Quay, lực đẩy và cân bằng vận tốc cuối cùng
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| buoyancy/ˈbɔɪənsi/ | The upward force from a fluid, equal to the weight of fluid displaced. |
| terminal speed/ˈtɜːmɪnl spiːd/ | vận tốc cuối cùng |
| parallel axis/ˈpærəlel ˈæksɪs/ | The rule I = I_cm + Md squared for an axis parallel to one through the centre of mass. |
Flip the block, change the waterline
- A wooden block with a stone glued on top floats with 60% of its volume under water. Turn it over so the stone hangs underneath, and less wood is submerged — though the weight is unchanged.
- GRE mechanics tests torque about a pivot, rolling energy, composite buoyancy 浮力 and terminal speed 终端速度.
Torque about a pivot
- $\tau = rF\sin\phi = I\alpha$, with $I$ about the pivot. A uniform rod pivoted at one end: $I = \tfrac13 ML^2$ (parallel axis 平行轴: $\tfrac1{12}ML^2 + M(L/2)^2$).
- At angle $\theta$ from vertical: $\tau = Mg\tfrac{L}{2}\sin\theta$, so $\alpha = \dfrac{3g\sin\theta}{2L}$.
A uniform rod 1.5 m long, pivoted at one end, is released horizontal. Initial angular acceleration in rad/s² (take $g = 10$)?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Torque $mgL/2$ about the end, with $I = mL^2/3$, gives $\alpha = 3g/(2L) = 30/3 = 10$ rad/s².
Rolling without slipping
- $v = R\omega$. With $I = \beta MR^2$: $K = \tfrac12(1 + \beta)Mv^2$; rotational fraction $\dfrac{\beta}{1 + \beta}$.
- Down a height $h$: $v^2 = \dfrac{2gh}{1 + \beta}$ — independent of $M$ and $R$. Smaller $\beta$ wins the race.
Against a solid disk
Predict rolling races from β alone.
Rotational fraction of kinetic energy for a rolling solid sphere ($\beta = 2/5$), three decimal places?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Rotational energy is $\tfrac12 \beta mv^2$ out of a total $\tfrac12(1 + \beta)mv^2$, so the fraction is $\beta/(1 + \beta) = (2/5)/(7/5) = 2/7 = 0.286$.
A solid sphere, a solid disk and a hoop roll from rest down the same slope. The finishing order is
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Acceleration down the slope is $g\sin\theta/(1 + \beta)$, so the smallest $\beta$ wins: sphere (2/5), disk (1/2), hoop (1).
Static friction does no work on a wheel rolling without slipping on a fixed surface.
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
True. The contact point is momentarily at rest, so the static friction force acts through zero displacement and dissipates no energy.
Buoyancy of a composite
- Buoyancy = weight of all displaced fluid. Floating: total displaced volume $\times\rho_{\text{water}}$ = total mass.
Stone volume $0.10V$, on top: wood displaces $0.60V$. Stone underneath: it displaces $0.10V$ itself, so the wood needs only $0.50V$ submerged. Total displaced volume is still $0.60V$.
Terminal speed
- Terminal means zero acceleration: drag balances weight. Quadratic drag $Cv^2 = Mg$: $v_t = \sqrt{Mg/C}$, so doubling mass multiplies $v_t$ by $\sqrt2$.
- Linear drag $bv = Mg$ gives $v_t \propto M$. Include buoyancy as $(M - \rho_f V)g$ if it matters.
With quadratic drag (same $C$), a ball 4 times heavier has terminal speed multiplied by
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
At terminal speed $mg = Cv^2$, so $v \propto \sqrt{m}$; four times the mass gives twice the speed.
Recap
- Use $I$ about the pivot; $\alpha = \tau/I$.
- Rolling: $v^2 = 2gh/(1 + \beta)$; rotational share $\beta/(1+\beta)$.
- Buoyancy counts every submerged part; terminal speed depends on the drag law.
- buoyancy: The upward force from a fluid, equal to the weight of fluid displaced.
- parallel axis: The rule I = I_cm + Md squared for an axis parallel to one through the centre of mass.