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USAPhO 1999, Полуфинал — Задача 5. B1

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1999 Semi-Final Exam

Условие

Two diagrams of an upside-down V made of two rods of length L, each at angle θ to the horizontal. Case 1: the left end is hinged to a vertical wall/fixed point at the bottom and the right end rests on the horizontal surface. Case 2: both ends rest on the horizontal surface.

Consider two uniform rods each of mass $M$ and length $L$ , hinged together to form an upside-down “vee” shape whose angle can vary. The rods are released from rest when their angle θ with the horizontal is 45°. All hinges are frictionless and have negligible mass.

Case 1: The left end is also hinged at a fixed point at the bottom. The right-hand end of the configuration slides on the horizontal surface without friction.

Case 2: Both ends can slide frictionlessly along the surface.

(The sub-parts of this problem may be worked in any order.)

a. Find the upward force $F_N$ that the horizontal surface exerts on the right hand side of the Case 1 rods just after release, when θ = 45°. [10 т.]

b. Repeat Part (a) for Case 2. [10 т.]

c. Find the angular velocity of the Case 1 rods as a function of θ. $0 < \theta < 45^{\circ}$ [10 т.]

d. Repeat Part (c) for the configuration of Case 2. [10 т.]

(contributed by Leaf Turner)

Отговори

Решение

B1a. For θ = 45°, both $\cos\theta = 1/\sqrt{2}$ and $\sin\theta = 1/\sqrt{2}$ . The relation between the magnitude of the angular acceleration α and the magnitude of the linear acceleration a is $a = \alpha L/2$ with $a_x = a_y = \alpha L/\left(2\sqrt{2}\right)$ . Using Iα = Στ for the rotation about the hinge at the lower end of the left rod

\frac{ML^2\alpha}{3} = \frac{MgL}{2\sqrt{2}} - \frac{VL}{\sqrt{2}} - \frac{HL}{\sqrt{2}} \qquad (a.1)

Applying Newton's second law to the right rod

Ma_x = \sum F_x \qquad \frac{M3\alpha L}{2\sqrt{2}} = H \qquad (a.2)

Ma_y = \sum F_y \qquad \frac{M\alpha L}{2\sqrt{2}} = Mg + V - F_N \qquad (a.3)

The net torque about the center of mass of the right rod is

I\alpha = \sum\tau \qquad \frac{ML^2\alpha}{12} = \frac{F_N L}{2\sqrt{2}} + \frac{VL}{2\sqrt{2}} - \frac{HL}{2\sqrt{2}} \qquad (a.4)

Equations (a.1), (a.2), (a.3), and (a.4) form a set of four simultaneous equations in the four unknowns $H$ , $V$ , $\alpha$ , and $F_N$ . Solving them simultaneously for $F_N$ , yields $F_N = \frac{7}{10}Mg$

b. By symmetry V = 0. Applying Newton's second law to the right rod

Ma_x = \sum F_x \qquad \frac{M\alpha L}{2\sqrt{2}} = H \qquad (b.1)

Ma_y = \sum F_y \qquad \frac{M\alpha L}{2\sqrt{2}} = Mg - F_N \qquad (b.2)

The net torque about the center of mass

I\alpha = \sum\tau \qquad \frac{ML^2\alpha}{12} = \frac{F_N L}{2\sqrt{2}} - \frac{HL}{2\sqrt{2}} \qquad (b.3)

Solving these three equations in unknowns $H$ , $\alpha$ and $F_N$ . Solving them simultaneously for $F_N$ , yields

F_N = \frac{5}{8}Mg

c. Considering both rods as the system so we don't need to include the work done by the forces H and V,

PE_{Li} + PE_{Ri} = PE_{Lf} + PE_{Rf} + KE_{Lf} + KE_{Rf} \qquad (c.1)

Initially the center of mass of each rod is at a height $(L/2)\sin\theta_i$ where $\theta_i = 45^{\circ}$

PE_{Li} + PE_{Ri} = Mg\frac{L}{2}\sin\theta_i + Mg\frac{L}{2}\sin\theta_i = MgL\sin\theta_i

Finally,

PE_{Lf} + PE_{Rf} = Mg\frac{L}{2}\sin\theta + Mg\frac{L}{2}\sin\theta = MgL\sin\theta

For each side

KE_f = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 \qquad \text{with} \qquad I = \frac{1}{12}ML^2.

For the left rod $v_L = \omega L/2$ , so

KE_{Lf} = \frac{1}{2}M\left(\frac{\omega L}{2}\right)^2 + \frac{1}{2}\left(\frac{1}{12}ML^2\right)\omega^2 = \frac{1}{6}ML^2\omega^2.

For the right rod $v_{Ry} = \omega(L/2)\cos\theta$ , but $v_{Rx} = 3\,\omega(L/2)\sin\theta$ , so

KE_{Rf} = \frac{1}{2}M\left(\frac{\omega L}{2}\right)^2\left(\cos^2\theta + 9\sin^2\theta\right) + \frac{1}{2}\left(\frac{1}{12}ML^2\right)\omega^2 = ML^2\omega^2\left(\frac{1}{6} + \sin^2\theta\right)

Substituting all of the above into (c.1):

MgL\sin\theta_i = MgL\sin\theta + ML^2\omega^2\left(\frac{2}{6} + \sin^2\theta\right)

\omega^2 = \frac{3g}{L}\frac{\left(\sin\theta_i - \sin\theta\right)}{\left(1 + 3\sin^2\theta\right)} = \frac{3g}{L}\frac{\left(1/\sqrt{2} - \sin\theta\right)}{\left(1 + 3\sin^2\theta\right)}.

d. By symmetry, V = 0. The top of the “vee” doesn't move horizontally so H does no work. $v = \omega L/2$ for both rods. Applying energy conservation to either rod

PE_i = PE_f + KE_f \qquad (d.1)

with

PE_i = Mg\frac{L}{2}\sin\theta_i \qquad \text{and} \qquad PE_f = Mg\frac{L}{2}\sin\theta \qquad (d.2)

For either side

KE_f = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 \qquad \text{with} \qquad I = \frac{1}{12}ML^2.

With $v_L = \omega L/2$ , so

KE_f = \frac{1}{2}M\left(\frac{\omega L}{2}\right)^2 + \frac{1}{2}\left(\frac{1}{12}ML^2\right)\omega^2 = \frac{1}{6}ML^2\omega^2. \qquad (d.3)

Combining equations (d.1), (d.2), and (d.3), yields

MgL\sin\theta_i = MgL\sin\theta + \frac{2}{6}ML^2\omega^2

\omega^2 = \frac{3g}{L}\left(\sin\theta_i - \sin\theta\right) = \frac{3g}{L}\left(1/\sqrt{2} - \sin\theta\right)

Free-body diagrams of the two rods of Case 1: left rod with hinge forces at the top (V up, H left), acceleration components a_x, a_y and a at its centre, weight Mg, and reaction arrows at the lower hinge; right rod with V down and H right at the top (2a_x), a_y and 3a_x at its centre, weight Mg and normal force F_N at the lower end.

Free-body diagram of the right rod in Case 2: force H at the top pointing right, acceleration labels 2a_y, a_x, a_y and 2a_x, weight Mg at the centre and normal force F_N at the lower end.

Автори

The 1999 Semi-Final Examination questions were written by the coaches of the United States Physics Team. The coaches are: Academic Director, Dr. Mary Mogge – Professor of Physics at California State Polytechnic University, Pomona, CA; Senior Coach, Dr. Leaf Turner – Physicist in the Theoretical Division of Los Alamos National Laboratory, Los Alamos, NM; Dr. Warren Turner – Physics Teacher at St. Paul's School, Concord, NH. The coaches would also like to thank former academic director, Dr. Larry Kirkpatrick for his many helpful comments.

Оригинал в Архива: Semifinal 1999.pdf · официални решения: Semifinal 1999 Solutions.pdf