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 uniform disk ($I=\frac{1}{2}MR^{2}$) of mass $8.0\,\mathrm{kg}$ can rotate without friction on a fixed axis. A string is wrapped around its circumference and is attached to a $6.0\,\mathrm{kg}$ mass. The string does not slip. What is the tension in the cord while the mass is falling? 
A. $20.0\,\mathrm{N}$
B. $24.0\,\mathrm{N}$
C. $34.3\,\mathrm{N}$
D. $60.0\,\mathrm{N}$
E. $80.0\,\mathrm{N}$
参考答案: B
本题详细解析:
Let $m = 6.0\,\mathrm{kg}$ (falling mass) and $M = 8.0\,\mathrm{kg}$ (disk).
1. For the falling mass $m$, Newton's second law (with down as positive) is: $mg - T = ma$.
2. For the disk, Newton's second law for rotation is: $\tau = TR = I\alpha$.
Since the string doesn't slip, $a = \alpha R$. The rotational inertia is $I = \frac{1}{2}MR^2$.
Substituting: $TR = (\frac{1}{2}MR^2)(a/R) \implies T = \frac{1}{2}Ma$.
3. Substitute $T$ from (2) into (1): $mg - \frac{1}{2}Ma = ma$.
Solve for $a$: $mg = a(m + \frac{1}{2}M)$.
$a = mg / (m + \frac{1}{2}M) = (6.0)g / (6.0 + \frac{1}{2}(8.0)) = 6g / (6+4) = 0.6g$.
4. Now find $T$ using $T = \frac{1}{2}Ma$: $T = \frac{1}{2}(8.0)(0.6g) = 4(0.6g) = 2.4g$.
Using $g \approx 10\,\mathrm{m/s^2}$, $T = 2.4(10) = 24.0\,\mathrm{N}$. (Using $g=9.8\,\mathrm{m/s^2}$ gives $T = 23.52\,\mathrm{N}$, which is closest to 24.0 N).
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