7. 质量为 $m 、$ 长 $l$ 的均匀细杆可绕其一端 $O$ 的水平轴转动, 另一端与质量为 $m$ 的小球固连。系统从水平位置由静止转过 $ heta$ 角时, 求系统的角速度动能和力矩所作的功。

Question

7. 质量为 $m 、$ 长 $l$ 的均匀细杆可绕其一端 $O$ 的水平轴转动, 另一端与质量为 $m$ 的小球固连。系统从水平位置由静止转过 $	heta$ 角时, 求系统的角速度动能和力矩所作的功。

Accepted Answer

∻Solution∻ ‖ a ⋮ Identify the system: A uniform rod of mass $m$ and length $l$ is hinged at one end $O$ and has a mass $m$ attached at the other end. ‖ ‖ b ⋮ Determine the moment of inertia $I$ of the system about the pivot point $O$. For the rod: $I_{ ext{rod}} = \frac{1}{3}ml^2$. For the mass at the end of the rod: $I_{ ext{mass}} = ml^2$. Total $I = I_{ ext{rod}} + I_{ ext{mass}} = \frac{1}{3}ml^2 + ml^2 = \frac{4}{3}ml^2$. ‖ ‖ c ⋮ Apply conservation of mechanical energy. Initially, the potential energy is $U = mgl\left(1 - \cos heta ight)$ (for the rod's center of mass) + $mgl\left(2 - \cos heta ight)$ (for the mass at the end). ‖ ‖ d ⋮ At the $ heta$ angle, all potential energy has been converted into rotational kinetic energy: $K = \frac{1}{2}I\omega^2$. ‖ ‖ e ⋮ Set the initial potential energy equal to the final kinetic energy: $mgl\left(1 - \cos heta ight) + mgl\left(2 - \cos heta ight) = \frac{1}{2}\left(\frac{4}{3}ml^2 ight)\omega^2$. ‖ ‖ f ⋮ Solve for the angular velocity $\omega$: $\omega = \sqrt{\frac{6g\left(3 - 2\cos heta ight)}{4l}}$. ‖ ‖ g ⋮ Calculate the work done by the torque: Work $W$ is equal to the change in kinetic energy, which is $W = K = \frac{1}{2}I\omega^2$. ‖ ‖ h ⋮ Substitute the value of $I$ and $\omega$ into the expression for work: $W = \frac{1}{2}\left(\frac{4}{3}ml^2 ight)\left(\sqrt{\frac{6g\left(3 - 2\cos heta ight)}{4l}} ight)^2$. ‖ ‖ i ⋮ Simplify to find the work done by the torque: $W = \frac{1}{2}\left(\frac{4}{3}ml^2 ight)\left(\frac{6g\left(3 - 2\cos heta ight)}{4l} ight) = mgl\left(3 - 2\cos heta ight)$. ‖ ∻Answer∻ ⚹ The system's angular velocity is $\omega = \sqrt{\frac{6g\left(3 - 2\cos heta ight)}{4l}}$ and the work done by the torque is $W = mgl\left(3 - 2\cos heta ight)$. ⚹ ∻Key Concept∻ ⚹ Conservation of mechanical energy and calculation of moment of inertia for a system of rigid bodies. ⚹ ∻Explanation∻ ⚹ The problem involves converting potential energy into rotational kinetic energy as the system rotates about the pivot. By equating the initial potential energy to the final kinetic energy, we can solve for the angular velocity. The work done by the torque is equal to the change in kinetic energy.⚹

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