Newton's Second Law in Rotational Form
Introduced| English |
|---|
| rotational inertia/rəʊˈteɪʃənl ɪˈnɜːʃə/ |
| angular acceleration/ˈæŋɡjʊlə əkˌseləˈreɪʃn/ |
| Newton's second law for rotation/ˈnjuːtnz ˈsekənd lɔː fɔː rəʊˈteɪʃn/ |
The recipe for spinning something up
- A strong twist spins a wheel up quickly; a gentle one barely stirs it.
- A light wheel responds eagerly; a massive flywheel resists.
- This is exactly like $F = ma$ -- but for turning.
- One equation ties torque, rotational inertia, and angular acceleration together.
Newton's second law for rotation
- Newton's second law for rotation 转动的牛顿第二定律 is:
- Net torque equals rotational inertia times angular acceleration 角加速度.
- Give it a torque, and it gains angular acceleration.
A net torque of $8\ \text{N}\cdot\text{m}$ acts on a wheel with $I = 2\ \text{kg}\cdot\text{m}^2$. Its angular acceleration (in rad/s^2)?
$\alpha = \tau/I = 8/2 = 4\ \tfrac{\text{rad}}{\text{s}^2}$.
The rotational form of Newton's second law is tau = I times ____.
$\tau_{net} = I\alpha$ -- net torque equals rotational inertia times angular acceleration.
What net torque gives a $5\ \text{kg}\cdot\text{m}^2$ wheel an angular acceleration of $3\ \tfrac{\text{rad}}{\text{s}^2}$ (in N·m)?
$\tau = I\alpha = 5 \times 3 = 15\ \text{N}\cdot\text{m}$.
The perfect analogy
- Compare with $F = ma$: torque plays the role of force.
- Rotational inertia 转动惯量 $I$ plays the role of mass.
- Angular acceleration $\alpha$ plays the role of acceleration.
In $\tau = I\alpha$, rotational inertia $I$ plays the role of which quantity in $F = ma$?
Torque ~ force, $I$ ~ mass, $\alpha$ ~ acceleration. So $I$ is the rotational mass.
Select all correct analogies between $F = ma$ and $\tau = I\alpha$.
Force ~ torque and mass ~ $I$; acceleration maps to angular acceleration, not torque.
Bigger inertia, gentler response
- For a fixed torque, a larger $I$ gives a smaller $\alpha$.
- A massive, spread-out wheel is hard to spin up or slow down.
- That is why flywheels resist changes in their rotation.
Newton's second law for rotation
Torque produces angular acceleration in proportion to the rotational inertia: torque = I times alpha.
For the same net torque, a wheel with a larger rotational inertia gets a smaller angular acceleration.
$\alpha = \tau/I$, so a larger $I$ means a smaller $\alpha$ -- a gentler response.
Translation and rotation together
- A mass on a string over a pulley couples the two worlds.
- Write $F = ma$ for the hanging mass and $\tau = I\alpha$ for the pulley.
- Link them with the rope tension and $a = \alpha r$, then solve together.
A net torque of $12\ \text{N}\cdot\text{m}$ acts on a wheel with rotational inertia $I = 3\ \text{kg}\cdot\text{m}^2$.
- Angular acceleration: $\alpha = \dfrac{\tau}{I} = \dfrac{12}{3} = 4\ \tfrac{\text{rad}}{\text{s}^2}$.
- Double the rotational inertia and the same torque gives only $2\ \tfrac{\text{rad}}{\text{s}^2}$.
You cannot use $F = ma$ for the pulley itself. A pulley with mass has rotational inertia, so the rope tensions on its two sides are different -- that difference provides the net torque $\tau = I\alpha$. Treating it as massless (equal tensions) is a common error.
Newton's second law for rotation is $\tau_{net} = I\alpha$ -- the rotational twin of $F = ma$, with torque for force, rotational inertia for mass, and angular acceleration for acceleration. A bigger $I$ means a gentler response, and coupled mass-pulley systems solve $F = ma$ and $\tau = I\alpha$ together.