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Physics 1

Rigid body dynamics: torque, moment of inertia and rolling

Torque and lever arm, the proof of τ = Iα, moments of inertia and the parallel axis theorem, rotational energy, angular momentum, rolling without slipping and equilibrium.

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So far we have treated bodies as points. A wheel, a propeller or a motor’s rotor, however, spin, and describing them needs a new set of quantities: torque, moment of inertia, angular momentum. The good news is that each one matches a quantity we already know from straight-line motion, and the equations have the same form.

The rigid body and rotational kinematics

A rigid body is a system in which the distances between points do not change. In a rotation about a fixed axis all points turn with the same angular velocity \omega and the same angular acceleration \alpha = d\omega/dt; a point at distance r from the axis has speed v = \omega r. The laws of uniformly accelerated rotation are identical to those of straight-line motion: \omega = \omega_0 + \alpha t, \theta = \theta_0 + \omega_0 t + \tfrac12\alpha t^2.

Torque

The same force turns a door much more if you apply it far from the hinges. The turning effect of a force is measured by its torque (moment) about a point O:

    \[ \vec\tau = \vec r\times\vec F,\qquad \tau = r\,F\sin\varphi = b\,F \]

Torque of a force: position vector, force and lever arm
Torque depends only on the component of the force perpendicular to \vec r. Equivalently, it is the force times the lever arm b, the distance from O to the force’s line of action.

The lever arm b = r\sin\varphi is the distance between the pivot and the line of action of the force. A force pointing at the pivot has zero lever arm and turns nothing. The unit is the newton-metre (N·m), dimensionally the same as the joule, but a different quantity: that is why torque is never written in joules.

Moment of inertia and the rotation equation

Proof: \tau = I\alpha

Split the body into many elements of mass m_i at distance r_i from the axis. Each element moves on a circle with tangential acceleration a_i = r_i\,\alpha. By the second law the tangential component of the force on it is F_i = m_i r_i\alpha, and its torque about the axis is \tau_i = r_i F_i = m_i r_i^2\alpha. Adding over all elements, the torques of the internal forces cancel in pairs (action and reaction along the same line), leaving:

    \[ \tau_{ext} = \Big(\sum_i m_i\,r_i^2\Big)\,\alpha = I\,\alpha \]

The quantity I = \sum m_i r_i^2 (or \int r^2\,dm for a continuous body) is the moment of inertia: it plays the role of mass in rotation. It depends not only on how much mass there is, but on how far it is from the axis.

Body (axis through the centre)Moment of inertia
thin ring or hollow cylinderM R^2
solid disc or cylinder\tfrac12 M R^2
solid sphere\tfrac25 M R^2
thin hollow sphere\tfrac23 M R^2
thin rod of length L\tfrac{1}{12} M L^2

The parallel axis theorem

If you know I_{cm} about an axis through the centre of mass, about a parallel axis at distance d

    \[ I = I_{cm} + M\,d^2 \]

Proof. Put the origin at the centre of mass and the shifted axis at (d, 0). For each element r_i'^2 = (x_i - d)^2 + y_i^2 = x_i^2 + y_i^2 - 2d\,x_i + d^2. Summing with weights m_i: the first term gives I_{cm}, the last gives Md^2, and the middle term -2d\sum m_i x_i is zero because \sum m_i x_i = M\,x_{cm} = 0. For example, a rod turning about one end has I = \tfrac1{12}ML^2 + M(L/2)^2 = \tfrac13 ML^2.

Rotational kinetic energy

Each element has energy \tfrac12 m_i v_i^2 = \tfrac12 m_i r_i^2\omega^2. Adding up:

    \[ K_{rot} = \tfrac12\,I\,\omega^2 \]

This is exactly how a flywheel stores energy. A 100 kg steel disc of radius 0.3 m at 6000 rpm (628 rad/s) has I = 4.5\ \text{kg·m}^2 and K = \tfrac12\cdot 4.5\cdot 628^2 \approx 0.89 MJ, about 0.25 kWh.

Angular momentum

For a point mass the angular momentum is \vec L = \vec r\times\vec p. For a rigid body spinning about a symmetry axis L = I\,\omega. Differentiating gives the rotational counterpart of \vec F = d\vec p/dt:

    \[ \vec\tau_{ext} = \frac{d\vec L}{dt} \]

If the external torque is zero, angular momentum is conserved. A skater pulling in her arms reduces I and, since I\omega stays constant, spins faster. A neutron star, the collapsed core of a star, spins hundreds of times per second for the same reason.

TranslationRotation
position xangle \theta
velocity vangular velocity \omega
mass mmoment of inertia I
force Ftorque \tau
F = m\,a\tau = I\,\alpha
K = \tfrac12 m v^2K = \tfrac12 I\omega^2
p = m\,vL = I\,\omega

Rolling without slipping

A wheel rolling without slipping has its contact point at rest relative to the ground, instant by instant. This gives the constraint

    \[ v_{cm} = \omega\,R \]

Wheel rolling without slipping: contact point at rest, centre at v, top at 2v
The motion is the sum of a translation at v_{cm} and a rotation about the centre. At the contact point P the two velocities cancel; at the top they add up to 2v_{cm}.

The total kinetic energy is that of the centre of mass plus that of rotation: K = \tfrac12 M v_{cm}^2 + \tfrac12 I_{cm}\omega^2.

Proof: who reaches the bottom of the slope first?

A body of mass M, radius R and moment of inertia I rolls down from a height h, starting at rest. Static friction at the contact point does no work (the point is at rest), so mechanical energy is conserved:

    \[ M g h = \tfrac12 M v^2 + \tfrac12 I\frac{v^2}{R^2} \quad\Rightarrow\quad v = \sqrt{\frac{2gh}{1 + I/(MR^2)}} \]

The further the mass is from the axis, the more energy goes into rotation and the slower the body. With I/(MR^2) equal to 2/5 for the solid sphere, 1/2 for the solid cylinder and 1 for the ring, in a race down the same slope the sphere wins, then the cylinder, with the ring last. Neither mass nor radius matters, only shape.

Equilibrium of a rigid body

A rigid body is in equilibrium if both the sum of the forces and the sum of the torques about any point are zero:

    \[ \sum\vec F = 0,\qquad \sum\vec\tau = 0 \]

Example. A 4 m, 50 kg beam rests on supports at its two ends and carries a 100 kg mass 1 m from support A. Torques about A: R_B\cdot 4 = 50 g\cdot 2 + 100 g\cdot 1, so R_B = 50 g ≈ 490 N. Sum of vertical forces: R_A = 150 g - 50 g = 100 g ≈ 981 N. Choosing the pivot on an unknown force removes it from the equation: that is the trick that makes these problems quick.

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