力矩与转动动力学

Why is a door handle on the side away from the hinges, and why does a tightrope walker carry a long pole? Both are about turning: how hard a force turns something, and how hard that something resists being turned. This unit gives each idea a number.

Rotation is described by angular quantities that mirror the linear ones:

• angular displacement 角位移 $\theta$ (in radians 弧度),
• angular velocity 角速度 $\omega=\dfrac{\Delta\theta}{\Delta t}$,
• angular acceleration 角加速度 $\alpha=\dfrac{\Delta\omega}{\Delta t}$.

For constant $\alpha$, the rotational kinematic equations 转动运动学方程 have the same form as the linear ones, with $\theta,\omega,\alpha$ replacing $x,v,a$: $\omega=\omega_0+\alpha t$, $\theta=\omega_0 t+\tfrac12\alpha t^2$, and $\omega^2=\omega_0^2+2\alpha\theta$.

Worked example. A wheel starts from rest and speeds up uniformly to $30\ \text{rad/s}$ in $6.0\ \text{s}$. Find its angular acceleration and the total angle turned:

A point at radius $r$ from the axis of rotation 转轴 has linear quantities tied to the angular ones:

Points farther from the axis move faster. This link lets you switch between "how fast the wheel spins" and "how fast a point on its rim moves."

Worked example. A bicycle wheel of radius $0.35\ \text{m}$ spins at $12\ \text{rad/s}$. A point on the rim (and so the bike) moves at $v=r\omega=0.35\times12=4.2\ \text{m/s}$. A point halfway to the axis moves at half that speed.

Torque 力矩 is the rotational effect of a force – how effectively it turns an object about an axis:

where $r_\perp$ is the moment arm 力臂 (the perpendicular distance from the axis to the force's line of action 作用线). A force applied farther out, or more perpendicular, produces more torque. Torque has a sign (clockwise 顺时针 vs counterclockwise 逆时针).

Worked example. You push with $20\ \text{N}$ at the end of a $0.30\ \text{m}$ wrench. Perpendicular to the wrench the torque is $\tau=rF=0.30\times20=6.0\ \text{N m}$. If you push at $60^{\circ}$ to the wrench instead, only the perpendicular part counts: $\tau=rF\sin 60^{\circ}=0.30\times20\times0.87=5.2\ \text{N m}$ – which is why you push at a right angle for the most turning effect.

Rotational inertia 转动惯量 (moment of inertia 惯性矩) $I$ measures how hard it is to change an object's rotation – the rotational version of mass. It depends on both the mass and how far that mass sits from the axis: mass spread farther out gives a larger $I$. For a point mass 质点 $I=mr^2$; for extended bodies, standard formulas are provided (a hoop is $mR^2$, a solid disk $\tfrac12 mR^2$). This is why a figure skater spins faster when she pulls her arms in – she reduces $I$.

Equal masses, different distributions: the further the mass sits from the axis, the larger I

Worked example (hoop or disk?). A hoop 圆环 and a solid disk 实心圆盘 each have mass $2.0\ \text{kg}$ and radius $0.50\ \text{m}$. The same net torque of $1.0\ \text{N m}$ acts on each. The hoop has $I=MR^2=0.50\ \text{kg m}^2$, so $\alpha=1.0/0.50=2.0\ \text{rad/s}^2$. The disk has $I=\tfrac12MR^2=0.25\ \text{kg m}^2$, so $\alpha=4.0\ \text{rad/s}^2$. Same mass and same size, but the disk speeds up twice as fast, because its mass sits closer to the axis.

An object is in rotational equilibrium 转动平衡 when the net torque 合力矩 is zero, so its angular velocity stays constant. For a balanced (static) object, both the net force and the net torque are zero. Choosing the axis at an unknown force's location removes it from the torque equation – a useful trick for beam and ladder problems.

Worked example. A $30\ \text{kg}$ child sits $2.0\ \text{m}$ from the pivot 支点 of a seesaw. Where must a $40\ \text{kg}$ child sit on the other side to balance it? Set the clockwise torque equal to the anticlockwise torque (the $g$'s cancel):

The heavier child sits closer to the pivot – less distance, same torque.

Net torque produces angular acceleration, in direct analogy with $F=ma$:

So a larger net torque, or a smaller rotational inertia, gives a larger angular acceleration. Solve rotation problems just like translation 平动 problems, with $\tau\leftrightarrow F$, $I\leftrightarrow m$, and $\alpha\leftrightarrow a$.

Worked example. A net torque of $12\ \text{N m}$ acts on a wheel with rotational inertia $I=3.0\ \text{kg m}^2$. Its angular acceleration is $\alpha=\tau/I=12/3.0=4.0\ \text{rad/s}^2$ – the exact rotational twin of $a=F/m$.

Many AP problems join a falling object to a turning one. Write Newton's second law for each object, then link them with $a=r\alpha$.

Worked example (a block on a pulley). A $2.0\ \text{kg}$ block hangs from a light string wrapped around a pulley, a uniform disk of mass $4.0\ \text{kg}$ and radius $0.10\ \text{m}$. The block is released from rest. Find its acceleration and the tension 张力 in the string.

• Pulley: $I=\tfrac12(4.0)(0.10)^2=0.020\ \text{kg m}^2$.
• Block (down positive): $mg-T=ma$.
• Pulley: $TR=I\alpha=I\dfrac{a}{R}$, so $T=\dfrac{I}{R^2}a=\dfrac{0.020}{0.010}a=2.0a$.
• Add the two equations: $mg=(m+2.0)a$, so $19.6=4.0a$ and $a=4.9\ \text{m/s}^2$. Then $T=2.0(4.9)=9.8\ \text{N}$.

The block falls at half of $g$. The tension is less than the block's weight ($19.6\ \text{N}$), as it must be for the block to accelerate down. A heavier pulley would mean a slower fall.

Static problems need both conditions: zero net force and zero net torque. Take torques about one support to remove its unknown force.

Worked example. A uniform 均匀的 plank $4.0\ \text{m}$ long with mass $20\ \text{kg}$ rests on supports at its two ends. A $50\ \text{kg}$ person stands $1.0\ \text{m}$ from the left end. Find the force from each support.

• Torques about the left support: $F_R(4.0)=(20)(9.8)(2.0)+(50)(9.8)(1.0)=392+490=882$, so $F_R=220\ \text{N}$.
• Forces: $F_L+F_R=(70)(9.8)=686\ \text{N}$, so $F_L=470\ \text{N}$.

The person stands nearer the left support, so that support carries more. The plank's weight acts at its center, $2.0\ \text{m}$ from either end.

Start such a problem with a force diagram 力图. It is like a free-body diagram, but each force is drawn at the point where it acts on the rigid object, and the arrow lengths show the relative sizes. Every lever arm can then be read straight off the drawing.

Answer each before reading the answers below. Take $g=9.8\ \text{m/s}^2$.

1. A wheel speeds up uniformly from rest to $20\ \text{rad/s}$ in $4.0\ \text{s}$. Find its angular acceleration and how many revolutions 转数 it makes.
2. A $15\ \text{N}$ force pushes a door $0.40\ \text{m}$ from the hinge. Find the torque when the force is perpendicular to the door, and when it is at $30°$ to the door.
3. A $25\ \text{kg}$ child sits $1.6\ \text{m}$ from a seesaw pivot. Where must a $40\ \text{kg}$ child sit to balance?
4. A disk with $I=0.50\ \text{kg m}^2$ starts from rest. A net torque of $2.0\ \text{N m}$ acts for $3.0\ \text{s}$. Find its final angular velocity.
5. A student says: "Two wheels with the same mass and radius will speed up equally under the same torque." Correct the student.

1. $\alpha=20/4.0=5.0\ \text{rad/s}^2$. $\theta=\tfrac12(5.0)(4.0)^2=40\ \text{rad}=40/2\pi=6.4$ revolutions.
2. Perpendicular: $(0.40)(15)=6.0\ \text{N m}$. At $30°$: $(0.40)(15)\sin30°=3.0\ \text{N m}$.
3. $(25)(1.6)=(40)d$, so $d=1.0\ \text{m}$.
4. $\alpha=2.0/0.50=4.0\ \text{rad/s}^2$, so $\omega=(4.0)(3.0)=12\ \text{rad/s}$.
5. Rotational inertia depends on where the mass is, not only how much. A hoop has twice the $I$ of a solid disk with the same mass and radius, so under the same torque it speeds up only half as fast.

• Torque $\tau=Fr_\perp$ uses the perpendicular distance from the pivot; a force through the pivot gives zero torque.
• For balance, set clockwise torque = anticlockwise torque; choosing the pivot at an unknown force removes it from the equation.
• Use the rotational analogues 类比量: $\tau\leftrightarrow F$, $I\leftrightarrow m$, $\alpha\leftrightarrow a$, so $\sum\tau=I\alpha$ mirrors $\sum F=ma$.
• Rotational inertia depends on how far the mass sits from the axis, not just its amount — a hoop resists spinning more than a disc of equal mass.
• Convert angles to radians and link linear to angular with $v=r\omega$, $a_t=r\alpha$.

Common mistakes

• Using the distance along the force's line of action, or the length of the rod, when the force is at an angle. Torque uses the perpendicular distance $r_\perp$ (or $rF\sin\theta$).
• Leaving angles in degrees or revolutions in $v = r\omega$ and the rotational kinematic equations. Use radians; one revolution is $2\pi\ \text{rad}$.
• Setting the string tension equal to the hanging weight when a massive pulley turns. The block accelerates, so $T < mg$, and the tension on the two sides of a massive pulley can differ.
• Forgetting the weight of a uniform beam. It acts at the beam's center and must appear in the torque equation.
• Checking only $\sum\tau = 0$ for a static object. A static object also needs $\sum F = 0$; use it to find the second support force.
• Thinking a larger mass always means a larger rotational inertia. The distribution matters: $I$ grows with $r^2$.