第二秒内。易错 （What is the magnitude of the average velocity of the particle during the second second?）
单位！！！
错误分析（你的失误）
在最后一步，你误将 M + m 写成了 M，导致动量守恒方程错误：
$$ \begin{aligned}& M v \cos \alpha = m v'\end{aligned} $$
于是错误地解出：
$$ \begin{aligned}& v' = \frac{M v \cos \alpha}{m}\end{aligned} $$
错误原因：忽略了炮弹的质量 m。发射前，炮弹和枪车一起运动，总质量应为 M + m，而不是只有枪车的质量 M。
教训：做动量守恒题时，仔细确认系统的总质量，尤其是在系统分离前后（发射前后）。
Please calculate the result completely.(differential equation)
Pay attention to how to set the unknowns.
Introduction to the basic concepts of angular momentum conservation.
https://grok.com/share/bGVnYWN5_19e5d057-e57f-41bd-9f5c-a3363af4116b
quick review
$$ \begin{aligned}&\text{The moment of inertia } I \text{ is defined as } I = \int r^2 \, dm \text{ because it quantifies} \\&\text{how mass distribution affects resistance to rotational acceleration.} \\&\text{Rotational Dynamics:} \\&\text{In linear motion, } F = m a \text{ shows mass resists acceleration.} \\&\text{In rotation, torque } \tau = I \alpha \text{ involves moment of inertia } I \text{ and} \\&\text{angular acceleration } \alpha. \\&\text{Why } r^2\text{? Consider a mass } dm \text{ at distance } r \text{ from the axis:} \\&\text{Linear speed } v = r \omega, \text{ where } \omega \text{ is angular velocity.} \\&\text{Kinetic energy of } dm: \\&dK = \frac{1}{2} (dm) v^2 = \frac{1}{2} (dm) (r \omega)^2 = \frac{1}{2} (dm) r^2 \omega^2 \\&\text{Total rotational kinetic energy:} \\&K = \int dK = \int \frac{1}{2} r^2 \omega^2 \, dm = \frac{1}{2} \omega^2 \int r^2 \, dm \\&\text{Since } K = \frac{1}{2} I \omega^2, \text{ we have:} \\&I = \int r^2 \, dm \\&\text{The } r^2 \text{ term shows mass farther from the axis contributes more,} \\&\text{since } (2r)^2 = 4r^2 \text{ quadruples the contribution.} \\&\text{Physical Intuition:} \\&\text{Mass far from the axis needs more torque due to higher speed } v \\&\text{and larger circular path, increasing kinetic energy and resistance.} \\&\text{Thus, } I = \int r^2 \, dm \text{ defines moment of inertia, tied to} \\&\text{rotational energy and dynamics.}\end{aligned} $$
The particularly error - prone part is that when calculating the moment of inertia, it is always in integral form, and the integration is carried out with respect to dm. So it is not at the center - of - mass position which is at the 1/2 mark.
$$ \begin{aligned} &\text{Problem: Calculate the moment of inertia of three helicopter rotor blades,} \\ &\text{each 3.75 m long, 160 kg, about the rotation axis, and the torque} \\ &\text{to reach 5.0 rev/s in 8.0 s.} \\ &\text{Step 1: Moment of Inertia of Three Rotor Blades} \\ &\text{Each blade is a thin rod, length } L = 3.75 \, \text{m}, \text{ mass } m = 160 \, \text{kg}. \\ &\text{Moment of inertia of a rod about its end:} \\ &I_{\text{rod}} = \frac{1}{3} m L^2 \\ &\text{Derivation: For a rod, } I = \int_0^L x^2 \left( \frac{m}{L} \right) dx = \frac{m}{L} \cdot \frac{L^3}{3} = \frac{m L^2}{3} \\ &\text{For one blade:} \\ &I_{\text{rod}} = \frac{1}{3} (160) (3.75)^2 = \frac{1}{3} (160) (14.0625) = 750 \, \text{kg m}^2 \\ &\text{Total for three blades:} \\ &I_{\text{total}} = 3 \times 750 = 2250 \, \text{kg m}^2 \\ &\text{Step 2: Angular Acceleration} \\ &\text{Final angular velocity: } \omega = 5.0 \times 2\pi = 31.416 \, \text{rad/s} \\ &\text{Angular acceleration: } \alpha = \frac{\omega}{t} = \frac{31.416}{8.0} = 3.927 \, \text{rad/s}^2 \\ &\text{Step 3: Torque} \\ &\tau = I \alpha = 2250 \times 3.927 = 8835.75 \, \text{N m} \approx 8836 \, \text{N m} \\ &\text{Final Answer:} \\ &\boxed{I = 2250 \, \text{kg m}^2, \ \tau = 8836 \, \text{N m}} \end{aligned} $$