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Torque $\tau = rF\sin\theta$ and the lever arm, rotational inertia $\int r^2\,dm$ for rods and disks, the parallel-axis theorem, $\tau_{\text{net}} = I\alpha$, and static equilibrium.
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By the end of this lesson you will be able to compute torques and rotational inertias, and use $\tau_{\text{net}} = I\alpha$ with the no-slip condition to solve coupled rotation problems.
The last lesson described rotation with an angle and its derivatives. From Physics 1 you know torque as force times lever arm and the idea of rotational inertia. From calculus you can integrate over a length and an area, and you know the cross product. This lesson makes torque and rotational inertia precise and writes Newton's second law for rotation, which the next lessons build on for angular momentum and rolling.
| Term | What it means |
|---|---|
| Torque | $\vec{\tau} = \vec{r} \times \vec{F}$, of size $rF\sin\theta$; the turning effect of a force about an axis. |
| Lever arm | $r_\perp = r\sin\theta$, the perpendicular distance from the axis to the force's line of action. |
| Rotational inertia | $I = \sum mr^2 = \int r^2\,dm$, resistance to angular acceleration about an axis. |
| Parallel-axis theorem | $I = I_{\text{cm}} + Md^2$ for an axis a distance $d$ from a parallel axis through the center of mass. |
| Second law for rotation | $\tau_{\text{net}} = I\alpha$ about a fixed axis. |
| Static equilibrium | No net force and no net torque about any point. |
| Radius of gyration | $k = \sqrt{I/M}$, the radius at which all the mass would give the same inertia. |
A force turns a body about an axis according to its torque,
$$\tau = rF\sin\theta = Fr_\perp,$$
where $r$ is the distance from the axis to the point of application and $r_\perp = r\sin\theta$ is the lever arm. As a vector, $\vec{\tau} = \vec{r} \times \vec{F}$, pointing along the axis by the right-hand rule. A body resists angular acceleration according to its rotational inertia,
$$I = \sum m_ir_i^2 = \int r^2\,dm,$$
with $r$ measured from the axis. Mass far from the axis counts heavily, so $I$ depends on both the mass and where it lies. Newton's second law for rotation about a fixed axis is
$$\tau_{\text{net}} = I\alpha.$$
It follows from $F = ma$ for each piece: a piece at radius $r$ has tangential acceleration $r\alpha$, needs force $mr\alpha$ and so torque $mr^2\alpha$; adding over pieces gives $I\alpha$.
Another way: picture
Push a door near its hinges and it barely moves; push at the handle and it swings easily. Same force, longer lever arm, more torque. Now try spinning a baseball bat about its handle and then about its middle: the first is much harder, because more of the bat's mass lies far from the handle. Torque is how hard you twist; rotational inertia is how the body's mass is arranged against the twist.
Another way: steps
Only the part of a force perpendicular to the line from the axis turns the body. A force pointed straight at the axis, or straight away from it, has zero torque however large it is. Two equivalent pictures give the same number: the perpendicular component $F\sin\theta$ times the distance $r$, or the full force times the lever arm $r\sin\theta$.
Torques turning the body one way are positive and the other way negative; add them with their signs. Gravity acts on a body as if all its weight were at the center of mass, so its torque about a pivot is $Mg$ times the horizontal distance from the pivot to the center of mass. That single fact solves most balance problems, from seesaws to cranes.
For a continuous body, divide it into pieces $dm$ at distance $r$ from the axis and integrate $r^2\,dm$. For a thin uniform rod of mass $M$ and length $L$ about its center, $dm = (M/L)\,dx$ and $I = \int_{-L/2}^{L/2}x^2(M/L)\,dx = \tfrac{1}{12}ML^2$. About one end the limits become $0$ to $L$ and $I = \tfrac{1}{3}ML^2$.
For a solid disk, slice it into rings of radius $r$, width $dr$ and mass $dm = (M/\pi R^2)2\pi r\,dr$. Each ring contributes $r^2\,dm$, and $I = \int_0^R2Mr^3/R^2\,dr = \tfrac{1}{2}MR^2$. A hoop of the same mass and radius has $MR^2$, twice as much, because all its mass is at the rim. Choosing slices that are all at one distance from the axis is the whole art.
If $I_{\text{cm}}$ is known about an axis through the center of mass, the inertia about any parallel axis a distance $d$ away is
$$I = I_{\text{cm}} + Md^2.$$
The proof expands $(x + d)^2$ under the integral; the cross term $2d\int x\,dm$ vanishes because $x$ is measured from the center of mass. The theorem shows that the smallest rotational inertia of any body, for a given direction of axis, is about the axis through its center of mass. A rod about its end has $\tfrac{1}{12}ML^2 + M(L/2)^2 = \tfrac{1}{3}ML^2$, confirming the direct integral.
Many problems couple a rotating body with one moving in a line: a bucket on a rope around a drum, an Atwood machine with a heavy pulley, a yo-yo. Write $F = ma$ for each translating body and $\tau = I\alpha$ for each rotating one, and link them with the no-slip condition $a = R\alpha$. The rope's tension appears in both and is eliminated.
The result always has the form of an effective mass: for a bucket of mass $m$ on a uniform drum of mass $M$, $a = mg/(m + \tfrac{1}{2}M)$. The drum behaves like an extra mass of $I/R^2$ added to the bucket's inertia but not to its weight. A heavier or larger-radius drum slows the fall; a hollow drum of the same mass slows it more than a solid one.
A body at rest, and not starting to turn, has zero net force and zero net torque. Because the net force is zero, the torque can be computed about any point, and a clever choice removes unknowns: take torques about the point where an unknown force acts, and that force drops out.
For a ladder leaning against a smooth wall, taking torques about the foot of the ladder eliminates the floor's two forces, leaving the wall's push balanced against the ladder's weight. Engineers checking a crane's stability, or a sign hanging from a strut, do exactly this. The torque condition is what makes a structure stand, and failing it is what tips a crane over.
Checking an answer. Units: torque in N m, inertia in kg m², $\alpha$ in rad/s². An accelerating load on a rope over a massive pulley must accelerate more slowly than it would over a massless one. Rotational inertia must be at least the value about the center of mass. And a body's inertia must lie between that of all its mass at the axis, zero, and all its mass at its farthest point.
Two wheels of the same mass and radius, one a solid disk and one a hoop, respond very differently to the same torque: the hoop has twice the inertia and so half the angular acceleration. Racing bicycle wheels put as little mass as possible at the rim, and flywheels meant to store energy put as much as possible there.
The radius of gyration $k = \sqrt{I/M}$ captures shape in one length: a hoop has $k = R$, a disk $k = 0.71R$, a solid sphere $k = 0.63R$. Figure skaters change their radius of gyration by pulling in their arms, which the angular momentum lesson turns into a spin that speeds up. Divers and gymnasts do the same with their bodies in the air.
Every engine and motor is rated by the torque it delivers, and every gearbox trades torque against rotation rate. A car engine producing $300$ N m at the crankshaft drives the wheels through gears that multiply the torque in low gear, giving the large force at the road needed to start moving, and multiply it less in high gear, where speed matters more.
Power links the two: $P = \tau\omega$, the rotational form of $P = Fv$. An electric motor that delivers $300$ N m at $300$ rad/s puts out $90$ kW. Wrench makers print torque settings in N m on lug-nut specifications for the same reason: the bolt's clamping force depends on the torque applied, not on the force alone.
A playground merry-go-round is close to a uniform disk: a $200$ kg platform of radius $2.5$ m has $I = \tfrac{1}{2}MR^2 = 625$ kg m². A parent pushing tangentially at the rim with $100$ N gives a torque of $250$ N m and an angular acceleration of $0.40$ rad/s². After five seconds it turns at $2.0$ rad/s, about one revolution every three seconds.
Pushing halfway in, at $1.25$ m, halves the torque and the acceleration. Four children of $30$ kg sitting at the rim add $4 \times 30 \times 2.5^2 = 750$ kg m², more than doubling the inertia. Playground safety standards limit how fast merry-go-rounds may be spun, and some modern designs include a governor, a brake that increases the retarding torque as the speed rises.
A flywheel stores energy as rotation, and its rotational inertia decides how much it can hold at a given speed. Grid-storage flywheels, such as those in a plant in Stephentown, New York, that help steady the grid's frequency, spin steel or carbon-fiber rotors at thousands of rpm in vacuum chambers to cut air drag.
Designers put mass near the rim, where $r^2$ weights it most, making the rotor as close to a hoop as its strength allows. The torque from the motor-generator determines how quickly energy flows in or out: charging with a torque of $\tau$ at angular speed $\omega$ adds power $\tau\omega$. Because a flywheel can deliver full power in seconds, the grid uses such plants to correct small frequency swings faster than any power station could.
Mass is a property of a body alone, so it is natural to think of rotational inertia the same way. It is not: $I$ depends on where the axis is. A rod's inertia about its end is four times its inertia about its center, and a door is far harder to swing about its handle edge than about its hinges only because of where the axis sits. Every value of $I$ must name its axis.
A second error is to think a larger force always gives a larger torque. A force aimed at the axis gives none at all; only the lever arm times the force counts.
A $40$ N push is applied $0.80$ m from a door's hinges, at $60°$ to the door. Find the lever arm.
$r_\perp = 0.80\sin 60° = 0.80 \times 0.866 = 0.69\ \text{m}$
The perpendicular distance to the line of the push.
Find the torque.
$\tau = 40 \times 0.69 = 27.7\ \text{N m}$
Force times lever arm.
Find the torque of a push straight toward the hinges.
$\tau = rF\sin 0° = 0$
No lever arm.
Find the best angle.
$\theta = 90° \Rightarrow \tau = 0.80 \times 40 = 32\ \text{N m}$
A perpendicular push uses the whole distance.
Find the door's angular acceleration with $I = 8.0$ kg m².
$\alpha = \dfrac{32}{8.0} = 4.0\ \text{rad/s}^2$
$\tau = I\alpha$ for the perpendicular push.
Slice a uniform disk of mass $M$ and radius $R$ into rings of radius $r$ and width $dr$. Write the mass of one ring.
$dm = \dfrac{M}{\pi R^2} \times 2\pi r\,dr = \dfrac{2M}{R^2}r\,dr$
Mass per area times the ring's area.
Write the ring's contribution.
$dI = r^2\,dm = \dfrac{2M}{R^2}r^3\,dr$
All of the ring is at distance $r$.
Integrate from the center to the rim.
$I = \dfrac{2M}{R^2}\displaystyle\int_0^Rr^3\,dr = \dfrac{2M}{R^2} \cdot \dfrac{R^4}{4}$
Power rule.
Simplify the result.
$I = \tfrac{1}{2}MR^2$
Half that of a hoop of the same mass and radius.
Evaluate for a $2.0$ kg, $0.30$ m disk.
$I = \tfrac{1}{2} \times 2.0 \times 0.09 = 0.090\ \text{kg m}^2$
Substitute.
A $4.0$ kg bucket hangs from a rope on a $12$ kg solid cylindrical windlass of radius $0.10$ m. Write the bucket's equation.
$4.0 \times 9.8 - T = 4.0a$
Downward positive.
Write the windlass's equation.
$T \times 0.10 = \tfrac{1}{2} \times 12 \times 0.10^2 \times \dfrac{a}{0.10}$
$\tau = I\alpha$ with $\alpha = a/R$.
Simplify for the tension.
$T = 6.0a$
Half the cylinder's mass times $a$.
Solve for the acceleration.
$39.2 = 10a \Rightarrow a = 3.92\ \text{m/s}^2$
Add the equations.
Find the tension.
$T = 6.0 \times 3.92 = 23.5\ \text{N}$
Less than the bucket's weight of $39.2$ N.
Find the windlass's angular acceleration.
$\alpha = \dfrac{3.92}{0.10} = 39.2\ \text{rad/s}^2$
The rope does not slip.
Apply the second law for rotation.
$\alpha = \dfrac{12}{3.0} = 4.0\ \text{rad/s}^2$
$\tau = I\alpha$.
Integrate for the angular velocity.
$\omega = \alpha t = 4.0 \times 2.0$
Constant angular acceleration from rest.
Evaluate the angular velocity.
A uniform rod has rotational inertia $9$ kg m² about an axis through its center, perpendicular to it. What is its rotational inertia about a parallel axis through one end?
Complete the worked solution: a force of $44$ N pushes on a wrench $0.4$ m from a bolt, at $90°$ to the wrench, with $\sin90° = 1$. Find the lever arm in m, the torque in N m, and the angular acceleration it would give a wheel of rotational inertia $2$ kg m², in rad/s².
Multiply the distance by the sine of the angle.
$r_\perp = r\sin\theta =$ l
The perpendicular distance from the axis to the force's line.
Multiply the force by the lever arm.
$\tau = Fr_\perp =$ q
Only the perpendicular part of the force turns the bolt.
Divide the torque by the rotational inertia.
$\alpha = \dfrac{\tau}{I} =$ a
Newton's second law for rotation.
Check the best angle.
$\text{torque is largest when the push is perpendicular}$
Then the lever arm is the full distance.
Match each body and axis to its rotational inertia.
| $MR^2$ | $\tfrac{1}{2}MR^2$ | $\tfrac{1}{12}ML^2$ | $\tfrac{1}{3}ML^2$ | |
|---|---|---|---|---|
| hoop about its axis | ||||
| solid disk about its axis | ||||
| rod about its center | ||||
| rod about one end |
A uniform rod has mass $6$ kg and length $2$ m. Fill in its rotational inertia about its center, about one end, and the parallel-axis term $M(L/2)^2$, all in kg m².
| value | |
|---|---|
| about the center (kg m²) | |
| about one end (kg m²) | |
| parallel-axis term (kg m²) |
A $4$ kg flywheel has rotational inertia $5$ kg m² about an axis through its center of mass. Write its inertia about a parallel axis a distance $d$ away, in kg m², as a formula in $d$ (meters).
Answer:
A $2$ kg bucket hangs from a rope wound around a well's windlass, a uniform solid cylinder of mass $16$ kg turning on frictionless bearings. With $g = 10$ m/s², what is the bucket's acceleration as it falls, in m/s²?
Answer: m/s² for the bucket
At a Chicago park, a parent pushes tangentially on the rim of a merry-go-round, a uniform disk of mass $200$ kg and radius $2.5$ m, with a force of $124$ N. Ignoring friction, what is its angular acceleration, in rad/s²?
Answer: rad/s² for the merry-go-round
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A $6$ kg flywheel has rotational inertia $3$ kg m² about an axis through its center of mass. Write its inertia about a parallel axis a distance $d$ away, in kg m², as a formula in $d$ (meters).
Answer:
You can use torque and rotational inertia. Explain to someone why a door is easy to open at the handle and hard to open near the hinges.
19. Your turn: a torque of $12$ N m acts on a wheel with $I = 3.0$ kg m². Find its angular acceleration and its speed after $2.0$ s from rest., step 3
$\omega = 8.0\ \text{rad/s}$
About $1.3$ revolutions per second.