10 - Angular Momentum and Rolling Motion

A structured guide to angular momentum, torque, rolling constraints, rotational energy, and the analysis of systems that combine translation with rotation.

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Rotational motion can be described with quantities that parallel linear momentum and energy. The central ideas are , , , and .

For a particle with linear momentum $p=mv$ , the magnitude about a selected axis is

L=r_\perp p=r_\perp mv=rp\sin\phi.

The result depends on the axis. The same particle can have zero about one axis and nonzero about another.

For a rigid object rotating about a fixed axis,

L=I\omega.

The is

I=\sum_i m_i r_i^2,

so mass located farther from the axis contributes disproportionately. Common center-of-mass values include:

Thin hoop or hollow cylinder: $I=MR^2$
Solid cylinder or disk: $I=\frac12 MR^2$
Solid sphere: $I=\frac25 MR^2$
Thin spherical shell: $I=\frac23 MR^2$
Point mass at distance $R$ : $I=MR^2$

Takeaway: Always identify the axis before calculating or .

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changes according to

\sum\tau=\frac{dL}{dt}.

For a rigid object with constant , this becomes

\sum\tau=I\alpha.

Over a time interval with constant net , the is

\Delta L=\tau_{\text{net}}\Delta t.

To apply these relationships, choose a positive rotational direction, such as counterclockwise, and keep signs consistent. , , and must all be evaluated about the same axis.

Before using conservation, follow this sequence:

Define the system.
Select the axis.
Identify external torques about that axis.
Decide whether those torques are zero or negligible during the relevant interval.
Write initial and final with consistent signs.

Internal torques can redistribute among parts of a system, but they cancel when the total system is considered. External torques can change the system’s total .

Takeaway: A net external is the mechanism that changes a system’s .

applies when the net external about the chosen axis is zero or negligible:

\sum\tau_{\text{ext}}=0\quad\Longrightarrow\quad L_i=L_f.

For a single rigid object whose changes,

I_i\omega_i=I_f\omega_f.

Thus, reducing $I$ increases the magnitude of $\\omega$ , while increasing $I$ decreases the magnitude of $\\omega$ .

For example, when a student pulls their arms inward on a frictionless rotating platform, the student-platform system changes from $I_i$ to $I_f$ . The final angular speed is

\omega_f=\frac{I_i}{I_f}\omega_i.

If $I_f$ is one-half of $I_i$ , then the final angular speed is twice the initial angular speed. The increase in does not contradict conservation: internal work converts chemical energy into .

Takeaway: A changing can change angular speed without changing total , provided external is negligible.

Rolling Motion and Contact Conditions

Rolling motion combines translation of an object’s center of mass with rotation about its center of mass. For , the is

v_{\text{CM}}=R\omega,\qquad a_{\text{CM}}=R\alpha,\qquad s=R\theta.

These relations are geometric constraints and apply only when there is no sliding at the contact point.

For a wheel rolling at center-of-mass speed $v_{\\text{CM}}$ , the top of the wheel moves at speed $2v_{\\text{CM}}$ relative to the ground. The instantaneous contact point has zero velocity relative to the ground because the center-of-mass velocity and the contact point’s tangential velocity cancel there.

is different. At the contact point, the object and surface slide relative to each other, so

v_{\text{CM}}\ne R\omega.

Kinetic friction acts during slipping and can transform mechanical energy into thermal energy. Do not impose the rolling-without-slipping constraint during this motion.

generally involves . Its direction is determined by the tendency of the contact point to slip. In ideal rolling, can provide without doing net work because the point of application has no displacement relative to the surface.

Takeaway: Check whether the contact point slips before using any relation involving $R\\omega$ .

Rotational Energy in Rolling

A rolling rigid object has both translational and :

K=K_{\text{trans}}+K_{\text{rot}}=\frac12 Mv_{\text{CM}}^2+\frac12 I_{\text{CM}}\omega^2.

For , substitute $\omega=\frac{v_{\text{CM}}}{R}$ :

K=\frac12 Mv_{\text{CM}}^2+\frac12 I_{\text{CM}}\left(\frac{v_{\text{CM}}}{R}\right)^2.

For an object starting from rest and rolling down a vertical height $h$ with negligible energy losses,

Mgh=\frac12 Mv_{\text{CM}}^2+\frac12 I_{\text{CM}}\omega^2.

If $I_{\\text{CM}}=\\beta MR^2$ , the final center-of-mass speed is

v_{\text{CM}}=\sqrt{\frac{2gh}{1+\beta}}.

The mass cancels in this ideal expression, but shape still matters through $\\beta$ . Examples are:

Solid sphere: $\beta=\frac25$ , giving $v_{\text{CM}}=\sqrt{\frac{10gh}{7}}$
Solid cylinder: $\beta=\frac12$ , giving $v_{\text{CM}}=\sqrt{\frac{4gh}{3}}$
Hoop: $\beta=1$ , giving $v_{\text{CM}}=\sqrt{gh}$

A smaller $\\beta$ means that a smaller fraction of the gravitational energy is stored as rotation, so the object generally reaches the bottom with greater speed.

Takeaway: Include both translational and whenever an object rolls.

Analyzing Combined Translational-Rotational Systems

Combined translational-rotational systems are solved by connecting independent physical principles. Choose the system and then write the relevant force, , energy, and rolling equations.

For a rolling object on an incline, the Newton– method uses

\sum F_x=Ma_{\text{CM}},

\sum\tau_{\text{CM}}=I_{\text{CM}}\alpha,

and, for ,

a_{\text{CM}}=R\alpha.

The force equation determines the center-of-mass acceleration, the equation describes angular acceleration, and the connects them.

For a hanging mass $m$ attached to a pulley of radius $R$ and $I$ , with a non-slipping string and a frictionless axle, take downward as positive for the mass:

mg-T=ma.

The pulley equation is

TR=I\alpha,

and the string constraint is

a=R\alpha.

Therefore,

T=\frac{I}{R^2}a,

mg-\frac{I}{R^2}a=ma.

Solving gives

a=\frac{mg}{m+I/R^2}.

The pulley’s rotational inertia reduces the acceleration below $g$ , because gravitational energy is divided between the mass’s translational kinetic energy and the pulley’s .

Problem-solving checklist:

Select a system and coordinate directions.
Identify the axis for and .
Use the correct about that axis.
Include every relevant form of kinetic energy.
Apply a only when there is no slipping.
Check signs, units, and limiting behavior.

Takeaway: The strongest solutions connect equations rather than treating translation and rotation as separate problems.

Angular Momentum and Moment of Inertia

Torque and Angular Impulse

Conservation of Angular Momentum

Rolling Motion and Contact Conditions

Rotational Energy in Rolling

Analyzing Combined Translational-Rotational Systems

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