So far we treated bodies as point particles. Real bodies, however, are extended - they have size and shape and can rotate. A cricket ball hit with a bat can spin, a spinning top keeps rotating, and a wheel rolls along a road. To describe such systems, we need the concept of the centre of mass and the dynamics of rotation.
This chapter introduces the centre of mass of a system of particles, which is the point where the entire mass of the system may be assumed to be concentrated. We then study torque, the rotational analogue of force, and angular momentum, the rotational analogue of linear momentum. The law of conservation of angular momentum is one of the most powerful principles in physics, explaining why a spinning skater speeds up when she pulls in her arms.
Finally we introduce the moment of inertia, which plays the role of mass in rotational motion, and derive the equations of rotational motion. The chapter closes with rolling motion, which combines translation and rotation.
The centre of mass of a system of particles is that point which moves as though all the mass of the system were concentrated there and all external forces were applied there. For a system of particles with masses m1, m2, ..., mn at positions r1, r2, ..., rn, the position of the centre of mass is:
R_cm = (m1r1 + m2r2 + ... + mn*rn) / (m1 + m2 + ... + mn)
For the x-coordinate of the centre of mass of two particles:
x_cm = (m1x1 + m2x2) / (m1 + m2)
If the two masses are equal, the centre of mass is exactly midway between them. The centre of mass depends on the distribution of mass; for a uniform symmetric body such as a sphere or a rod, it lies at the geometric centre.
The centre of mass of an isolated system moves with uniform velocity even if internal forces act within the system, because internal forces cannot change the motion of the centre of mass. This is why a projectile exploding into fragments has its centre of mass continuing along the original parabolic path.
Torque is the rotational analogue of force. When a force F is applied at a point whose position vector from the axis of rotation is r, the torque is the cross product:
tau = r x F
The magnitude of the torque is:
tau = r F sin theta
where theta is the angle between r and F. The SI unit of torque is N m. Torque depends on the perpendicular distance of the line of action of the force from the axis, called the lever arm or moment arm. A longer lever arm produces a larger torque for the same force, which is why door handles are placed far from the hinges and wrenches are long.
The net torque acting on a body determines its angular acceleration. Just as F = ma for linear motion, the rotational analogue is:
tau = I alpha
where I is the moment of inertia and alpha is the angular acceleration.
The moment of inertia of a body is a measure of its resistance to a change in its rotational state, just as mass resists changes in linear motion. For a system of particles, the moment of inertia about an axis is the sum of the products of the masses and the squares of their perpendicular distances from the axis:
I = m1r1^2 + m2r2^2 + ...
The SI unit of moment of inertia is kg m^2. The moment of inertia depends on the mass of the body, its distribution about the axis, and the location and orientation of the axis. For a given body, the moment of inertia is smallest about an axis through its centre of mass.
The parallel axis theorem states that the moment of inertia about any axis parallel to an axis through the centre of mass is:
I = I_cm + M d^2
where d is the distance between the two parallel axes. The perpendicular axis theorem states that for a planar lamina lying in the x-y plane, the moment of inertia about the z-axis equals the sum of the moments of inertia about the x and y axes:
Iz = Ix + Iy
The equations of rotational motion with constant angular acceleration alpha are the exact analogues of the linear equations. With theta as the angular displacement, omega as the angular velocity, and omega0 as the initial angular velocity:
omega = omega0 + alpha t
theta = omega0 t + (1/2) alpha t^2
omega^2 = omega0^2 + 2 alpha theta
The linear variables at a distance r from the axis are related to the angular variables by v = omega r and a = alpha r. The kinetic energy of a rotating body is:
K = (1/2) I omega^2
which is the rotational analogue of (1/2)mv^2.
The angular momentum of a particle is the moment of its linear momentum. If a particle of mass m moves with velocity v at position r, its angular momentum about a point is:
L = r x p = r x (m v)
For a rigid body rotating about a fixed axis, the angular momentum is:
L = I omega
where I is the moment of inertia and omega the angular velocity. The SI unit of angular momentum is kg m^2/s.
The rotational analogue of Newton's second law is:
tau = dL/dt
If the net external torque on a system is zero, its angular momentum is conserved:
L = I omega = constant
This is the law of conservation of angular momentum. It explains why a spinning figure skater speeds up when she pulls her arms in (I decreases, omega increases) and why a diver tucks his body to spin faster in the air. The Earth's spin axis is fixed in space because there is no significant external torque acting on it.
A rigid body is in mechanical equilibrium when both its linear and angular accelerations are zero. This requires two conditions. First, the vector sum of all external forces must be zero:
Sum of F = 0
Second, the sum of all external torques about any axis must be zero:
Sum of tau = 0
These two conditions are used to solve problems involving static bodies, such as beams supported at two points, ladders leaning against walls, and seesaws. In such problems we take moments about a convenient point so that the unknown forces through that point produce zero torque and drop out of the equations.
Rolling motion combines rotation and translation. When a wheel rolls without slipping, the speed of its centre of mass is related to its angular velocity by:
v_cm = omega * R
where R is the radius of the wheel. Rolling without slipping requires static friction at the point of contact. The kinetic energy of a rolling body is the sum of its translational and rotational kinetic energies:
K = (1/2) M v_cm^2 + (1/2) I omega^2
For a solid sphere rolling down an inclined plane, the acceleration is less than g sin theta because part of the potential energy converts into rotational kinetic energy. Bodies with smaller moments of inertia roll faster down the incline; for example, a hollow sphere and a solid sphere of the same mass and radius behave differently, with the solid sphere reaching the bottom first.
| Quantity | Linear | Rotational |
|---|---|---|
| Mass | m | I (moment of inertia) |
| Force | F | tau (torque) |
| Velocity | v | omega |
| Acceleration | a | alpha |
| Momentum | p = mv | L = I omega |
| Kinetic energy | (1/2)mv^2 | (1/2)I omega^2 |
| Second law | F = ma | tau = I alpha |
| Body | Moment of Inertia |
|---|---|
| Ring about centre, radius R | M R^2 |
| Disc about centre, radius R | (1/2) M R^2 |
| Solid sphere | (2/5) M R^2 |
| Rod about centre | (1/12) M L^2 |
In this chapter we extended mechanics to extended bodies and rotation. We located the centre of mass, the point at which the whole mass acts, and saw that it moves uniformly in an isolated system. We introduced torque as the rotational analogue of force and the moment of inertia as the rotational analogue of mass, and wrote the rotational equations omega = omega0 + alpha t and tau = I alpha. Angular momentum L = I omega obeys its own conservation law when no external torque acts, explaining phenomena from spinning skaters to spinning tops. We stated the two conditions of rigid-body equilibrium and finally combined translation and rotation in rolling motion. These ideas form the basis for the study of gravitation, where central forces and angular momentum play a central role.