Torque and Rotational Dynamics

Angular displacement is measured in radians. Torque depends on force, distance from the axis, and orientation. Rotational inertia is defined relative to an axis; the same object can have different rotational inertia about different axes.

For a rigid body rotating about a fixed axis, all points share angular displacement, angular velocity, and angular acceleration. Points farther from the axis travel longer arcs and have greater tangential speed at the same angular speed. Use radians in the standard linear–angular relationships.

Tangential acceleration changes speed, while radial acceleration changes direction. An object can have radial acceleration even when angular acceleration is zero. For constant angular acceleration, rotational kinematic equations mirror the constant-linear-acceleration equations.

Torque magnitude is force times perpendicular lever arm, equivalently rF sinθ. The angle is between the position vector from the axis and the force. A force directed through the axis produces zero torque about that axis even if its magnitude is large.

Choose an axis and sign convention before summing torques. A convenient pivot can eliminate unknown-force torques, but it does not eliminate those forces from the translational force equations. For static equilibrium, both net force and net torque must be zero.

Moment of inertia measures resistance to angular acceleration about a specified axis. Mass farther from the axis contributes more strongly because each point contribution is mr². Two objects with the same mass and radius can have different moments of inertia if their mass is distributed differently.

Changing the axis changes the moment of inertia. Use a supplied or appropriate shape formula for the actual axis. Treating an extended object as a point mass at its center generally misses its rotational inertia about that center.

A force on a rigid body can accelerate the center of mass and create angular acceleration about it. A rolling or pulley problem may require both ΣF = ma and Στ = Iα. A no-slip constraint connects the linear and angular accelerations.

For a massive pulley, tensions on its two sides can differ because a nonzero net torque is needed to accelerate it. Assuming equal tensions while also giving the pulley rotational inertia can contradict the model. Distinguish ideal massless pulleys from rotating objects with appreciable inertia.

Balancing a beam compares clockwise and counterclockwise torques. Place the beam’s own weight at its center of mass when the gravitational field is uniform. If the pivot is not at that center, the beam’s weight can contribute a torque.

A useful experiment varies one lever arm or load and measures the other needed for balance. Account for the mass of hangers and the actual perpendicular distance, not just a distance printed along a tilted beam. Repeated trials help distinguish a systematic offset from random measurement variation.

For constant angular acceleration, rotational kinematics has equations analogous to constant linear acceleration. Use angular variables consistently: θ for angle, ω for angular velocity, and α for angular acceleration. Tangential speed at radius r is rω, so points on a rigid object share angular speed but not necessarily linear speed.

Tangential acceleration is rα and changes speed around the circle. Radial acceleration is rω² and changes velocity direction. A rotating object can have zero angular acceleration while points away from the axis still have radial acceleration. The word “constant rotation” therefore does not mean zero acceleration of every point.

Torque magnitude is rF sinθ, where θ is the angle between the position vector from the axis and the force. Equivalently, multiply force by the perpendicular distance from the axis to its line of action. A force aimed directly through the axis produces zero torque even when its magnitude is large.

Choose clockwise or counterclockwise as positive and sum signed torques about one axis. In static equilibrium, both net force and net torque must be zero. Choosing an axis through an unknown force can eliminate that force’s torque and simplify a calculation, but it does not remove the force from the separate translational equilibrium equations.

Rotational inertia sums contributions mr² for point masses or uses the appropriate extended-body model. Moving mass farther from the axis increases its contribution quadratically. Two objects with the same mass and outer radius can have different rotational inertias if one concentrates more mass near the rim.

In τ_net=Iα, a given net torque produces a smaller angular acceleration when rotational inertia is larger. For a pulley with inertia, different tensions on its two sides can produce the net torque needed to accelerate its rotation. Treating those tensions as equal would predict zero torque and contradict the assumed angular acceleration.

Torque depends on force and its perpendicular lever arm about the chosen axis. A force applied far from a pivot can still produce zero torque if its line of action passes through the pivot. Use rF sin θ when θ is the angle between the position vector and force, or use force times perpendicular distance.

Choose a positive rotational sense and keep torque signs consistent. For static equilibrium, both net force and net torque must vanish. Choosing a pivot through an unknown support force can eliminate that force from the torque equation, but it does not make the force physically absent.

A force diagram and a torque diagram answer related but different questions. The same force can have different torques about different axes. State the axis whenever comparing torques or moments of inertia.

Rotational inertia weights mass by squared distance from the axis. Two objects with equal total mass and outer radius can have different inertias if one places more mass far from the axis. Larger inertia requires larger net torque for the same angular acceleration under the fixed-axis rigid-body model.

Angular position, velocity and acceleration describe orientation and its time change. For a point fixed at radius r on a rigid body, tangential speed is rω and tangential acceleration is rα. Radial acceleration rω² changes velocity direction even if angular speed is constant. Points at different radii share angular speed but not linear speed.

Constant-angular-acceleration equations mirror linear kinematics only when that condition holds. Do not infer constant angular acceleration merely because an object rotates. Determine net torque and whether the inertia and force geometry stay constant.

A massive pulley can require a net torque to accelerate, so tensions on its two sides need not be equal. Write translational equations for the connected masses and a rotational equation for the pulley. A no-slip cord relates linear acceleration to angular acceleration through a=rα.

The idealization of a massless pulley removes its rotational inertia and can allow equal tension in an otherwise ideal arrangement. The two models should not be mixed. State whether bearing friction, cord mass and slipping are neglected.

A useful check is the limit of vanishing pulley inertia: the massive-pulley result should approach the ideal result when other assumptions match. A larger pulley inertia generally reduces the acceleration for a given driving imbalance because some input energy or torque accelerates rotation.

Rotational inertia measures resistance to angular acceleration about a specified axis. Moving the same mass farther from that axis increases its contribution through mr². Two objects with equal total mass can have different rotational inertias because their mass distributions differ.

Torque is also axis-dependent. Use the perpendicular distance from the pivot to the force’s line of action, or rF sin θ. A force applied far from the pivot can still produce zero torque if its line of action passes through the pivot. Choose a sign convention for clockwise and counterclockwise effects.

For rotational dynamics about an appropriate fixed axis, net torque equals Iα. For static equilibrium, both net force and net torque must vanish. Satisfying only one condition can allow translation or rotation, so a balanced-torque calculation is not the entire equilibrium problem.

In an equilibrium problem, choose a pivot that removes unknown forces whose lines of action pass through it. This reduces algebra but does not eliminate those forces from the force-balance equations. After finding one support force from torque, use vertical or horizontal balance to find the others.

For a beam, include its weight at its center of mass, not automatically at a geometric center if mass is nonuniform. Distinguish a force’s application point from its perpendicular lever arm. A slanted applied force may have only one component producing torque about the chosen pivot.

Check the resulting forces for physical plausibility. A support capable only of pushing cannot supply a tensile force without additional hardware. A negative result may indicate that the assumed contact cannot be maintained, rather than merely an algebraic sign to ignore.