $I_0\omega_0=I\prime\omega\prime$ (applying it) (1)

Let’s calculate $I\prime$: it has two contributions, one for the damaged disc missing the chip, and one for the chip. When the chip comes loose,

$=I_0$. (3)

Hence, $\omega_0=\omega\prime$. (1)