2007美国US F=MA物理竞赛 (id: fef1f3027)

本题目来源于试卷: 2007美国US F=MA物理竞赛，类别为 美国F=MA物理竞赛

[单选题]
A simplified model of a bichyj01klz g itlwruf.d )as,w;3 ,8rs/yc ry c*1qdl32pskq r7z-,sy v,hle of mass M has two tires that each comes into contact with the ground at a point. The wheelbase of this bicycle (the distance between the points of contact with the ground) is w, and the center of mass of the bicycle is located midway between the tires and a height h above the ground. The bicycle is moving to the right, but slowing down at a constant rate. The acceleration has a magnitude a. Air resistance may be ignohs r 1q k3d-*l,vyz7r cq,sp2yred. Assume, instead, that the coefficient of sliding friction between each tire and the ground is different: $\mu_1$ for the front tire and $\mu_2$ for the rear tire. Let $\mu_1 = 2\mu_2$. Assume that both tires are skidding: sliding without rotating. What is the maximum value of $a$ so that both tires remain in contact with the ground?

A. $\frac{wg}{h}$
B. $\frac{wg}{3h}$
C. $\frac{2wg}{3h}$
D. $\frac{hg}{2w}$
E. none of the above


参考答案:  E


本题详细解析:
The maximum decelerationb.z *u 37 cqs+.vuxfn/47ef;z.gfb 6o wcgovs $a$ is limited by two possibilities: 1. **Tipping (Lifting):** The rear tire lifts ($N_2=0$). As calculated in problem 29, this happens when $a = \frac{wg}{2h}$. This value depends only on geometry ($w, h$) and gravity ($g$), not on the friction coefficients. 2. **Skidding:** The required friction $f = Ma$ exceeds the maximum available friction $f_{max} = \mu_1 N_1 + \mu_2 N_2$. $Ma \le \mu_1 N_1 + \mu_2 N_2$. Substitute $\mu_1 = 2\mu_2$ and the expressions for $N_1$ and $N_2$ from the Q29 analysis: $Ma \le (2\mu_2) \left[ \frac{1}{2}Mg(1 + \frac{2ah}{wg}) \right] + (\mu_2) \left[ \frac{1}{2}Mg(1 - \frac{2ah}{wg}) \right]$ $a \le \mu_2 g (1 + \frac{2ah}{wg}) + \frac{1}{2}\mu_2 g (1 - \frac{2ah}{wg})$ $a \le \mu_2 g + \frac{2a\mu_2 h}{w} + \frac{1}{2}\mu_2 g - \frac{a\mu_2 h}{w}$ $a \le \frac{3}{2}\mu_2 g + \frac{a\mu_2 h}{w}$ $a (1 - \frac{\mu_2 h}{w}) \le \frac{3}{2}\mu_2 g \implies a \le \frac{\frac{3}{2}\mu_2 g}{1 - \mu_2 h/w}$. The maximum deceleration $a$ is the *minimum* of these two values (the tipping value and the skidding value). Since the answer depends on $\mu_2$ in a way not listed in options (a)-(d), the correct choice is (e) none of the above.

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