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Physics M
Posted 3 months ago

7. 质量为 mm 、ll 的均匀细杆可绕其一端 OO 的水平轴转动, 另一端与质量为 mm 的小球固连。系统从水平位置由静止转过 θ\theta 角时, 求系统的角速度动能和力矩所作的功。
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Answer from Sia
Posted 3 months ago
Identify the system: A uniform rod of mass mm and length ll is hinged at one end OO and has a mass mm attached at the other end
Determine the moment of inertia II of the system about the pivot point OO. For the rod: Irod=13ml2I_{\text{rod}} = \frac{1}{3}ml^2. For the mass at the end of the rod: Imass=ml2I_{\text{mass}} = ml^2. Total I=Irod+Imass=13ml2+ml2=43ml2I = I_{\text{rod}} + I_{\text{mass}} = \frac{1}{3}ml^2 + ml^2 = \frac{4}{3}ml^2
Apply conservation of mechanical energy. Initially, the potential energy is U=mgl(1cosθ)U = mgl\left(1 - \cos\theta\right) (for the rod's center of mass) + mgl(2cosθ)mgl\left(2 - \cos\theta\right) (for the mass at the end)
At the θ\theta angle, all potential energy has been converted into rotational kinetic energy: K=12Iω2K = \frac{1}{2}I\omega^2
Set the initial potential energy equal to the final kinetic energy: mgl(1cosθ)+mgl(2cosθ)=12(43ml2)ω2mgl\left(1 - \cos\theta\right) + mgl\left(2 - \cos\theta\right) = \frac{1}{2}\left(\frac{4}{3}ml^2\right)\omega^2
Solve for the angular velocity ω\omega: ω=6g(32cosθ)4l\omega = \sqrt{\frac{6g\left(3 - 2\cos\theta\right)}{4l}}
Calculate the work done by the torque: Work WW is equal to the change in kinetic energy, which is W=K=12Iω2W = K = \frac{1}{2}I\omega^2
Substitute the value of II and ω\omega into the expression for work: W=12(43ml2)(6g(32cosθ)4l)2W = \frac{1}{2}\left(\frac{4}{3}ml^2\right)\left(\sqrt{\frac{6g\left(3 - 2\cos\theta\right)}{4l}}\right)^2
Simplify to find the work done by the torque: W=12(43ml2)(6g(32cosθ)4l)=mgl(32cosθ)W = \frac{1}{2}\left(\frac{4}{3}ml^2\right)\left(\frac{6g\left(3 - 2\cos\theta\right)}{4l}\right) = mgl\left(3 - 2\cos\theta\right)
The system's angular velocity is ω=6g(32cosθ)4l\omega = \sqrt{\frac{6g\left(3 - 2\cos\theta\right)}{4l}} and the work done by the torque is W=mgl(32cosθ)W = mgl\left(3 - 2\cos\theta\right).
Key Concept
Conservation of mechanical energy and calculation of moment of inertia for a system of rigid bodies.
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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