Rotational dynamics

A spinning flywheel, a turning grinding stone at a mill in Bhaktapur, the rotor of a Kulekhani turbine — none of these move from place to place, yet they clearly have motion and energy. To describe them we need a second set of mechanics, built on angle instead of distance. Every idea you learned in Grade 11 has a rotational twin: displacement becomes angle, mass becomes moment of inertia, force becomes torque. Learn the dictionary and most of this unit writes itself.

1.1 Equation of angular motion; Relation between linear and angular kinematics

When a rigid body rotates about a fixed axis, every particle of it moves in a circle centred on that axis. Different particles travel different distances in the same time, but all of them sweep the same angle. That is why angle, not distance, is the natural variable.

The angular displacement \theta is measured in radians, defined by

where s is the arc length and r the radius. The angular velocity and angular acceleration are then

with units rad s⁻¹ and rad s⁻².

If \alpha is constant, integrating twice gives the equations of angular motion, exactly parallel to the linear ones:

and the angle turned in the n^{\text{th}} second is \theta_n = \omega_0 + \frac{\alpha}{2}(2n-1).

Because s = r\theta and r is fixed for a given particle, differentiating gives the link between the linear and angular pictures:

Here a_t is the tangential acceleration (it changes the speed) and a_c is the centripetal acceleration (it changes the direction). The resultant linear acceleration has magnitude a = \sqrt{a_t^{2} + a_c^{2}}.

Linear quantity	Rotational quantity	Connection
Displacement s (m)	Angular displacement \theta (rad)	s = r\theta
Velocity v (m s⁻¹)	Angular velocity \omega (rad s⁻¹)	v = r\omega
Acceleration a (m s⁻²)	Angular acceleration \alpha (rad s⁻²)	a_t = r\alpha
Mass m (kg)	Moment of inertia I (kg m²)	I = \sum m r^{2}
Force F (N)	Torque \tau (N m)	\tau = rF\sin\phi
Momentum p = mv	Angular momentum L = I\omega	L = rp\sin\phi

1.2 Kinetic energy of rotation of rigid body

Think of the rigid body as a collection of particles of masses m_1, m_2, \ldots at perpendicular distances r_1, r_2, \ldots from the axis. All share the same \omega, so particle i has speed v_i = r_i\omega.

Compare \frac{1}{2}mv^{2}: the moment of inertia plays exactly the role that mass plays in translation. A body that is also moving forward (a rolling wheel) has total kinetic energy \frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2}.

1.3 Moment of inertia; Radius of gyration

Moment of inertia measures rotational inertia — the opposition a body offers to a change in its angular velocity. Unlike mass it is not a single number for a body: it depends on (i) the total mass, (ii) how that mass is distributed, and (iii) which axis is chosen. Mass placed far from the axis counts much more, because the distance is squared.

Two theorems shorten most calculations:

• Parallel axis theorem. I = I_{cm} + Md^{2}, where I_{cm} is about a parallel axis through the centre of mass and d is the separation.
• Perpendicular axis theorem (plane laminae only). I_z = I_x + I_y, the z-axis being perpendicular to the lamina.

Body (mass M)	Axis	Moment of inertia	Radius of gyration k
Thin rod, length L	⊥ through centre	ML^{2}/12	L/\sqrt{12}
Thin rod, length L	⊥ through one end	ML^{2}/3	L/\sqrt{3}
Ring, radius R	central, ⊥ to plane	MR^{2}	R
Disc, radius R	central, ⊥ to plane	MR^{2}/2	R/\sqrt{2}
Solid cylinder, radius R	own axis	MR^{2}/2	R/\sqrt{2}
Solid sphere, radius R	diameter	2MR^{2}/5	R\sqrt{2/5}
Hollow sphere, radius R	diameter	2MR^{2}/3	R\sqrt{2/3}

The radius of gyration k is the distance from the axis at which the whole mass could be concentrated as a single point without changing the moment of inertia:

1.4 Moment of inertia of a uniform rod

1.5 Torque and angular acceleration for a rigid body

A force applied to a body that is free to rotate produces a turning effect called torque (or moment of force). If the force \vec{F} acts at a point whose position vector from the axis is \vec{r}, and \phi is the angle between them,

The quantity d = r\sin\phi is the moment arm. Torque is largest when the force is applied perpendicular to \vec{r} and as far from the axis as possible — which is why a door handle is fitted at the edge, not next to the hinge. The SI unit is N m; torque is a vector, \vec{\tau} = \vec{r}\times\vec{F}.

1.6 Work and power in rotational motion

When a torque \tau turns a body through a small angle d\theta, the force F = \tau/r acts through an arc ds = r\,d\theta, so the work done is

For a constant torque, W = \tau\theta. The power is the rate of doing this work:

the exact analogue of P = Fv. And since \tau = I\alpha = I\,\omega\,d\omega/d\theta,

which is the work–energy theorem for rotation: the work done by the resultant torque equals the change in rotational kinetic energy.

1.7 Angular momentum; conservation of angular momentum

Differentiating L = I\omega for a rigid body (I constant about a fixed axis),

So torque is the rate of change of angular momentum, just as force is the rate of change of linear momentum. Written this way the result is more general: it holds even when I changes.

Everyday consequences, all examined regularly:

• An ice skater or a dancer spins faster on folding the arms in.
• A diver tucks to somersault quickly, then stretches out to slow the spin before entering the water.
• A planet moves faster when nearer the Sun; since gravity is a central force its torque about the Sun is zero. This is Kepler's second law.
• A spinning top or a bicycle wheel resists being tipped over, because tipping it would change the direction of \vec{L}.

Chapter summary

• For constant \alpha: \omega = \omega_0 + \alpha t, \theta = \omega_0 t + \frac{1}{2}\alpha t^{2}, \omega^{2} = \omega_0^{2} + 2\alpha\theta; and s = r\theta, v = r\omega, a_t = r\alpha, a_c = r\omega^{2}.
• Rotational kinetic energy E_k = \frac{1}{2}I\omega^{2}, where I = \sum m_i r_i^{2} = \int r^{2}dm is the moment of inertia.
• Radius of gyration: I = Mk^{2}, so k = \sqrt{I/M}. Parallel axis: I = I_{cm} + Md^{2}; perpendicular axis (laminae): I_z = I_x + I_y.
• Uniform rod of mass M, length L: I = ML^{2}/12 about the centre and ML^{2}/3 about one end.
• Torque \tau = rF\sin\phi and \tau = I\alpha, the rotational form of F = ma.
• Work W = \tau\theta, power P = \tau\omega, and W = \frac{1}{2}I\omega_2^{2} - \frac{1}{2}I\omega_1^{2}.
• Angular momentum L = I\omega with \tau = dL/dt. If \tau = 0 then I_1\omega_1 = I_2\omega_2: angular momentum is conserved, but kinetic energy need not be.

Practice questions

Group A — Multiple choice (1 mark each)

1. The moment of inertia of a uniform rod of mass M and length L about an axis through one end and perpendicular to it is [1]aML^{2}/12bML^{2}/3cML^{2}/2dML^{2}
2. The rotational analogue of mass is [1]atorquebangular velocitycmoment of inertiadangular momentum
3. A body of moment of inertia 2\ \text{kg m}^{2} rotates at 3\ \text{rad s}^{-1}. Its rotational kinetic energy is [1]a3\ \text{J}b6\ \text{J}c9\ \text{J}d18\ \text{J}
4. A skater spinning with arms outstretched folds her arms in. Her angular velocity increases because [1]aher angular momentum increasesban external torque acts on hercher moment of inertia decreases while L is constantdher kinetic energy is conserved
5. The power delivered by a torque \tau at angular velocity \omega is [1]a\tau\thetab\tau\omegacI\omegad\tau/\omega

Answers to Group A

1. (b) — integrating x^{2}dm from 0 to L gives ML^{2}/3.

2. (c) — I appears in \tau = I\alpha exactly where m appears in F = ma.

3. (c) — E_k = \frac{1}{2}(2)(3)^{2} = 9\ \text{J}.

4. (c) — no external torque, so I\omega is fixed and \omega must rise.

5. (b) — P = dW/dt = \tau\,d\theta/dt = \tau\omega.

Group B — Short answer (5 marks each)

1. Define moment of inertia and radius of gyration. On what factors does the moment of inertia of a body depend? [5]
2. Derive the relation \tau = I\alpha for a rigid body rotating about a fixed axis. [5]
3. A flywheel of mass 50\ \text{kg} in the form of a uniform disc of radius 0.40\ \text{m} rotates at 300 rev min⁻¹. Calculate its kinetic energy and the constant torque needed to stop it in 10\ \text{s}. [5]
4. A disc of moment of inertia 2\ \text{kg m}^{2} rotating at 10\ \text{rad s}^{-1} is suddenly coupled to a stationary coaxial disc of moment of inertia 3\ \text{kg m}^{2}. Find the common angular velocity and the loss of kinetic energy. [5]

Answers to Group B

1. Outline: state I = \sum m_i r_i^{2} (unit kg m², dimensions [ML^{2}]) and k = \sqrt{I/M}. I depends on the total mass, on the distribution of that mass about the axis, and on the position and direction of the axis itself.

2. Outline: for particle i, F_i = m_i r_i\alpha, so \tau_i = m_i r_i^{2}\alpha; sum over all particles (internal torques cancel in pairs) to get \tau = (\sum m_i r_i^{2})\alpha = I\alpha.

3. I = \frac{1}{2}MR^{2} = \frac{1}{2}(50)(0.40)^{2} = 4\ \text{kg m}^{2}; \omega = 2\pi(300)/60 = 31.42\ \text{rad s}^{-1}. E_k = \frac{1}{2}(4)(31.42)^{2} = 1974\ \text{J}. Retardation \alpha = 31.42/10 = 3.142\ \text{rad s}^{-2}, so \tau = I\alpha = 4 \times 3.142 = 12.6\ \text{N m}.

4. Angular momentum is conserved: 2 \times 10 = (2+3)\omega, so \omega = 4\ \text{rad s}^{-1}. E_1 = \frac{1}{2}(2)(10)^{2} = 100\ \text{J}, E_2 = \frac{1}{2}(5)(4)^{2} = 40\ \text{J}; loss = 60\ \text{J}, dissipated as heat by friction between the coupled discs.

Group C — Long answer (8 marks each)

1. (a) Define moment of inertia and radius of gyration. [3](b) Derive an expression for the moment of inertia of a uniform rod of mass M and length L about an axis through its centre and perpendicular to its length, and hence, using the parallel axis theorem, about an axis through one end. [5]
2. (a) Show that the kinetic energy of a rigid body rotating about a fixed axis is \frac{1}{2}I\omega^{2}. [3](b) A flywheel of mass 100\ \text{kg} and radius of gyration 0.50\ \text{m} rotates at 600 rev min⁻¹. A constant retarding torque brings it to rest in 40\ \text{s}. Find its moment of inertia, its initial kinetic energy, the retarding torque, and the number of revolutions it makes before stopping. [5]

Answer to Group C question 2(b)

I = Mk^{2} = 100 \times (0.50)^{2} = 25\ \text{kg m}^{2}.

\omega_0 = 2\pi(600)/60 = 20\pi = 62.83\ \text{rad s}^{-1}, so E_k = \frac{1}{2}(25)(62.83)^{2} = 4.93 \times 10^{4}\ \text{J}.

\alpha = (0 - 62.83)/40 = -1.571\ \text{rad s}^{-2}, so the retarding torque is |\tau| = I|\alpha| = 25 \times 1.571 = 39.3\ \text{N m}.

\theta = \frac{1}{2}(\omega_0 + 0)t = \frac{1}{2}(62.83)(40) = 1257\ \text{rad}, so the number of revolutions = 1257/2\pi = 200.