Rotation, buoyancy and terminal-speed balances
| English | Español |
|---|---|
| moment of inertia/ˈməʊmənt ɒv ɪˈnɜːʃə/ | moment of inertia |
| terminal speed | terminal speed |
A decision before an answer
- Turning a floating block upside down can change how much wood is submerged even though the total weight stays the same.
- Your goal: Use pivot torque and rotational inertia consistently.
Read the relationship
- Torque about a fixed pivot is r×F; its magnitude uses the perpendicular lever arm. For a uniform rod of length L pivoted at one end, gravity acts at its centre L/2 and the moment of inertia is ML²/3. If its angle θ is measured from vertical, gravitational torque magnitude is Mg(L/2)sinθ, so angular-acceleration magnitude is 3g sinθ/(2L). Define the positive rotation direction before assigning a sign. Using the centre-of-mass inertia ML²/12 without the parallel-axis correction gives a wrong acceleration.
- Partition rolling kinetic energy with the no-slip constraint.
For a uniform rod pivoted at one end, gravitational angular-acceleration magnitude at angle θ from vertical is:
Divide Mg(L/2)sinθ by ML²/3.
Use the defining rule
- An object rolling without slipping satisfies v=Rω. Write total kinetic energy as ½Mv²+½Iω². With I=βMR², the rotational fraction is β/(1+β); for a solid disk β=1/2 and the fraction is 1/3. A hoop has β=1 and fraction 1/2. Static friction can supply the torque required for rolling without dissipating energy at an instantaneously stationary contact on a fixed surface. A sliding object does not satisfy the no-slip relation automatically.
- Balance buoyancy and drag for composite bodies.
For a solid sphere I=2MR²/5 rolling without slipping, the rotational fraction of total K is:
Use β/(1+β)=(2/5)/(7/5)=2/7.
Check the conditions
- For a floating composite, buoyancy equals the weight of displaced fluid, summed over every submerged part. Let a wood block have volume V and a dense stone volume Vs. If the stone is above the water, only submerged wood contributes: ρwater g fV equals total weight. If the stone is attached below and fully submerged, its displaced volume reduces the wood’s required submerged fraction to f−Vs/V. The stone does not cease to weigh anything; the water now supplies some of its support. Check whether the assumed orientation and flotation are physically possible.
- Balance buoyancy and drag for composite bodies.
A solid disk rolls without slipping: I=MR²/2 gives rotational K=¼Mv², translational K=½Mv² and rotational fraction 1/3. A wood–stone composite initially displaces 0.60V through wood alone. If the stone has volume 0.10V and is then fully underwater, the required submerged wood volume is 0.50V. Total displaced volume remains 0.60V.
Identical-shape balls have quadratic drag and masses M and 9M, with negligible buoyancy. Their terminal-speed ratio is ____.
The ratio is sqrt(9).
Apply the task format
- Terminal speed means zero acceleration, not zero speed or zero force. With quadratic drag C v² and a downward weight Mg, neglecting buoyancy gives vt=sqrt(Mg/C). Two otherwise identical balls share the drag coefficient C; doubling mass then multiplies terminal speed by sqrt(2). If buoyancy matters, replace Mg by (M−ρfluid V)g. A linear-drag model instead gives a different mass scaling. State the specified drag law and compare force balances, rather than transferring a formula between models.
- Balance buoyancy and drag for composite bodies.
Use the inertia about the chosen pivot. In flotation, include the stone’s displaced volume as well as its weight. Terminal speed requires a specified drag law.
Which answer fits this case?
Use pivot torque and rotational inertia consistently
At terminal speed the drag force vanishes.
Drag balances the effective weight; it is generally nonzero.
Keep the distinctions
- moment of inertia 转动惯量 — Mass-weighted squared perpendicular distance from a specified rotation axis.
- terminal speed 终端速度 — Steady speed at which opposing forces balance.
- Use pivot torque and rotational inertia consistently.
- Partition rolling kinetic energy with the no-slip constraint.
- Balance buoyancy and drag for composite bodies.
Match each term with its precise meaning in this lesson.
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
The order follows the stated process; check each stage before the next.