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]
Two wheels with fixed hubs, each havi
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;.so3vvrr -scsjc7s6 ass of $1\,\mathrm{kg}$, start from rest, and forces are applied as shown. Assume the hubs and spokes are massless, so that the rotational inertia is $I=mR^{2}$. In order to impart identical angular accelerations about their respective hubs, how large must $F_{2}$ be?
A. $0.25\,\mathrm{N}$
B. $0.5\,\mathrm{N}$
C. $1\,\mathrm{N}$
D. $2\,\mathrm{N}$
E. $4\,\mathrm{N}$
参考答案: D
本题详细解析:
We use Newton's second law for rot7yqeg7 j;a,5y bhhyr h,tu-;5jmx 7ixation, $\tau = I\alpha$.
We want $\alpha_1 = \alpha_2$.
Wheel 1: $R_1 = 0.5\,\mathrm{m}$, $F_1 = 1\,\mathrm{N}$, $m_1 = 1\,\mathrm{kg}$.
$\tau_1 = F_1 R_1 = (1\,\mathrm{N})(0.5\,\mathrm{m}) = 0.5\,\mathrm{N\cdot m}$.
$I_1 = m_1 R_1^2 = (1\,\mathrm{kg})(0.5\,\mathrm{m})^2 = 0.25\,\mathrm{kg\cdot m^2}$.
$\alpha_1 = \tau_1 / I_1 = 0.5 / 0.25 = 2\,\mathrm{rad/s^2}$.
Wheel 2: $R_2 = 1\,\mathrm{m}$, $m_2 = 1\,\mathrm{kg}$.
We need $\alpha_2 = \alpha_1 = 2\,\mathrm{rad/s^2}$.
$I_2 = m_2 R_2^2 = (1\,\mathrm{kg})(1\,\mathrm{m})^2 = 1\,\mathrm{kg\cdot m^2}$.
The required torque is $\tau_2 = I_2 \alpha_2 = (1\,\mathrm{kg\cdot m^2})(2\,\mathrm{rad/s^2}) = 2\,\mathrm{N\cdot m}$.
The force $F_2$ is $\tau_2 = F_2 R_2 \implies F_2 = \tau_2 / R_2 = (2\,\mathrm{N\cdot m}) / (1\,\mathrm{m}) = 2\,\mathrm{N}$.
