Free AP Physics 1 Torque and Rotational Dynamics

AP Physics 1 · Torque and Rotational Dynamics · free lesson

Torque

The idea, then worked examples solved step by step, then practice questions whose numbers change every time you try them.

The idea

What makes a rigid system turn

A force can change how an object moves along a line, and a force can also change how it turns. Push a door close to its hinges and it hardly moves. Push the same amount at the handle and it swings. Push the door straight toward the hinges, along the door, and it does not turn at all. What matters is not only how large the force is, but where it acts and in which direction.

This lesson is about a rigid system: an object, or a group of objects, whose parts keep their places relative to each other as it turns. A door, a wheel, a wrench and a meter stick are rigid systems. They turn about an axis of rotation: a line, such as the line through the hinges, that stays fixed while the system turns. The point where a force acts on the system is its point of application.

Definition.

The position vector $\vec r$ points from the axis of rotation to the point of application of a force. The line of action of a force is the straight line through the point of application along the force. The lever arm $r_\perp$ is the perpendicular distance from the axis of rotation to the line of action of the force.

Take a wrench with its axis at the left end. A force of $30\,\mathrm{N}$ acts at a point $0.4\,\mathrm{m}$ from the axis, at an angle of $30$ degrees to the wrench. The picture shows the position vector (blue), the force (red) and its line of action (dashed gray, extended backward). The green segment is the lever arm: it runs from the axis to the line of action and meets it at a right angle.

The lever arm is shorter than $r$ . Here $r$ = $0.4\,\mathrm{m}$ and the lever arm is $0.200\,\mathrm{m}$ . If the force were at a right angle to the wrench, the lever arm would be the whole of $r$ .

The torque of a force

Definition.

Torque is the measure of how strongly a force turns a rigid system about an axis. It results only from the component of the force that is perpendicular to the position vector $\vec r$ . The component of the force along the position vector, pointing at the axis or away from it, has no turning effect.

Rule.

Magnitude of the torque. For a force $F$ applied at a distance $r$ from the axis, with $\theta$ the angle between the force vector and the position vector from the axis to the point of application,

\begin{aligned} \tau &= rF\sin\theta \end{aligned}

The unit of torque is the newton-meter (N·m).

There are two ways to read the same product, and they give the same number.

$\tau = r\,(F\sin\theta)$ : the distance times the perpendicular component of the force.
$\tau = (r\sin\theta)\,F$ : the lever arm times the whole force.

The components are not extra forces. The next picture shows the same force as its two components, $F_p$ perpendicular to $\vec r$ and $F_a$ along it: the red dashed arrow is perpendicular to $\vec r$ and turns the wrench, and the orange dashed arrow lies along it and does not.

\begin{aligned} F_p &= F\sin(\theta^\circ) \\ &= 30\,\mathrm{N} \times \sin(30^\circ) \\ &= 15.0\,\mathrm{N} \\ F_a &= F\cos(\theta^\circ) \\ &= 30\,\mathrm{N} \times \cos(30^\circ) \\ &= 26.0\,\mathrm{N} \\ \tau &= rF_p \\ &= 0.4\,\mathrm{m} \times 15\,\mathrm{N} \\ &= 6.00\,\mathrm{N{\cdot}m} \end{aligned}

The perpendicular component is $15.0\,\mathrm{N}$ and the component along $\vec r$ is $26.0\,\mathrm{N}$ . The torque is $0.4\,\mathrm{m}$ times $15.0\,\mathrm{N}$ , which is $6.00\,\mathrm{N{\cdot}m}$ . The other reading gives the same: the lever arm $0.200\,\mathrm{m}$ times the whole force $30\,\mathrm{N}$ is $6.00\,\mathrm{N{\cdot}m}$ .

How the torque changes with the angle

With $r$ and $F$ fixed, the torque depends on the angle through $\sin\theta$ . The animation turns the force about its point of application from 0 to 180 degrees. The lever arm (green) grows from zero to $r$ and shrinks again, and the torque reading follows it.

The same thing as a graph, for the force of $30\,\mathrm{N}$ at $0.4\,\mathrm{m}$ :

The torque is zero at 0 and at 180 degrees, when the line of action passes through the axis. It is largest at 90 degrees, where it is $rF$ = $12.0\,\mathrm{N{\cdot}m}$ . The torque at 30 degrees equals the torque at 150 degrees, since $\sin 30^\circ = \sin 150^\circ$ .

Rule.

Special cases. The torque is greatest, $\tau = rF$ , when the force is perpendicular to the position vector ( $\theta = 90^\circ$ , lever arm equal to $r$ ). The torque is zero when the force is along the position vector ( $\theta = 0^\circ$ or $180^\circ$ ) and when the force acts at the axis ( $r = 0$ ). In both cases the lever arm is zero.

Same force, different place

With the force perpendicular to the wrench, $\tau = rF$ : the torque is in proportion to the distance from the axis. A force of $30\,\mathrm{N}$ gives a torque three times as large at a point three times as far out. This is why door handles are placed far from the hinges, and why a long wrench loosens a bolt that a short one cannot.

The graph is a straight line through the origin, and its slope is the force, $30\,\mathrm{N}$ . A longer lever arm gives the same torque with a smaller force: to get $12.0\,\mathrm{N{\cdot}m}$ with a lever arm of $0.8\,\mathrm{m}$ takes only $15\,\mathrm{N}$ .

Force diagrams

Definition.

A force diagram is like a free-body diagram: it shows the forces on one rigid system, with arrows whose lengths show the relative magnitudes and whose directions are the forces' directions. Unlike a free-body diagram, it also shows where each force is applied relative to the axis of rotation. It is the diagram used to find the torques on a rigid system.

In a free-body diagram all the arrows start from one dot, since only the net force matters. For torques the point of application matters, so each arrow starts where its force acts, and the axis is marked. Draw each force whole, and not its components.

A light rod of length $1\,\mathrm{m}$ can turn about a fixed axis at its left end. Three forces act on it: $20\,\mathrm{N}$ upward at $0.6\,\mathrm{m}$ from the axis, $15\,\mathrm{N}$ downward at $0.9\,\mathrm{m}$ from the axis, and $25\,\mathrm{N}$ at $0.3\,\mathrm{m}$ from the axis, at an angle of $150$ degrees to the position vector (up, and tilted back toward the axis). The pivot also pushes on the rod, but at the axis itself.

This is a force diagram. The three arrows start at the points where the forces act, and their lengths are in proportion to the forces. The pivot's force is not drawn because it acts at the axis: with $r = 0$ it has no torque, whatever its size.

Clockwise and counterclockwise

A torque about a given axis turns the system one of two ways, clockwise or counterclockwise, and this course describes the direction of a rotation in no other way. The direction of torque as a vector in three dimensions is beyond this course. Here a counterclockwise torque is taken as positive and a clockwise one as negative. For a force at an angle $\theta$ from the position vector, with $\theta$ between 0 and 180 degrees, the force turns the system counterclockwise when it points to the left of $\vec r$ as seen from the axis, and clockwise when it points to the right.

Take the three forces on the rod one at a time. The $20\,\mathrm{N}$ force pushes up on the right of the axis, so it turns the rod counterclockwise. It is at a right angle to the rod.

\begin{aligned} \tau &= x_1F_1 \\ &= 0.6\,\mathrm{m} \times 20\,\mathrm{N} \\ &= 12.0\,\mathrm{N{\cdot}m} \end{aligned}

Its torque is $12.0\,\mathrm{N{\cdot}m}$ , counterclockwise. The $15\,\mathrm{N}$ force pushes down on the right of the axis, so it turns the rod clockwise. It is also at a right angle to the rod.

\begin{aligned} \tau &= x_2F_2 \\ &= 0.9\,\mathrm{m} \times 15\,\mathrm{N} \\ &= 13.5\,\mathrm{N{\cdot}m} \end{aligned}

Its torque is $13.5\,\mathrm{N{\cdot}m}$ , clockwise, which is $-13.5$ N·m as a signed number. The $25\,\mathrm{N}$ force is at $150$ degrees to the rod, so only part of it is perpendicular to the rod, and it points up to the left of $\vec r$ , so it turns the rod counterclockwise.

\begin{aligned} \tau &= x_3F_3\sin(\phi^\circ) \\ &= 0.3\,\mathrm{m} \times 25\,\mathrm{N} \times \sin(150^\circ) \\ &= 3.75\,\mathrm{N{\cdot}m} \end{aligned}

Its torque is $3.75\,\mathrm{N{\cdot}m}$ , counterclockwise.

The bars show the torque of each force about the axis, with its sign: up is counterclockwise and down is clockwise. The $15\,\mathrm{N}$ force has the largest torque, $13.5$ N·m, although it is not the largest force: it acts farthest from the axis. The $25\,\mathrm{N}$ force, the largest force, has the smallest torque, because it acts close to the axis and at a poor angle. Two torques cancel only when they are equal in size and opposite in sense.

Warning.

Four traps. Torque is about an axis: say which one, because the same force has a different torque about a different axis. The distance in $\tau = rF\sin\theta$ is measured from the axis to the point where the force acts, not to the end of the object. The angle is the one between the force and the position vector, so a force at right angles to the object gives $\sin\theta = 1$ , while a force along it gives zero, however large it is. And torque is in N·m: the unit is the same as that of energy, but a torque is not an energy, so do not write it in joules.

Worked examples

Five examples of the kinds the AP exam asks on this topic: a torque from a force at an angle, a "what changes if" on a door, the torques of the forces on a rod from a force diagram, a force found from a torque, and a claim justified.

Example.

Example 1. A wrench pulled at an angle

A mechanic pulls on a wrench with a force of $80\,\mathrm{N}$ at a point $0.25\,\mathrm{m}$ from the center of a bolt. The force makes an angle of $65$ degrees with the wrench. Find the torque about the bolt and the lever arm.

Solution

The angle is the one between the force and the position vector, which here runs along the wrench from the bolt. Use $\tau = rF\sin\theta$ .

\begin{aligned} \tau &= rF\sin(\theta^\circ) \\ &= 0.25\,\mathrm{m} \times 80\,\mathrm{N} \times \sin(65^\circ) \\ &= 18.1\,\mathrm{N{\cdot}m} \\ r_p &= r\sin(\theta^\circ) \\ &= 0.25\,\mathrm{m} \times \sin(65^\circ) \\ &= 0.227\,\mathrm{m} \end{aligned}

Result.

The torque is $18.1\,\mathrm{N{\cdot}m}$ and the lever arm $r_p$ is $0.227\,\mathrm{m}$ . The same torque follows from the lever arm times the force, $18.1\,\mathrm{N{\cdot}m}$ . The force is $80\,\mathrm{N}$ but only its perpendicular part, $72.5\,\mathrm{N}$ , turns the bolt. At a right angle the torque would be $20.0\,\mathrm{N{\cdot}m}$ .

Example.

Example 2. A door

A door is $0.9\,\mathrm{m}$ wide. A person pushes at right angles to the door with a force of $12\,\mathrm{N}$ , first at the handle, at the outer edge, and then at a point $0.15\,\mathrm{m}$ from the hinges. How do the two torques about the hinges compare? What force at the second point gives the torque the push at the handle gave?

Solution

The push is perpendicular to the door both times, so $\sin\theta = 1$ and $\tau = rF$ .

\begin{aligned} \tau &= wF \\ &= 0.9\,\mathrm{m} \times 12\,\mathrm{N} \\ &= 10.8\,\mathrm{N{\cdot}m} \end{aligned}

At the handle the torque is $10.8\,\mathrm{N{\cdot}m}$ . Near the hinges:

\begin{aligned} \tau &= xF \\ &= 0.15\,\mathrm{m} \times 12\,\mathrm{N} \\ &= 1.80\,\mathrm{N{\cdot}m} \\ F_2 &= \frac{wF}{x} \\ &= \frac{0.9\,\mathrm{m} \times 12\,\mathrm{N}}{0.15\,\mathrm{m}} \\ &= 72.0\,\mathrm{N} \end{aligned}

Result.

The torque at the handle is $10.8$ N·m and near the hinges it is $1.80\,\mathrm{N{\cdot}m}$ , which is the fraction $0.167$ of the first, the same as the ratio of the distances. To get $10.8$ N·m at $0.15\,\mathrm{m}$ from the hinges takes $72.0\,\mathrm{N}$ , $6.00$ times the force. The same push does less near the hinges because its lever arm is shorter, so a door is hard to open there.

Example.

Example 3. The torque of each force on a rod

A light rod can turn about an axis at its left end. A force of $18\,\mathrm{N}$ pushes down on it $0.4\,\mathrm{m}$ from the axis. A force of $24\,\mathrm{N}$ pushes up $0.8\,\mathrm{m}$ from the axis. A force of $30\,\mathrm{N}$ pulls $0.6\,\mathrm{m}$ from the axis, directed $30$ degrees below the rod's direction, down and to the right. Draw the force diagram. Find the torque of each force about the axis, with its sense, and say which force has the greatest torque.

Solution

Each arrow starts at the point where its force acts, and its length shows the force. A downward push to the right of the axis turns the rod clockwise. An upward push there turns it counterclockwise. The third force points to the right of the position vector, below the rod, so it turns the rod clockwise.

\begin{aligned} \tau &= x_1F_1 \\ &= 0.4\,\mathrm{m} \times 18\,\mathrm{N} \\ &= 7.20\,\mathrm{N{\cdot}m} \end{aligned}

The $18\,\mathrm{N}$ force has a torque of $7.20\,\mathrm{N{\cdot}m}$ , clockwise.

\begin{aligned} \tau &= x_2F_2 \\ &= 0.8\,\mathrm{m} \times 24\,\mathrm{N} \\ &= 19.2\,\mathrm{N{\cdot}m} \end{aligned}

The $24\,\mathrm{N}$ force has a torque of $19.2\,\mathrm{N{\cdot}m}$ , counterclockwise.

\begin{aligned} \tau &= x_3F_3\sin(\phi^\circ) \\ &= 0.6\,\mathrm{m} \times 30\,\mathrm{N} \times \sin(30^\circ) \\ &= 9.00\,\mathrm{N{\cdot}m} \end{aligned}

Result.

The $30\,\mathrm{N}$ force has a torque of $9.00\,\mathrm{N{\cdot}m}$ , clockwise. The $24\,\mathrm{N}$ force, which is not the largest, has the greatest torque, because its lever arm is the longest: it acts $0.8\,\mathrm{m}$ from the axis at a right angle. The $30\,\mathrm{N}$ force, the largest, loses part of its effect since only the fraction $0.500$ of it is perpendicular to the rod.

Example.

Example 4. The force needed

A bolt needs a torque of $14\,\mathrm{N{\cdot}m}$ to turn. A wrench pulls at a point $0.35\,\mathrm{m}$ from the bolt. What force is needed if the pull is at right angles to the wrench? What force is needed if the pull is 40 degrees from the wrench's direction?

Solution

Solve $\tau = rF\sin\theta$ for $F$ .

\begin{aligned} F &= \frac{\tau}{r\sin(\theta^\circ)} \\ &= \frac{14\,\mathrm{N{\cdot}m}}{0.35\,\mathrm{m} \times \sin(90^\circ)} \\ &= 40.0\,\mathrm{N} \end{aligned}

At a right angle the force is $40.0\,\mathrm{N}$ . At 40 degrees:

\begin{aligned} F &= \frac{\tau}{r\sin(\theta^\circ)} \\ &= \frac{14\,\mathrm{N{\cdot}m}}{0.35\,\mathrm{m} \times \sin(40^\circ)} \\ &= 62.2\,\mathrm{N} \end{aligned}

Result.

At a right angle the force is $40.0\,\mathrm{N}$ . At 40 degrees it is $62.2\,\mathrm{N}$ , which is $1.56$ times as much, since the sine of 40 degrees is $0.643$ . A force at a smaller angle must be larger because less of it is perpendicular to the wrench.

Example.

Example 5. A claim justified

A student pushes on a rod, pivoted at one end, at a point $0.5\,\mathrm{m}$ from the pivot. In trial 1 the force is $20\,\mathrm{N}$ , at $90$ degrees to the rod. In trial 2 the force is $30\,\mathrm{N}$ , at 30 degrees to the rod. The student claims: "Trial 2 has the larger torque, because its force is larger." Is the claim right?

Solution

\begin{aligned} \tau &= rF_1\sin(\theta^\circ) \\ &= 0.5\,\mathrm{m} \times 20\,\mathrm{N} \times \sin(90^\circ) \\ &= 10.0\,\mathrm{N{\cdot}m} \end{aligned}

The torque in trial 1 is $10.0\,\mathrm{N{\cdot}m}$ . In trial 2:

\begin{aligned} \tau &= rF_2\sin(\theta^\circ) \\ &= 0.5\,\mathrm{m} \times 30\,\mathrm{N} \times \sin(30^\circ) \\ &= 7.50\,\mathrm{N{\cdot}m} \end{aligned}

Result.

The torque is $10.0$ N·m in trial 1 and $7.50\,\mathrm{N{\cdot}m}$ in trial 2. The claim is wrong: the larger force gives the smaller torque. Torque depends on the perpendicular component of the force, and in trial 2 only $15.0\,\mathrm{N}$ of the $30\,\mathrm{N}$ is perpendicular to the rod, less than the $20\,\mathrm{N}$ of trial 1, whose force is all perpendicular.

Practice

Answer AP-style questions on torque. Each answer is checked, with the full solution, and every try brings new numbers. Practice needs a free account: it keeps your progress.

Practise torque with new numbers