Friction, Torque and Rotational Motion

Friction is a contact force that opposes relative slipping, or the tendency to slip, between touching surfaces. The relevant motion is between the surfaces, not simply the motion of the whole object. Gravity can act without contact, but surface friction requires contact.

Why friction can push you forward

During ordinary walking, your foot pushes backward against the ground. The ground's static friction pushes your foot forward. Friction therefore helps propel you rather than always opposing your forward motion. A car's driven tyre also needs traction: frictional grip between the tyre and road.

A pencil mark depends on contact and friction with paper. Friction also helps prevent an object from slipping from your hand. Friction is useful when grip is needed, but unwanted friction can produce heat and wear.

No. Static friction adjusts to oppose slipping, up to a maximum possible value. Static means the contacting surfaces are not sliding relative to each other. The whole body can still move, as a wheel does when rolling without slipping.

For the usual dry-contact model, fₛ ≤ μₛN. The maximum is fₛ,max = μₛN. Here, N is the normal force, acting perpendicular to the surface. The coefficient μₛ describes the contacting pair and has no unit. Limiting friction is this maximum static friction, just before slipping begins.

A box that does not slide

Suppose N = 100 N and μₛ = 0.4. The maximum static friction is 40 N. Push horizontally with 25 N, with no other horizontal forces. Static friction supplies 25 N in the opposite direction. It does not automatically become 40 N. The box stays at rest because the horizontal forces balance. A push above 40 N exceeds the available static friction in this model.

The normal force is not always equal to weight. That equality applies on a horizontal surface when no other vertical forces act.

Kinetic friction acts when the touching surfaces slide relative to each other. In the usual dry-contact model, its magnitude is approximately fₖ = μₖN. Here, μₖ is the kinetic-friction coefficient for the contacting pair. For many ordinary surfaces, kinetic friction is smaller than maximum static friction. This comparison concerns the maximum, not every possible value of static friction.

Compare pushing with sliding

For the box above, let μₖ = 0.3 after sliding starts. Kinetic friction is then approximately 0.3 × 100 = 30 N. A steady 50 N push leaves a net horizontal force of 20 N. That net force accelerates the box; friction does not cancel every push.

Rolling resistance describes the resistance to rolling in real wheels and surfaces. Deformation and other losses can oppose rolling even without gross sliding. Wheels and ball bearings often reduce resistance compared with dragging parts across each other. Rolling resistance is not universally smaller for every material and surface.

Lubricants such as oil and grease reduce direct rubbing between moving solid surfaces. Ball bearings replace much of the sliding contact with rolling contact. Neither method eliminates every frictional loss.

Keep useful friction where grip is needed

Lubricate machine parts where friction causes unwanted wear. Do not treat grip surfaces as parts that should always be made slippery. On ice, the available grip can be small. Short, careful steps can reduce the horizontal force demanded from the ground. Slipping occurs when the required grip exceeds the available static friction. Small steps do not guarantee safety on every icy surface.

Can motion continue without friction?

A body already moving on an ideal frictionless horizontal surface can continue at constant velocity. Gravity and the upward normal force balance when no other vertical forces act. Without a net horizontal force, the body has no horizontal acceleration. A wheel can also continue rolling if its initial motion already matches rolling without slipping. Lack of traction prevents ordinary grip-based propulsion; it does not forbid motion itself.

Torque is the turning effect of a force about a chosen axis. Its magnitude is τ = Fd, where d is the perpendicular distance from the axis to the force's line of action. Equivalently, τ = rF sinθ. Here, r is the distance to the force application point. The angle θ is between that position direction and the force. The International System of Units (SI) unit of torque is the newton metre, written N·m.

Calculate the door's turning effect

Push with 20 N perpendicular to a door, 0.8 m from its hinge axis. The torque is 20 × 0.8 = 16 N·m. The same push only 0.2 m from the hinge gives 4 N·m. The farther handle therefore gives more turning effect for the same perpendicular force.

A force directed straight towards the hinge axis has zero perpendicular lever arm. Its torque about that axis is zero, even when the force is large. Torque depends on force direction as well as force magnitude and distance.

A lever is a rigid bar that turns around a pivot, called the fulcrum. In rotational balance, clockwise and anticlockwise torques cancel. A longer effort arm can balance a larger load acting through a shorter arm.

Balance a wheelbarrow

In a wheelbarrow, the wheel acts as the fulcrum. The load sits between the wheel and the handles. This arrangement is called a second-class lever.

Consider an ideal lever with a 300 N load acting 0.3 m from the pivot. Let the effort act through a perpendicular arm of 0.9 m. Balancing torques gives effort × 0.9 = 300 × 0.3. The required effort is therefore 100 N. The effort is smaller, but the effort point moves farther during the same small turn.

For the simple calculation, neglect the lever's own weight and friction. Real wheelbarrows also require effort to move their own weight and overcome resistance. Balanced torques alone do not prove complete equilibrium. A stationary lever must also have balanced forces.

Moment of inertia measures resistance to changes in rotation about a specified axis. It depends on total mass, mass distribution and the chosen axis. For separate point masses, I = Σmr². The symbol Σ means add the contribution from every mass. Each r is the perpendicular distance of that mass from the rotation axis. The SI unit is kilogram metre squared, written kg·m².

Move the same mass farther outward

A 2 kg point mass at 0.5 m contributes 2 × 0.5² = 0.5 kg·m². Move that mass to 1 m from the same axis. Its contribution becomes 2 × 1² = 2 kg·m². Doubling the distance makes the contribution four times larger.

For a rigid body rotating about a fixed axis, net torque about that axis equals Iα. Here, α is angular acceleration: the rate of change of angular velocity. Its unit is radians per second squared, written rad/s². This relation assumes the body's moment of inertia about that axis remains constant. For the same net torque, a larger moment of inertia gives smaller angular acceleration.

Compare a solid sphere and a disc

A uniform solid sphere has I = ⅖MR² about a diameter. A uniform solid disc has I = ½MR² about its central axis perpendicular to its face. For equal mass M and radius R, the disc has the larger value about these specified axes. Different axes can change the comparison. Moment of inertia is not a property of mass alone.

Angular momentum is a measure of rotational motion about a chosen origin or axis. When net external torque is zero, the system's angular momentum remains constant. For rotation about a fixed principal axis, the angular momentum along that axis is L = Iω. Here, ω is angular velocity, measured in radians per second.

Keep the product constant

1. Imagine a skater with I = 10 kg·m² spinning at 2 rad/s.
2. Angular momentum about the spin axis is 10 × 2 = 20 kg·m²/s.
3. The skater pulls inward and reduces I to 5 kg·m².
4. With negligible external torque, the new angular speed is 20 ÷ 5 = 4 rad/s.

The speed doubles because the moment of inertia halves.

This example does not conserve rotational kinetic energy. Pulling inward requires muscular work. Rotational kinetic energy is ½Iω². It rises from 20 J to 40 J in this ideal example. Conserved angular momentum does not automatically mean conserved kinetic energy.

Linear momentum has a related condition: net external force must be zero. Do not replace external torque with external force when discussing angular momentum.

Unstable equilibrium means a small displacement creates a tendency to move farther from the original position. A ball balanced on top of a narrow vertical support provides an example. A small sideways shift can lower its centre of mass. Gravity then tends to carry it farther from the balanced position.

Compare a ball at the bottom of a bowl. A small displacement raises its centre of mass. Gravity tends to bring the ball back towards the bottom. That arrangement shows stable equilibrium.

A ball resting on an ideal horizontal surface can show neutral equilibrium. A small sideways displacement leaves its centre of mass at the same height. Equilibrium describes a balance, not a guarantee that every disturbance will be corrected.

A washing machine's rotating drum carries the clothes around a circular path. Following that path requires an inward force. Contact with the drum helps provide the clothes with that force. Water can pass through the drum's holes rather than remaining constrained with the clothes.

Picture the difference

An object does not follow a circle merely because nearby objects rotate. Without the required inward force, a moving object tends to follow a straight path. The drum's holes allow water to separate from the rotating clothes. The spin cycle therefore removes water without needing to evaporate all of it.

From a rotating viewpoint, the motion can be described using an outward centrifugal effect. From a stationary viewpoint, explain the separation through inertia and the forces constraining circular motion. Do not confuse this outward effect with the real inward force required for circular motion.

Expelling air can give the body a recoil in the opposite direction. For an ideal person-and-expelled-air system, total horizontal momentum remains constant when external horizontal forces are absent. The body's momentum changes because some mass leaves the body carrying momentum.

Distinguish recoil from grip

Suppose air is expelled horizontally forward. The body receives a small backward recoil. Ordinary attempts to run cannot obtain a horizontal push from truly frictionless ground. Jumping straight upward also supplies no horizontal impulse from an ideal horizontal surface.

Internal movements alone cannot change the horizontal motion of the entire isolated system's centre of mass. The person and expelled air together still obey momentum conservation. A sneeze does not create momentum from nothing. It redistributes momentum between the body and the air.

The Combined Defence Services examination is abbreviated CDS.

Recoil on frictionless ground

A person initially at rest cannot gain horizontal traction from ideal frictionless ground. Expelling air can give the body horizontal recoil when the air carries horizontal momentum. Distinguish the person's motion from the centre-of-mass motion of the person-and-air system. CDS-II 2017, Q16.

Balanced forces on a frictionless plane

On a horizontal plane, gravity can balance the normal force. If no other force acts, the net force is zero. An object already moving continues at constant velocity. A frictionless inclined plane is different: gravity can have a component along the slope. Do not infer zero acceleration from the word frictionless alone. CDS-II 2017, Q59.

Static versus kinetic friction

Identify whether the contact surfaces slip relative to each other. A moving wheel can have static friction when rolling without slipping. A sliding box has kinetic friction at its contact with the ground.

Torque direction

Measure the perpendicular distance from the axis to the force's line of action. The distance to the application point alone is not enough. A large force through the axis gives zero torque about that axis.