On a farm in Punjab, a farmer starts his diesel pump set by hand. A cord is wound round a small flywheel on the engine shaft. He braces a foot and gives one hard, steady pull. The flywheel whirls and the engine fires. His young son pulls the same length of cord gently, and the flywheel barely turns. The cord, the wheel and the length pulled were the same. Only the force was different.
A force that turns
A body that turns about a fixed axis needs a turning effect, and a force alone does not measure it. The chapter on Torque and Equilibrium defines that turning effect, the torque. Take a force \(F\) acting at a point a distance \(r\) from the axis, at an angle \(\theta\) to the radius. Its torque about the axis is \(\tau = rF\sin\theta\).
Only the part of the force at right angles to the radius, \(F\sin\theta\), can turn the body. The part along the radius points straight at the axis, or straight away from it, and turns nothing. The same torque can be read another way. It is \(F\) times its moment arm \(r\sin\theta\), the perpendicular distance from the axis to the line of the force.
One particle at a time
The mass of a body resists a change in its velocity, and its moment of inertia resists a change in its turning. The chapter on Rigid Bodies and Moment of Inertia found \(I = \sum mr^2\) from the kinetic energy of a spinning body. Now find what torque does to that turning. Apply Newton's second law to one particle, then add all the particles.
\[ F_t = m_i a_t = m_i r_i \alpha \]
Why. Newton's second law along the tangent. A particle at distance \(r_i\) from the axis has tangential acceleration \(a_t = r_i\alpha\), from Kinematics of Circular Motion.
\[ \tau_i = r_i F_t = m_i r_i^2 \alpha \]
Why. the torque of the net force on the particle about the axis. The radial part of the force passes through the axis and adds no torque.
\[ \sum_i \tau_i = \left(\sum_i m_i r_i^2\right)\alpha \]
Why. add the equations for all the particles. The body is rigid, so every particle has the same \(\alpha\), which can be taken outside the sum.
\[ \tau_{net} = I\alpha \]
Read it. the bracket is the moment of inertia about the axis. The internal forces add no torque, as the next topic shows, so only the external torques remain.
Fig 1.1 shows the whole idea on one particle. The radial part \(F_r\) only keeps the particle on its circle; it has no moment arm. The tangential part \(F_t\) changes the particle's speed along the circle, at the rate \(r\alpha\). Its torque is \(rF_t\), and that equals \(mr^2\alpha\). Every particle of the body obeys the same rule with the same \(\alpha\).
Take a rigid body turning about a fixed axis. The net external torque about that axis equals the moment of inertia about the same axis times the angular acceleration: \(\tau_{net} = I\alpha\).
Compare the law with \(F_{net} = ma\). Torque takes the place of force, the moment of inertia takes the place of mass, and \(\alpha\) takes the place of \(a\). A large torque gives a large angular acceleration. A large moment of inertia gives a small one. And a zero net torque means \(\alpha = 0\): the body keeps turning at a steady rate, or stays at rest.
A grindstone is a uniform disc of mass \(20\,\text{kg}\) and radius \(0.3\,\text{m}\), free to turn on a fixed axle. A worker pushes along the tangent at its rim with a steady force of \(12\,\text{N}\). Find its angular acceleration, in rad/s².
A motor applies a torque of \(10\,\text{N m}\) to a wheel of moment of inertia \(2\,\text{kg m}^2\). Friction in the bearings applies an opposing torque of \(4\,\text{N m}\). Find the angular acceleration of the wheel, in rad/s².
Why the internal forces cancel
Each particle of a body also feels forces from its neighbours. These internal forces hold the body together. Yet the law has no place for them. Newton's third law explains why. Particle A pulls particle B, and B pulls A with a force of the same size in the opposite direction. Both forces act along the line AB.
Study Fig 1.2. The two forces share one line, so they share one moment arm \(d\) about the axis. They have the same size \(F\). But they tend to turn the body in opposite senses: one anticlockwise, the other clockwise. So their torques are \(+Fd\) and \(-Fd\), and they cancel. Every internal pair cancels in the same way, whatever the shape of the body.
So only an external torque can change the rotation of a body. A body cannot set itself spinning by pushing on its own parts. The motor of a ceiling fan turns the blades by pushing on the rotor from outside it. The forces between the parts of one blade cancel in pairs.
In vector form, a mass element \(dm\) at position \(\vec r\) from the axis has tangential acceleration \(\vec\alpha\times\vec r\). The torque of the net force on it is \(dm\,\vec r\times(\vec\alpha\times\vec r) = dm\,[r^2\vec\alpha - (\vec r\cdot\vec\alpha)\vec r]\). For a fixed axis, \(\vec r\) is drawn perpendicular to the axis and \(\vec\alpha\) lies along it, so \(\vec r\cdot\vec\alpha = 0\). Adding the elements gives \(\vec\tau_{net} = I\vec\alpha\). The cancelling of internal torques uses the strong form of the third law: each pair acts along the line joining its two particles. The forces of mechanics obey it.
Can the forces between the particles of a rigid body give the body an angular acceleration about a fixed axis?
A rope on a wheel
The simplest way to apply a known torque is a rope wound round a wheel. The rope leaves the rim along the tangent. So the force is at right angles to the radius, and its moment arm is the radius \(R\) itself. The torque is \(FR\), however much rope has already been pulled off.
A uniform disc of mass \(4\,\text{kg}\) and radius \(0.5\,\text{m}\) turns freely on a fixed axle through its centre. A light rope wound round its rim is pulled with a steady force of \(10\,\text{N}\). Find the angular acceleration, and the angular velocity \(2\,\text{s}\) after it starts from rest.
The moment of inertia of the disc about its axle is \(I = \tfrac12 MR^2 = \tfrac12 \times 4 \times 0.5^2 = 0.5\,\text{kg m}^2\).
The rope pulls along the tangent, so \(\tau = FR = 10 \times 0.5 = 5\,\text{N m}\). Then \(\alpha = \tau/I = 5/0.5 = 10\,\text{rad/s}^2\).
The torque is constant, so \(\alpha\) is constant. From rest, \(\omega = \alpha t = 10 \times 2 = 20\,\text{rad/s}\).
In those \(2\,\text{s}\) the disc turns through \(\tfrac12\alpha t^2 = 20\,\text{rad}\), about three turns, and \(R\theta = 10\,\text{m}\) of rope is pulled off.
Try the three chips of Fig 1.3. Each graph is a straight line, because a steady torque gives a steady \(\alpha\), and \(\alpha\) is the slope of the \(\omega\)-\(t\) graph. Doubling the pull to \(20\,\text{N}\) doubles the torque, and the slope doubles to \(20\,\text{rad/s}^2\). The ring has all its mass at the rim, so \(I = MR^2 = 1\,\text{kg m}^2\), twice the disc's. The same \(10\,\text{N}\) pull then gives only \(5\,\text{rad/s}^2\).
A thin ring and a uniform disc have the same mass and the same radius. Each turns on a fixed axle, and the same torque acts on each. Which gains angular velocity faster?
The torque and the moment of inertia in \(\tau = I\alpha\) must be taken about the same axis. A uniform rod hinged at one end turns about the hinge. So its \(I\) is \(\tfrac13 ML^2\) about the end, not \(\tfrac{1}{12}ML^2\) about the centre. Mixing the two makes \(\alpha\) wrong by a factor of four.
A uniform rod of mass \(M\) and length \(L\) is hinged at one end. You take torques about the hinge to find its angular acceleration. Which moment of inertia must you use?
Newton's Principia of 1687 gave laws for the motion of bodies treated as points. It gave no general law for a spinning body. The Swiss mathematician Leonhard Euler supplied one. In a paper of 1776 he set out two laws for a body of any shape. The first ties the net force to the motion of the centre of mass. The second ties the net torque to the change in rotation. The law \(\tau = I\alpha\) of this lesson is the simplest case of that second law, for a fixed axis. Many engineers still call the pair Euler's laws of motion.
Now return to the pump set. Suppose its flywheel has \(I = 0.05\,\text{kg m}^2\) and the cord is wound at a radius of \(0.08\,\text{m}\). A steady pull of \(150\,\text{N}\) gives \(\tau = 12\,\text{N m}\) and \(\alpha = 240\,\text{rad/s}^2\). In a quarter of a second the flywheel reaches \(60\,\text{rad/s}\), about 570 turns a minute, while \(0.6\,\text{m}\) of cord comes off. A pull of \(50\,\text{N}\) gives a third of the torque. Over the same \(0.6\,\text{m}\) of cord, the flywheel reaches only about \(35\,\text{rad/s}\), too slow to start the engine.
- Taking torque and moment of inertia about different axes.Name the axis first. Take \(\tau\) and \(I\) about that one axis.
- Counting the forces between the parts of a body as torques.Internal torques cancel in pairs. Only external torques enter \(\tau_{net} = I\alpha\).
- Thinking that a zero net torque means the body is at rest.Zero net torque means \(\alpha = 0\): the body turns at a steady rate, which may be zero.
- Using the length of rope pulled off to find the torque.A rope along the tangent has moment arm \(R\), so \(\tau = FR\) at every instant.