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]
If a planet of radius8 )yd((aidx9s+ gbmyw-mlwa )ja g3ge6r1*ws mbk cbg2stok94 1fo(1 $R$ spins with an angular velocity $\omega$ about an axis through the North Pole, what is the ratio of the normal force experienced by a person at the equator to that experienced by a person at the North Pole? Assume a constant gravitational field $g$ and that both people are stationary relative to the planet and are at sea level.
A. $g/(R\omega^2)$
B. $R\omega^2/g$
C. $1-R\omega^2/g$
D. $1+g/(R\omega^2)$
E. $1+R\omega^2/g$
参考答案: C
本题详细解析:
Let the person's massgy/xk- x5t+tqg+ *rtb0yvy+ j be $m$.
1. At the North Pole: The person is on the axis of rotation, so they are not in circular motion ($v=0$).
The net force is zero. The normal force $N_{pole}$ balances gravity $mg$.
$N_{pole} - mg = 0 \implies N_{pole} = mg$.
2. At the Equator: The person is in uniform circular motion with radius $R$ and angular velocity $\omega$.
The net force must provide the required centripetal force $F_c = m a_c = m R\omega^2$, directed towards the center (down).
The forces are gravity ($mg$, down) and the normal force ($N_{eq}$, up).
The net force is $F_{net} = mg - N_{eq}$.
Set $F_{net} = F_c$: $mg - N_{eq} = mR\omega^2$.
Solve for $N_{eq}$: $N_{eq} = mg - mR\omega^2$.
3. Find the ratio:
$N_{eq} / N_{pole} = (mg - mR\omega^2) / (mg) = (mg/mg) - (mR\omega^2 / mg) = 1 - R\omega^2/g$.
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