Question

a uniform rod of mass $m = 2.0, kg$ and length $l = 1.5, m$ is pivoted at one end. a force of $f = 10, n$ is applied perpendicularly to the rod at its midpoint. what is the resulting angular acceleration of the rod?

Explanation:

Step1: 回忆转动定律和转动惯量公式

转动定律为\(\tau = I\alpha\)，其中\(\tau\)是力矩，\(I\)是转动惯量，\(\alpha\)是角加速度。对于一端固定的均匀细杆，转动惯量\(I = \frac{1}{3}mL^{2}\)。力矩\(\tau = rF\sin\theta\)，这里力\(F\)垂直于杆，\(\theta = 90^{\circ}\)，\(\sin\theta = 1\)，力的作用点在中点，所以\(r=\frac{L}{2}\)。

Step2: 计算力矩\(\tau\)

已知\(F = 10\ N\)，\(L = 1.5\ m\)，\(r=\frac{L}{2}=\frac{1.5}{2}\ m = 0.75\ m\)，根据\(\tau = rF\)（因为\(\sin90^{\circ}=1\)），可得\(\tau = 0.75\ m\times10\ N = 7.5\ N\cdot m\)。

Step3: 计算转动惯量\(I\)

已知\(m = 2.0\ kg\)，\(L = 1.5\ m\)，根据\(I=\frac{1}{3}mL^{2}\)，代入数值可得\(I=\frac{1}{3}\times2.0\ kg\times(1.5\ m)^{2}=\frac{1}{3}\times2.0\times2.25\ kg\cdot m^{2}=1.5\ kg\cdot m^{2}\)。

Step4: 由转动定律求角加速度\(\alpha\)

根据\(\tau = I\alpha\)，变形得\(\alpha=\frac{\tau}{I}\)，将\(\tau = 7.5\ N\cdot m\)，\(I = 1.5\ kg\cdot m^{2}\)代入，可得\(\alpha=\frac{7.5}{1.5}\ rad/s^{2}=5\ rad/s^{2}\)。

Answer:

\(5\ rad/s^{2}\)