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Physics C: Mechanics Unit 6: Torque and Rotational Dynamics

Torque is what changes rotation, just as force changes linear motion. This unit applies Στ = Iα to pulleys, rolling objects and pivoting rods, and uses Στ = 0 for objects in static equilibrium.

Covers: Torque as a cross product, Newton's second law for rotation, Static equilibrium, Pulleys and connected rotating systems, Rolling with forces and friction, Rotational work and power

Study guide

Torque measures how effectively a force turns an object about a point. It is the cross product τ = r × F, where r runs from the pivot to where the force acts. Its magnitude is rF sin θ, which you can read as r times the perpendicular force component, or as F times the lever arm (the perpendicular distance from the pivot to the force's line of action). In the xy-plane, τ_z = xF_y − yF_x, with positive meaning counterclockwise. Forces aimed through the pivot produce no torque, which is why you choose the pivot at the location of unknown forces.

For rotation about a fixed axis, Newton's second law becomes Στ = Iα. Solve problems by drawing a free-body diagram for each object, writing ΣF = ma for translating objects and Στ = Iα for rotating ones, and linking them with constraints such as a = αR for a string that does not slip. For a block hanging from a massive disk pulley, this gives a = mg/(m + ½M): the pulley acts like extra mass ½M. A pulley with mass also means the string tensions on each side differ, because their difference provides the torque.

An object is in static equilibrium when ΣF = 0 and Στ = 0. When the net force is zero, the net torque is the same about every point, so choose the pivot that eliminates the most unknowns, usually a hinge or support. Treat the weight of a uniform object as acting at its center. Classic problems include beams on two supports, signs held by a cable and a hinge, and ladders leaning against walls. Tipping problems compare the torque that would rotate an object about an edge with the force that would make it slide.

Rolling combines translation and rotation. When an object rolls without slipping, v = ωR and a = αR, and the contact point is momentarily at rest, so static friction does no work. On an incline, a = g sin θ/(1 + I/(mR²)), and friction must be large enough to supply the needed torque. When an object slides and spins at the wrong rate, such as a bowling ball released without spin, kinetic friction acts until rolling begins; angular momentum about the contact line is conserved, giving v = (5/7)v₀ for a sphere. Pivoting rods are another favorite: released from horizontal, a rod pivoted at one end has α = 3g/(2L), and its tip initially accelerates faster than g.

Work done by a torque is W = ∫τ dθ, which equals τΔθ for constant torque, with θ in radians. The rotational work–energy theorem says this work changes the rotational kinetic energy ½Iω². Power is P = τω, the rotational counterpart of Fv. Energy methods are often the fastest route to speeds, such as the tip speed of a swinging rod, while Στ = Iα gives accelerations and forces at a particular instant. Many hard problems need both: energy for ω, then forces from Newton's laws.

Key ideas

• Torque depends on force, distance from the pivot and the angle between them.
• Στ = Iα governs rotation about a fixed axis.
• A string that does not slip links a = αR between rotating and translating parts.
• Static equilibrium requires ΣF = 0 and Στ = 0, and any point can serve as the pivot.
• Rolling without slipping uses static friction, which does no work.
• Rolling acceleration down an incline is g sin θ/(1 + I/(mR²)).
• Rotational work τΔθ changes rotational kinetic energy, and power is τω.

Formulas and rules

Common mistakes

• Using cos θ instead of sin θ, or measuring θ from the wrong line.
• Assuming the tensions on both sides of a massive pulley are equal.
• Forgetting the rotational inertia of a pulley or rolling object.
• Placing a uniform object's weight anywhere other than its center.
• Using revolutions instead of radians in τΔθ.

Before the test, make sure you can

• Calculate torque from magnitudes and angles or from a cross product.
• Apply Στ = Iα to pulleys with mass, including unequal tensions.
• Solve static equilibrium problems for beams, hinged rods and ladders.
• Find the acceleration and required friction for rolling down an incline.
• Analyze a pivoted rod released from rest, including its pivot force.
• Use rotational work and power, and combine energy with torque methods.
• Decide whether an object tips or slides.

Words to know

Torque

The turning effect of a force about a point, τ = r × F.

Lever arm

The perpendicular distance from the pivot to the line of action of a force.

Line of action

The line along which a force acts, extended in both directions.

Rotational second law

Στ = Iα for rotation about a fixed axis.

Static equilibrium

A state with zero net force and zero net torque, so the object stays at rest.

Center of gravity

The point where an object's weight can be treated as acting; the center of mass in uniform gravity.

Pivot choice

In equilibrium problems, any point may be the pivot; choose one through unknown forces.

No-slip condition

A string or rolling object that does not slip, so a = αR.

Rolling friction (static)

The static friction that provides the torque for rolling acceleration and does no work.

Rotational work

Work done by a torque through an angle, W = ∫τ dθ.

Rotational power

Rate of rotational work, P = τω.

Tipping point

The condition where the torque about an edge just overcomes the restoring torque of weight.

Hinge force

The force a hinge exerts, which often has both horizontal and vertical components.

Sample practice questions

30 more questions like these, from level 1 to 5, are in the practice set.