foundation28 min studyLesson 6 in Physics

Rotation, torque, and angular momentum

Explain why force placement matters and connect rotational motion to angular momentum about a stated origin.

Before you begin

On this page

The question for today

What you’ll understand

Calculate a torque from its lever arm.
Distinguish mass from moment of inertia.
State the conditions for angular-momentum conservation.

Opening a door near its hinge feels different from pushing at its outer edge, even with the same force. Translational dynamics alone does not express that difference. Rotation requires information about where forces act and how mass is distributed. Torque and angular momentum capture those geometric facts while extending the conservation reasoning already used for collisions.

Angle links many points of a rigid body

For rotation about a fixed axis, every point of a rigid body shares the same angular displacement, angular velocity, and angular acceleration. Their linear speeds differ because points farther from the axis travel larger circles. Radians are the natural angular measure: arc length divided by radius. Convert degree values before using calculus formulas that assume radians.

Circular motion requires inward acceleration even at constant speed, because velocity changes direction. The name centripetal describes this acceleration component and the corresponding net-force component. It is not another force to add beside tension, friction, or gravity. Those actual interactions may provide the required inward force, depending on the apparatus and the point in the orbit.

v=r\omega,\qquad a_{\mathrm{tan}}=r\alpha,\qquad a_{\mathrm{rad}}=r\omega^2

For a point a perpendicular distance r from a fixed rotation axis, ω is angular speed in rad/s and α angular acceleration in rad/s². The radial acceleration points toward the axis; tangential acceleration changes speed.↔ Scroll wide equations horizontally.

Torque includes the lever arm

Torque about an origin is the cross product of the position from that origin and the applied force. Its magnitude depends on the perpendicular lever arm, not merely the distance to the point of contact. A force directed along a door toward the hinge can be large yet exert zero torque about the hinge. Changing the reference origin can change torque.

For a rigid body constrained to rotate around a fixed axis, moment of inertia weights mass by squared distance from that axis. Moving mass outward increases resistance to a given angular acceleration even when total mass is unchanged. A solid disk and a thin hoop therefore respond differently to the same torque. The simple scalar rotational law assumes the axis and rigid-body conditions are appropriate.

\boldsymbol\tau=\mathbf r\times\mathbf F,\qquad I=\sum_i m_i r_{\perp i}^{\,2},\qquad \tau_{\mathrm{axis}}=I\alpha

The last equality applies to a constant moment of inertia about a suitable fixed axis. I has units kg·m². Torque has units N·m, but it is a rotational vector quantity, not a scalar energy despite sharing energy's dimensions.↔ Scroll wide equations horizontally.

Conservation needs a specified axis or origin

A particle's angular momentum about an origin combines position and linear momentum. For an isolated system, internal exchanges can redistribute angular momentum while leaving its total unchanged. For rotation about a fixed axis, zero net external torque about that axis keeps the axial component constant. A nonzero external force can coexist with zero torque if its line of action passes through the origin.

When a spinning person pulls their arms inward, their moment of inertia decreases and their angular speed can increase with almost conserved angular momentum. Their rotational kinetic energy need not stay constant: muscles do work during the change. Angular momentum later becomes central to atomic states and spin, but quantum spin should not be imagined as a tiny rigid sphere literally rotating in space.

\mathbf L=\mathbf r\times\mathbf p,\qquad \frac{d\mathbf L_{\mathrm{tot}}}{dt}=\boldsymbol\tau_{\mathrm{ext}},\qquad L_{\mathrm{axis}}=I\omega

Use a fixed inertial origin for the displayed general derivative law. The scalar L = Iω concerns the component along an appropriate fixed axis; general three-dimensional rotation requires the inertia tensor.↔ Scroll wide equations horizontally.

Work it through

Start a disk turning

A uniform solid disk has mass $2.0\,\mathrm{kg}$ and radius $0.50\,\mathrm{m}$ . A tangential $4.0\,\mathrm{N}$ force acts at its rim. It starts from rest on a frictionless fixed axle. Find angular acceleration and angular speed after $3.0\,\mathrm{s}$ .

Use the solid-disk moment of inertia about its symmetry axis: $I=MR^2/2=(2.0)(0.50)^2/2=0.25\,\mathrm{kg\,m^2}$ . A hoop formula would describe a different mass distribution.
The force is perpendicular to the radius, so its torque magnitude is $\tau=RF=(0.50)(4.0)=2.0\,\mathrm{N\,m}$ . Thus $\alpha=\tau/I=8.0\,\mathrm{rad\,s^{-2}}$ .
Constant applied torque and constant I give constant angular acceleration. Therefore $\omega=\omega_0+\alpha t=0+(8.0)(3.0)=24\,\mathrm{rad\,s^{-1}}$ . The axle constrains translation but contributes no resisting torque in this model.

The result

The disk accelerates at $8.0\,\mathrm{rad\,s^{-2}}$ and reaches $24\,\mathrm{rad\,s^{-1}}$ after three seconds. The applied force supplies rotational energy throughout that interval.

The living notebook

Explore the model

Interactive lab

Interactive lab · vectors

Direction controls the effective push

Change one vector's angle and magnitude. With a lever arm drawn along +x, its y component is the force component producing torque. The lab illustrates geometry; it does not evolve a rigid body or compute an inertia tensor.

Vectorx / y projectionsTrace marker

x component: 1.393
y component: 0.975
Magnitude: 1.7

Angle35°Magnitude1.7

t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe component perpendicular to the lever arm produces the torque magnitude.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

Mass stays essentially unchanged, but moment of inertia decreases. With

I\omega

approximately constant, ω increases. Muscular work can change rotational kinetic energy, so angular-momentum conservation does not imply energy conservation for rotation alone.

Your turn

Try it yourself

A $10\,\mathrm{N}$ force is applied $0.30\,\mathrm{m}$ from an axis at an angle of $30^\circ$ to the radial lever arm. What is the torque magnitude? What happens if the force is radial?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use only the component of force perpendicular to the radius.

Compare with the solution

The torque is $\tau=rF\sin\theta=(0.30)(10)\sin30^\circ=1.5\,\mathrm{N\,m}$ . The perpendicular component is $5.0\,\mathrm{N}$ . A radial force has θ = 0 or π and therefore zero torque about this axis, though it can still affect other aspects of motion or the axle's reaction force.

Keep the idea

Take it with you

Torque depends on where and in which direction a force acts.
Moment of inertia depends on the rotation axis and mass distribution.
Zero external torque conserves angular momentum, not necessarily rotational kinetic energy.

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