Unit 5 Foundations: How Forces Create Rotation

Torque

What torque is (and what it isn’t)

Torque is the rotational “effectiveness” of a force. When you push on an object that can rotate about some point or axis (a hinge, a pivot, an axle), the same force can produce very different amounts of rotation depending on where and how you apply it. Torque captures that idea quantitatively.

A key mindset shift: a force by itself does not guarantee rotation. If you push on a door right at the hinge, you can push hard and still get almost no rotation. Torque depends on both the force and the geometry of where it’s applied.

Torque is not energy. Even though the unit looks similar (newton-meter), torque is not measured in joules. A joule is \text{N}\cdot\text{m} too, but it represents energy (a scalar), while torque represents a tendency to rotate (a vector/pseudovector). On the AP exam, treating torque as energy is a common conceptual trap.

Why torque matters

Torque is the rotational analog of force in linear motion:

• In translation, net force determines how velocity changes.
• In rotation, net torque determines how angular velocity changes.

Later in Unit 5, this links directly to rotational dynamics through:

\tau_{\text{net}} = I\alpha

Even before you use that relationship, torque is essential for static equilibrium problems, where an object is not accelerating linearly or angularly.

How torque works: lever arm and angle

To understand torque, focus on two ideas:

1. Distance from the pivot matters: forces applied farther from the axis usually create more torque.
2. Direction matters: only the component of the force that is perpendicular to the radius (line from pivot to where the force is applied) produces torque.

The basic magnitude formula

For a force F applied at a point a distance r from the pivot (where r is the distance from the axis to the point of application), and where \theta is the angle between the radius vector and the force, the torque magnitude is:

\tau = rF\sin\theta

• \tau is torque (SI unit \text{N}\cdot\text{m})
• r is the distance from pivot to point of application (meters)
• F is the force magnitude (newtons)
• \theta is the angle between \vec r and \vec F

This formula builds the “perpendicular component” idea into \sin\theta. If the force is applied directly along the radius (pushing straight toward or away from the pivot), then \theta = 0 or \theta = 180^\circ and \sin\theta = 0, so there is no torque.

Lever arm (moment arm) viewpoint

Another way you’ll often see torque handled is with the lever arm (also called moment arm) r_{\perp}, the perpendicular distance from the pivot to the force’s line of action. Then:

\tau = F r_{\perp}

This is especially helpful in messy geometry problems because you can draw the line of action of the force and measure the shortest distance to the pivot.

Direction (sign) of torque

Torque has a direction: it tends to rotate an object clockwise or counterclockwise in a plane.

In most AP Physics 1 problems:

• Choose counterclockwise as positive and clockwise as negative (or vice versa), and be consistent.
• Then compute signed torques and add them.

A typical sign convention approach:

• If the force tends to rotate the object counterclockwise about the chosen axis, \tau > 0.
• If it tends to rotate clockwise, \tau < 0.

You do not need vector cross products for most AP 1 plane-rotation problems, but it’s useful to know the compact vector definition:

\vec\tau = \vec r \times \vec F

Choosing a pivot (axis) strategically

In equilibrium and torque-balance questions, you usually have freedom to choose the pivot about which you sum torques. The physics doesn’t change, but the algebra can get much easier.

A powerful strategy is to choose a pivot through a point where unknown forces act (like a hinge reaction force). Then those unknown forces produce no torque because r = 0 for that point, so they disappear from the torque equation.

Torque and static equilibrium

For an object to be in static equilibrium (not accelerating linearly and not accelerating angularly), you need both:

\sum F_x = 0

\sum F_y = 0

\sum \tau = 0

The torque condition prevents rotation. The force conditions prevent translation.

A common misconception is thinking \sum \tau = 0 alone guarantees equilibrium. It only guarantees no angular acceleration about the chosen axis. An object can still translate if the net force is not zero.

Worked example 1: torque from a force at an angle

A student pushes on a wrench to loosen a bolt. The wrench length from bolt to hand is r = 0.25\ \text{m}. The student applies a force of F = 80\ \text{N} at an angle of \theta = 60^\circ relative to the wrench handle (so the force is not perfectly perpendicular).

Step 1: Identify the torque relationship.
Use:

\tau = rF\sin\theta

Step 2: Substitute values.

\tau = (0.25)(80)\sin 60^\circ

Step 3: Compute.
Since \sin 60^\circ \approx 0.866:

\tau \approx (20)(0.866) = 17.3\ \text{N}\cdot\text{m}

Interpretation: If the student instead pushed perfectly perpendicular (making \theta = 90^\circ), the torque would be larger for the same force because \sin 90^\circ = 1.

Worked example 2: torque balance for a uniform beam

A uniform horizontal beam of length L = 4.0\ \text{m} is supported at its left end by a hinge and held up at its right end by a vertical rope. The beam has mass m = 20\ \text{kg}. Find the tension in the rope when the beam is in static equilibrium.

Step 1: Draw a free-body diagram.
Forces on the beam:

• Weight mg acting at the center of mass, located at L/2 from the hinge.
• Rope tension T upward at the right end (distance L from the hinge).
• Hinge forces at the left end (unknown components), but we can eliminate them by choosing the hinge as the pivot.

Step 2: Choose pivot at the hinge.
Then hinge forces create zero torque.

Step 3: Write torque equilibrium about the hinge.
Take counterclockwise as positive.

• Tension tends to rotate CCW: positive torque +TL.
• Weight tends to rotate CW: negative torque -mg(L/2).

Set sum of torques to zero:

TL - mg\frac{L}{2} = 0

Step 4: Solve for T.

TL = mg\frac{L}{2}

Cancel L:

T = \frac{mg}{2}

Substitute m = 20\ \text{kg} and g = 9.8\ \text{m/s}^2:

T = \frac{(20)(9.8)}{2} = 98\ \text{N}

Common pitfall: Putting the weight at the end instead of the center. For a uniform beam, the center of mass is at the midpoint.

Notation reference (torque)

Exam Focus

• Typical question patterns:
  • Compute net torque about a pivot for several forces at different distances and angles, then decide rotation direction or equilibrium.
  • Static equilibrium of beams, ladders, or signs: use \sum \tau = 0 with a smart pivot choice.
  • Conceptual comparisons: “Where should you push to maximize torque?” or “Which force produces zero torque?”
• Common mistakes:
  • Using \tau = rF automatically without the perpendicular component (forgetting \sin\theta or lever arm).
  • Measuring r incorrectly (it must be from the axis to the point of application, not the whole object length unless appropriate).
  • Confusing \text{N}\cdot\text{m} for torque with joules and treating torque like energy.

Rotational Inertia

What rotational inertia is

Rotational inertia (also called moment of inertia) measures how hard it is to change an object’s rotational motion about a particular axis. It plays the same role in rotation that mass plays in translation.

If you’ve ever spun a light bicycle wheel versus a heavier one, you’ve felt this: the wheel with more of its mass far from the axis is harder to start spinning and harder to stop.

The key idea is not just “more mass means more inertia” but:

• Mass farther from the axis increases rotational inertia dramatically.

That’s why a figure skater spins faster when pulling arms in: the distribution of mass relative to the spin axis changes.

Why rotational inertia matters

Rotational inertia connects torque to angular acceleration:

\tau_{\text{net}} = I\alpha

• \tau_{\text{net}} is the net external torque about the axis
• I is rotational inertia about that axis
• \alpha is angular acceleration

This tells you two major things:

1. For a fixed torque, a larger I gives a smaller \alpha (harder to spin up).
2. For a fixed angular acceleration, a larger I requires a larger torque.

This is why tools like wrenches are long (bigger torque for the same force) and why flywheels are designed to have large rotational inertia (to resist changes in rotational speed).

How rotational inertia works: dependence on axis and mass distribution

Rotational inertia is not a single inherent property of an object the way mass is. It depends on the axis of rotation.

The same rod has a different moment of inertia if you rotate it about:

• its center,
• one end,
• an axis parallel but shifted.

So whenever you talk about I, you must specify “about what axis?”

Discrete masses: the core definition

For a system of point masses rotating about an axis, rotational inertia is:

I = \sum m_i r_i^2

• m_i is the mass of the ith particle
• r_i is that particle’s perpendicular distance to the axis

The square is crucial: doubling distance from the axis multiplies that mass’s contribution to I by 4.

Continuous objects: standard results you’re allowed to use

For solid objects, you typically use standard formulas (provided on the AP equation sheet or commonly known). Here are the most common ones used in AP Physics 1:

• M is total mass, R is radius, L is length.

Conceptual meaning: A hoop has more mass concentrated at radius R, so it has larger I than a disk of the same M and R.

Connecting torque, angular acceleration, and rotational inertia

Once you know I, rotational dynamics problems often look like the linear version:

• Linear: F_{\text{net}} = ma
• Rotational: \tau_{\text{net}} = I\alpha

But you must be careful about the “about an axis” nature of rotation. Forces may produce different torques depending on where they act, and I changes if the axis changes.

Another common misconception is thinking “bigger object means bigger I.” Size matters only through where the mass is relative to the axis. A large but very light ring could have a smaller I than a smaller but very massive disk—what matters is the combination of mass and radius squared.

Worked example 1: moment of inertia of point masses

Three small masses are attached to a lightweight rod (rod mass is negligible) and rotate about an axis through the rod’s left end, perpendicular to the rod.

• m_1 = 0.50\ \text{kg} at r_1 = 0.20\ \text{m}
• m_2 = 0.30\ \text{kg} at r_2 = 0.40\ \text{m}
• m_3 = 0.20\ \text{kg} at r_3 = 0.60\ \text{m}

Find the rotational inertia I.

Step 1: Use the definition for discrete masses.

I = \sum m_i r_i^2

Step 2: Compute each term.

I = (0.50)(0.20^2) + (0.30)(0.40^2) + (0.20)(0.60^2)

Compute squares:

• 0.20^2 = 0.040
• 0.40^2 = 0.160
• 0.60^2 = 0.360

So:

I = (0.50)(0.040) + (0.30)(0.160) + (0.20)(0.360)

I = 0.020 + 0.048 + 0.072 = 0.140\ \text{kg}\cdot\text{m}^2

Interpretation: Even though m_3 is the smallest mass, it contributes the most because it’s farthest from the axis.

Worked example 2: comparing shapes (hoop vs disk)

Two objects have the same mass M and radius R:

• Object A is a thin hoop.
• Object B is a solid disk.

Compare their rotational inertias about the central axis.

For the hoop:

I_A = MR^2

For the disk:

I_B = \frac{1}{2}MR^2

So:

\frac{I_A}{I_B} = \frac{MR^2}{\frac{1}{2}MR^2} = 2

Conclusion: The hoop has twice the rotational inertia of the disk about that axis. If you apply the same net torque to each, the disk will have twice the angular acceleration because:

\alpha = \frac{\tau_{\text{net}}}{I}

Real-world application: why pulling mass inward speeds rotation

Even without doing full angular momentum problems yet, rotational inertia helps you predict what happens when mass moves relative to an axis.

If the net external torque is small, systems often approximately conserve angular momentum, which implies that decreasing I tends to increase rotational speed. When a skater pulls arms in, mass moves closer to the axis, decreasing the r^2 contributions in I = \sum m r^2.

The important “feel” of the concept: mass near the axis is “cheap” in rotational inertia; mass far away is “expensive.”

Notation reference (rotational inertia)

Exam Focus

• Typical question patterns:
  • Compute I for a set of point masses, then use it in \tau_{\text{net}} = I\alpha.
  • Compare angular accelerations for different objects under the same applied torque by comparing their I values.
  • Use provided/formula-sheet moments of inertia (disk, hoop, rod) in multi-step rotation problems.
• Common mistakes:
  • Forgetting that I depends on the axis (using a rod-about-center formula when the axis is at an end).
  • Treating rotational inertia like mass (assuming it’s the same no matter how the object rotates).
  • Dropping the square in r^2 when computing I = \sum m r^2, which massively changes results.