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Physics 1

Rolling without slipping in depth: moments of inertia, incline, friction and bowling ball

Moments of inertia of disc and sphere proved, velocity field, inclined plane with forces and minimum friction, why static friction does no work, transition from sliding to rolling, moments about the contact point and centre of percussion.

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This page goes deeper into the lesson on rigid-body dynamics, focusing on rolling without slipping, the topic where most exam mistakes are made. We derive step by step the moments of inertia of the disc and the sphere, the velocity field of a rolling body, motion on an inclined plane using forces rather than energy, the minimum friction needed, why static friction does no work, the transition from sliding to rolling of a bowling ball, and the method of moments about the contact point, with the problem of a spool pulled by a string.

The moments of inertia, proved

Theorem. A uniform disc (or solid cylinder) of mass M and radius R, about its axis, has I = \frac{1}{2}MR^2.

Proof. Divide the disc into rings of radius r and width dr. Each ring has all its mass at distance r from the axis, so it contributes r^2dm. With surface density \sigma = M/(\pi R^2), dm = \sigma\cdot 2\pi r\,dr:

    \[ I = \int_0^R r^2\,\sigma\,2\pi r\,dr = 2\pi\sigma\frac{R^4}{4} = \frac{\pi R^4}{2}\cdot\frac{M}{\pi R^2} = \frac{1}{2}MR^2 \]

∎

Theorem. A uniform solid sphere, about a diameter, has I = \frac{2}{5}MR^2.

Proof. Cut the sphere into discs perpendicular to the axis, at height z and of thickness dz. The disc has radius r = \sqrt{R^2 - z^2} and mass dm = \rho\pi r^2dz, and by the previous theorem contributes \frac{1}{2}r^2dm = \frac{1}{2}\rho\pi(R^2 - z^2)^2dz. Integrating:

    \[ I = \frac{\rho\pi}{2}\int_{-R}^{R}(R^2 - z^2)^2dz = \frac{\rho\pi}{2}\left[R^4z - \frac{2R^2z^3}{3} + \frac{z^5}{5}\right]_{-R}^{R} = \frac{\rho\pi}{2}\cdot\frac{16R^5}{15} = \frac{8\pi\rho R^5}{15} \]

With M = \frac{4}{3}\pi R^3\rho, i.e. \rho\pi = 3M/(4R^3): I = \frac{8}{15}\cdot\frac{3M}{4R^3}R^5 = \frac{2}{5}MR^2. ∎

It is convenient to write I = kMR^2: k = 1 for the ring, 1/2 for the disc, 2/5 for the solid sphere, 2/3 for the hollow sphere. All of rolling depends on k.

The velocity field

A plane rigid body has at every instant velocities \vec v_P = \vec v_{cm} + \vec\omega\times\vec r_P, where \vec r_P is the position of the point relative to the centre of mass. In rolling without slipping the contact point has zero velocity, which gives v_{cm} = \omega R. It follows that, instant by instant, the body rotates about the contact point with angular velocity \omega: every point has a velocity perpendicular to the line joining it to the contact point and proportional to its distance. The centre has v_{cm}, the top 2v_{cm}, the points on the rim at mid-height \sqrt{2}v_{cm} at 45°.

Velocity field of a wheel rolling without slipping
The contact point P is at rest and every point of the wheel moves perpendicular to the line joining it to P, with speed proportional to the distance from P. On the right the profile along the vertical diameter: from zero at the bottom to twice the speed of the centre at the top.

The inclined plane with forces

The lesson found the speed with energy conservation. The equations of motion also give the acceleration and the friction force.

Body rolling on an inclined plane
Acting on the body are the weight M g at the centre, the normal reaction N and the static friction f at the contact point P, directed up the plane. The centre accelerates by a down the plane and the body has angular acceleration alpha.

Theorem. A body with I = kMR^2 rolling without slipping on a plane inclined at \theta has

    \[ a = \frac{g\sin\theta}{1 + k},\qquad f = \frac{k}{1 + k}\,Mg\sin\theta \]

and rolling is possible only if the coefficient of static friction satisfies \mu_s \ge \dfrac{k}{1 + k}\tan\theta.

Proof. Along the plane, for the centre of mass: Ma = Mg\sin\theta - f. Rotation about the centre of mass: the only force with a moment is friction, with lever arm R: I\alpha = fR. Rolling constraint: a = \alpha R, so f = Ia/R^2 = kMa. Substituting into the first: Ma(1 + k) = Mg\sin\theta, hence a and then f. Static friction cannot exceed \mu_sN = \mu_sMg\cos\theta: \frac{k}{1 + k}Mg\sin\theta \le \mu_sMg\cos\theta, hence the condition. ∎

Example. A solid sphere on a 30° plane: a = 4.905/1.4 = 3.50 m/s², and \mu_s \ge \frac{2}{7}\tan 30° = 0.165 is needed. With less friction the sphere slides while it rolls, and the analysis changes (section on the bowling ball). A ring (k = 1) needs more: \mu_s \ge \frac{1}{2}\tan 30° = 0.289.

Why friction does no work

In rolling without slipping friction is static and applied at the contact point. The power of a force is the dot product of the force and the velocity of its point of application: P = \vec f\cdot\vec v_P. The contact point has zero velocity, so P = 0 at every instant, and the work is zero. Friction does not remove energy: it redistributes it between translation and rotation, because its moment spins the body while its force slows the centre. This is why energy conservation can be used in rolling without slipping, and it can be checked: with the acceleration found, after a distance s = h/\sin\theta we have v^2 = 2as = 2gh/(1 + k), the same result as the lesson.

From sliding to rolling: the bowling ball

A bowling ball thrown onto the lane starts out sliding: speed v_0, almost zero angular velocity. Kinetic friction \mu_kMg, opposing the sliding, slows it down while spinning it faster and faster, until v = \omega R: from then on it rolls without slipping and friction vanishes.

Theorem. The final rolling speed is v_f = \dfrac{v_0}{1 + k}, independent of the coefficient of friction.

Proof. Choose as pole a fixed point O on the lane. Friction passes through the contact line, so it has zero moment about O; weight and normal reaction are equal and opposite and their moments cancel. The angular momentum about O is conserved. For a rigid body L_O = Mv_{cm}R + I\omega (moment of the momentum of the centre, with lever arm R, plus angular momentum about the centre). At the start L_O = Mv_0R; at the end Mv_fR + kMR^2(v_f/R) = Mv_fR(1 + k). Equating, v_f = v_0/(1 + k). ∎

For the sphere, v_f = \frac{5}{7}v_0. The energy dissipated during sliding is \frac{1}{2}Mv_0^2 - \frac{1}{2}Mv_f^2(1 + k) = \frac{1}{2}Mv_0^2\frac{k}{1 + k}: for the sphere 2/7 of the initial energy. The coefficient of friction decides only how long the sliding lasts (a complete numerical example is in the exercises).

Centre speed and rim speed of a sliding bowling ball
Kinetic friction, opposing the sliding, makes the centre speed decrease linearly and omega times R increase linearly. When the two lines meet, at time t star, the ball rolls without slipping and from then on its speed stays constant.

Moments about the contact point

A powerful method: in rolling without slipping one can write the moment equation about the contact point P, where friction and the normal reaction give no moment:

    \[ \tau_P = I_P\,\alpha,\qquad I_P = I_{cm} + MR^2 = (1 + k)MR^2 \]

(parallel-axis theorem, lesson). The equation holds because the contact point, although it changes over time, always lies at distance R from the centre of mass and moves parallel to its velocity: in this case the extra terms that appear in the moment equation about a moving point vanish. On the inclined plane, Mg\sin\theta\cdot R = (1 + k)MR^2\,a/R immediately gives a = g\sin\theta/(1 + k), without computing friction.

Example: the spool pulled by a string. A solid cylinder of mass M on a horizontal plane is pulled by a horizontal force F applied at a height h above the contact point (with a wound string or an axle). Moments about P: Fh = \frac{3}{2}MR^2\cdot a/R, so

    \[ a = \frac{2Fh}{3MR} \]

Friction then follows from the centre-of-mass equation, f = Ma - F = F\left(\frac{2h}{3R} - 1\right) (positive in the direction of F). Pulling at the centre (h = R) friction is -F/3, backwards; pulling at the top (h = 2R) it is +F/3, forwards, and the acceleration 4F/(3M) exceeds that of a body sliding without friction. Friction vanishes for h = \frac{3}{2}R: this is the centre of percussion, the point where a blow makes the body roll without slipping, the same that makes a shot “clean” in billiards and tennis.

Cylinder pulled by a horizontal force at two different heights
On the left the force is applied at the centre and friction at the contact point points backwards; on the right it is applied at the top and friction points forwards. For h equal to three halves of R friction vanishes.

Exercises on the topic, from basic to exam level, are on the exercise page linked at the top.

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