转动系统的能量与动量

Roll a full can of soup and an empty can down the same slope, side by side. One reaches the bottom clearly first. Which one, and why, is the kind of question this unit answers with energy and angular momentum, not with forces.

A spinning object has rotational kinetic energy 转动动能, the rotational twin of $\tfrac12 mv^2$:

An object that both moves and spins (like a rolling ball) has both translational kinetic energy 平动动能 and rotational kinetic energy, and its total is $K=\tfrac12 mv^2+\tfrac12 I\omega^2$, with $v$ the centre-of-mass 质心 speed and $I$ measured about the centre of mass.

A torque acting through an angular displacement does work 功, changing rotational kinetic energy:

This is the rotational form of $W=Fd$ and $P=Fv$, and it extends the work–energy theorem 动能定理 to rotation.

Worked example. A motor applies a steady torque of $8.0\ \text{N m}$ to a flywheel 飞轮 while it turns through $10\ \text{rad}$. The work done is $W=\tau\,\Delta\theta=8.0\times10=80\ \text{J}$, and if the flywheel started from rest this all becomes rotational kinetic energy.

Angular momentum 角动量 describes rotational motion. For a rigid body 刚体 rotating about a fixed principal axis, its component along that axis is:

A net torque acting over time delivers an angular impulse 角冲量 that changes it: $\tau_{\text{avg}}\,\Delta t=\Delta L$ – the rotational impulse–momentum theorem 冲量–动量定理.

Worked example (angular impulse). A motor applies a torque of $0.50\ \text{N m}$ for $4.0\ \text{s}$ to a wheel with $I=0.20\ \text{kg m}^2$, starting from rest. The angular impulse is $\tau\,\Delta t=2.0\ \text{kg m}^2/\text{s}$, so $\omega=\Delta L/I=2.0/0.20=10\ \text{rad/s}$.

A single small object moving in a straight line also has angular momentum about a chosen point: $L=mvr_\perp$, where $r_\perp$ is the perpendicular distance from the point to its line of motion. This is what lets a moving object set something spinning.

If the net external torque 合外力矩 on a system is zero, its total angular momentum is conserved 守恒:

So if $I$ decreases, $\omega$ increases to keep $L$ constant – this is why a spinning skater speeds up when pulling their arms in. It applies to collisions 碰撞 and explosions 爆炸 of rotating systems too.

Worked example. A skater spins at $2.0\ \text{rev/s}$ with rotational inertia $I_1=4.0\ \text{kg m}^2$. She pulls her arms in, dropping her rotational inertia to $I_2=1.6\ \text{kg m}^2$. With no external torque, angular momentum is conserved:

Her kinetic energy actually rises – the extra energy comes from the work her muscles do pulling her arms in against the outward pull.

Check it with numbers. Writing $K=\dfrac{L^2}{2I}$, kinetic energy is inversely proportional 成反比 to $I$ when $L$ is fixed. Before: $\tfrac12(4.0)(4.0\pi)^2=316\ \text{J}$. After: $\tfrac12(1.6)(10\pi)^2=790\ \text{J}$, which is $2.5$ times as much, the same factor by which $I$ fell.

Worked example (a collision that starts a spin). A uniform rod of mass $1.2\ \text{kg}$ and length $0.80\ \text{m}$ hangs from a pivot at its top end, at rest. A $0.10\ \text{kg}$ lump of clay moving horizontally at $5.0\ \text{m/s}$ hits the bottom end and sticks. Find the angular velocity just after.

• About the pivot, gravity and the pivot force exert no torque during the brief collision, so angular momentum is conserved.
• Before: only the clay, $L=mvr=(0.10)(5.0)(0.80)=0.40\ \text{kg m}^2/\text{s}$.
• After: $I=\tfrac13(1.2)(0.80)^2+(0.10)(0.80)^2=0.256+0.064=0.320\ \text{kg m}^2$.
• $\omega=L/I=0.40/0.320=1.25\ \text{rad/s}$.

Linear momentum is not conserved here: the pivot pushes on the rod during the collision. Angular momentum about the pivot is, because that push passes through the pivot and has no torque.

Rolling without slipping 纯滚动 links the translational and rotational motions: the contact point 接触点 is momentarily at rest, so

A rolling object's energy splits between translation and rotation, so on an incline 斜面 it accelerates more slowly than a frictionless sliding object – some energy goes into spin.

Worked example. For a solid disk ($I=\tfrac12 mR^2$) that rolls without slipping, what fraction of its kinetic energy is rotational? Using $v=R\omega$, the rotational part is $\tfrac12 I\omega^2=\tfrac12(\tfrac12 mR^2)\omega^2=\tfrac14 mv^2$, while the translational part is $\tfrac12 mv^2$. So the total is $\tfrac34 mv^2$ and the rotational share is $\tfrac{1/4}{3/4}=\tfrac13$. A hoop, with its mass farther out, stores half its energy in spin and rolls down a ramp even more slowly.

Release several round objects from rest at the top of the same ramp, height $h$. Energy conservation with $v=R\omega$ gives

Neither mass nor radius appears on its own: only the shape factor 形状因子 $I/(mR^2)$ matters. For $h=1.2\ \text{m}$:

Object	$I/(mR^2)$	Speed at the bottom
block sliding without friction	0	$4.9\ \text{m/s}$
solid sphere	2/5	$4.1\ \text{m/s}$
solid disk (full can)	1/2	$4.0\ \text{m/s}$
hoop (empty can, roughly)	1	$3.4\ \text{m/s}$

The full can wins against the empty one: less of its energy goes into spinning, so more goes into speed.

A satellite 卫星 in orbit 轨道 is in free fall 自由落体: gravity curves its path. For a circular orbit, gravity provides the centripetal force 向心力, so

So a larger orbit means a slower speed. For a circular orbit, angular momentum and mechanical energy 机械能 are both constant; for an elliptical 椭圆的 orbit, angular momentum is conserved (no torque about the planet) while speed varies – fastest when closest.

Worked example. Find the speed of a satellite in a low orbit just above the Earth, radius $r=6.4\times10^{6}\ \text{m}$, with $GM=4.0\times10^{14}\ \text{m}^3/\text{s}^2$:

about $7.9\ \text{km/s}$ – and notice that the satellite's mass cancels, so all low orbits share this speed.

Escape velocity 逃逸速度 is the launch speed that just lets an object leave for good: its total mechanical energy is exactly zero, so it slows to zero speed only at infinite distance. Setting $\tfrac12 mv^2 - \dfrac{GMm}{r}=0$ (using the general gravitational PE $U_g=-GMm/r$) and solving for $v$,

It is $\sqrt2$ times the circular-orbit speed at the same radius, and (like orbital speed) does not depend on the escaping object's mass.

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

1. A flywheel with $I=2.0\ \text{kg m}^2$ spins at $30\ \text{rad/s}$. Find its rotational kinetic energy.
2. A hoop and a solid disk of equal mass and radius roll from rest down the same ramp. Which reaches the bottom first? Explain without calculating.
3. A skater spinning at $1.5\ \text{rev/s}$ reduces her rotational inertia from $3.0$ to $1.2\ \text{kg m}^2$. Find her new rate of spin.
4. One satellite orbits at four times the radius of another. Compare their orbital speeds.
5. A student says: "Angular momentum is conserved when the skater pulls her arms in, so her kinetic energy is conserved too." Correct the student.

1. $K=\tfrac12(2.0)(30)^2=900\ \text{J}$.
2. The disk. A hoop has all its mass at the rim, so a larger share of its energy goes into rotation and less into speed. The disk's shape factor is $\tfrac12$, the hoop's is 1.
3. $\omega_2=\dfrac{3.0}{1.2}(1.5)=3.75\ \text{rev/s}$.
4. $v\propto1/\sqrt{r}$, so the outer satellite moves at half the speed.
5. With $L$ fixed, $K=L^2/2I$ rises as $I$ falls. Her muscles do work pulling her arms inward, and that work becomes the extra kinetic energy. Angular momentum is conserved; kinetic energy is not.

• Conserve angular momentum $L=I\omega$ when no external torque acts: a smaller $I$ (arms pulled in) gives a larger $\omega$.
• A rolling object splits its energy between $\tfrac12 mv^2$ and $\tfrac12 I\omega^2$, linked by $v=r\omega$ — so it accelerates down a ramp more slowly than a sliding one.
• For a circular orbit set gravity equal to the centripetal requirement: $v=\sqrt{GM/r}$, so a larger orbit is slower and the satellite's mass cancels.
• Pulling in raises the spin and the kinetic energy — the extra energy comes from the work done pulling inward; $L$ is unchanged.
• Watch which rotational quantity is conserved: $L$ (no torque) versus energy (no friction) are different conditions.

Common mistakes

• Leaving out rotational kinetic energy for a rolling object. Its total kinetic energy is $\tfrac12 mv^2 + \tfrac12 I\omega^2$, so it reaches the bottom of a ramp slower than a sliding block.
• Using $\omega$ in rev/s inside $\tfrac12 I\omega^2$. Change to rad/s first ($\times 2\pi$); in $I_1\omega_1 = I_2\omega_2$ any unit works, because it cancels.
• Saying a skater's kinetic energy stays the same when she pulls her arms in. Angular momentum stays the same; kinetic energy rises, supplied by the work her muscles do.
• Conserving linear momentum when an object hits a pivoted rod. The pivot exerts an external force; conserve angular momentum about the pivot instead.
• Thinking a heavier ball rolls down faster. For rolling from rest, only the shape factor $I/(mR^2)$ decides the speed.
• Writing that "there is no gravity" on an orbiting satellite. Gravity is the centripetal force; the satellite is in free fall.