空间中施力于物体边角时，平动与转动的能量分配比例问询
Great question—this is a classic rigid body dynamics problem, and since you’ve already nailed the core behavior of your system (off-center forces induce rotation + translation, any force causes translation), let’s work through the exact energy breakdown step by step.
Key Assumptions (Matching Your System)
- 100% energy transfer (no losses)
- Rigid body (rectangle doesn’t deform)
- We’ll assume you’re pushing perpendicular to the line connecting the rectangle’s center of mass (CoM) to the corner—this is the case that maximizes rotation, and it’s the most intuitive scenario for your question.
Step 1: Define Our Variables
Let’s use standard physics terms to keep this concrete:
- ( m ): Mass of the rectangle
- ( L, W ): Length and width of the rectangle
- ( I ): Moment of inertia of the rectangle about its CoM (for a rectangle, ( I = \frac{m(L^2 + W^2)}{12} ))
- ( J = F\Delta t ): Total impulse from your applied force ( F ) over time ( \Delta t ) (using impulse makes calculating velocities easier than forces alone)
- ( r ): Distance from the CoM to the corner you’re pushing: ( r = \frac{\sqrt{L^2 + W^2}}{2} )
Step 2: Calculate Translational Kinetic Energy
No matter where you apply the force, the CoM acceleration depends only on the net force (Newton’s second law for translation: ( a_{\text{CoM}} = \frac{F}{m} )). The final CoM velocity is:
[ v_{\text{CoM}} = \frac{J}{m} ]
Translational kinetic energy is:
[ K_{\text{trans}} = \frac{1}{2} m v_{\text{CoM}}^2 = \frac{J^2}{2m} ]
Step 3: Calculate Rotational Kinetic Energy
When pushing the corner, the force creates a torque around the CoM:
[ \tau = rF ]
(We use ( \tau = rF ) because we’re pushing perpendicular to ( r ), so the sine term in the torque formula is 1.)
Angular acceleration is ( \alpha = \frac{\tau}{I} ), so final angular velocity is:
[ \omega = \alpha \Delta t = \frac{rF\Delta t}{I} = \frac{rJ}{I} ]
Rotational kinetic energy is:
[ K_{\text{rot}} = \frac{1}{2} I \omega^2 ]
Substitute ( I = \frac{m(L^2 + W^2)}{12} ) and ( r^2 = \frac{L^2 + W^2}{4} ) into this equation:
[ K_{\text{rot}} = \frac{1}{2} \cdot \frac{m(L^2 + W^2)}{12} \cdot \left( \frac{rJ}{I} \right)^2 = \frac{3J^2}{2m} ]
Step 4: The Energy Split
Now compare the two energies:
- ( K_{\text{trans}} = \frac{J^2}{2m} )
- ( K_{\text{rot}} = 3 \cdot \frac{J^2}{2m} = 3K_{\text{trans}} )
Total energy ( K_{\text{total}} = K_{\text{trans}} + K_{\text{rot}} = 4K_{\text{trans}} ), so:
- 25% of the input energy goes to translational motion
- 75% goes to rotational motion
Important Caveat
This split only applies when pushing perpendicular to the line between the CoM and the corner. If you push along that line (toward or away from the CoM), the torque becomes zero—all energy goes to translation, and there’s no rotation. For any other angle, the split will be somewhere between 0% and 75% rotational energy.
内容的提问来源于stack exchange，提问作者Vít Gardoň
