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 bicyclexcg r2rq(0 ,sesdz8us7 m8 1kto v*ls7 of mass M has two tirelxk e(a *+.wwojy4v1c s that each comes into contact with the ground at a point. The wheelbase of this1*( ya .j wkxwo4c+evl 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 ignored. 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 decelerako)u pnv;/v7wu7gm 4 ytion $a$ 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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