Unit 6 Beta

Energy and Momentum of Rotating Systems: the one-page sheet

6.1 Rotational Kinetic Energy

An object turning about an axis has rotational kinetic energy K = ½Iω², the sum of ½mv² over all its pieces. It is a scalar in joules, is present even when the center of mass is at rest, grows with the square of ω and in proportion to the rotational inertia.

A turning object has rotational kinetic energy K = ½Iω², even if its center of mass is at rest.
It is the sum of ½mv² over every piece, with v = rω.
A scalar in joules; the direction of turning does not matter, because ω is squared.
K ∝ ω² at fixed I (doubling ω gives 4K) and K ∝ I at fixed ω.
K against ω² is a straight line with slope ½I.

each piece of mass m at distance r moves at v = rω adding the pieces gives ½(Σmr²)ω² the wheel’s kinetic energy is K = ½Iω² K becomes 4 times as large I and so K are larger at the same ω

rotational kinetic energy: The kinetic energy of an object turning about an axis: K = ½Iω², a scalar in joules. An object can have it while its center of mass is at rest.

6.2 Torque and Work

A torque that turns an object through an angle does work W = τΔθ (angle in radians), or the area under a torque-angle graph when the torque varies. The net work done on a turning object changes its rotational kinetic energy; a torque that opposes the turning does negative work.

W = τΔθ for a constant torque, with Δθ in radians: the partner of W = Fd.
For a changing torque, W is the area under the τ-θ graph.
Positive when the torque turns the object the way it is going; negative when it opposes.
The net work done on a turning object equals its change in rotational kinetic energy.
Convert revolutions to radians first: 1 rev = 2π rad.

it does work Fs on the wheel the work is Fs = (rF)θ = τΔθ the work is the area under the torque-angle graph its rotational kinetic energy increases by that amount it does negative work and takes kinetic energy out

work done by a torque: The work done on a turning object by a torque: W = τΔθ for a constant torque, with Δθ in radians. Positive when the torque is in the direction of turning.
torque-angle graph: A graph of torque against the angle turned. The area under it is the work done by the torque.

6.3 Angular Momentum and Angular Impulse

Angular momentum is L = Iω for a rigid object turning about a fixed axis, and L = rmv sin θ for a point object about a chosen point. A net torque acting over a time changes it: angular impulse τΔt = ΔL, the area under a τ-t graph, and the net torque is the slope of an L-t graph.

L = Iω for a rigid object about a fixed axis; unit kg·m²/s.
A point object has L = rmv sin θ about a chosen point: momentum × perpendicular distance from the point to its line of motion. Change the point, change L.
A puck in straight-line motion keeps the same L about any fixed point, and has L = 0 about points on its path.
Angular impulse τΔt = ΔL; it is the area under a τ-t graph.
Net torque = slope of the L-t graph (and = Iα when I is constant).

it has angular momentum L = Iω it has angular momentum L = rmv sin θ = mv × (perpendicular distance) about that point it delivers an angular impulse τΔt the angular momentum changes by exactly that much: ΔL = τΔt L stays the same, and the L-t graph is flat

angular momentum: The rotational partner of momentum. For a rigid object turning about a fixed axis, L = Iω, in kg·m²/s; its sign gives the direction of turning.
angular momentum of a point object: For a point object moving past a chosen point, L = rmv sin θ: its momentum times the perpendicular distance from the point to its line of motion. It depends on the point chosen.
angular impulse: A torque times the time it acts, τΔt (the area under a torque-time graph), in N·m·s. It equals the change in angular momentum: τΔt = ΔL.

6.4 Conservation of Angular Momentum

A system’s total angular momentum stays constant when the net external torque on it is zero. When the system changes shape, its rotational inertia changes and its angular velocity changes the opposite way (I₁ω₁ = I₂ω₂), while kinetic energy does not stay constant. Around a pivot, angular momentum about the pivot is conserved in a collision even though linear momentum is not.

L is constant when the net external torque is zero. State the system and this condition every time.
Choose the system so that the forces that change each part are internal.
Shape change at constant L: I₁ω₁ = I₂ω₂. Pulling mass in speeds up the spin.
Kinetic energy is not conserved in these changes: pulling in raises K; sticking collisions lower it.
A pivot pushes during a collision, so linear momentum changes, but its torque about the pivot is zero, so L about the pivot is conserved.

no angular impulse comes from outside, so the total L stays the same they trade angular momentum, but the total is unchanged I falls, so ω = L/I rises K = L²/(2I) rises; internal work (muscles) supplies it L about the pivot is conserved (the pivot force has no torque about it) even though linear momentum is not

net external torque: A torque on a system from an object outside it. Only the net external torque can change a system’s total angular momentum.
conservation of angular momentum: The total angular momentum of a system stays constant when the net external torque on it is zero. The parts can trade angular momentum, but the total does not change.
changing rotational inertia: When a system changes shape with no net external torque (a skater pulling in her arms), its rotational inertia changes and its angular velocity changes the opposite way so that L = Iω stays the same.

6.5 Rolling

An object rolling without slipping has v_cm = Rω, and its kinetic energy is ½Mv² + ½Iω². Static friction makes it spin without doing work, so mechanical energy is conserved: from rest down a height h, v = √(2gh/(1 + I/MR²)), which depends on shape only. When an object slips, v_cm ≠ Rω and kinetic friction dissipates energy until rolling sets in.

Rolling without slipping: v_cm = Rω and Δx_cm = RΔθ.
Total kinetic energy = ½Mv_cm² + ½Iω².
Static friction makes a rolling object spin but does no work: mechanical energy is conserved.
Rolling race: v = √(2gh/(1 + I/MR²)). Mass and radius cancel; smaller I/MR² wins (sphere, disk, hoop). A frictionless sliding block beats them all.
Slipping: v_cm ≠ Rω; kinetic friction dissipates energy and changes v_cm and ω until they match.

its center moves one circumference per turn, so v_cm = Rω its kinetic energy is ½Mv² + ½Iω² Mgh is shared between moving and spinning: v = √(2gh/(1 + I/MR²)) more energy goes into spinning, so it moves more slowly v_cm ≠ Rω, kinetic friction dissipates energy, and the motion changes until rolling sets in

rolling without slipping: Rolling in which the contact point does not slide on the surface, so the center moves forward one circumference per turn: Δx_cm = RΔθ and v_cm = Rω. Static friction may act but does no work.
translational plus rotational kinetic energy: The kinetic energy of a rolling object: K = ½Mv_cm² + ½Iω², the translational kinetic energy of its center of mass plus its rotational kinetic energy about the center.
rolling while slipping: Rolling in which the contact point slides, so v_cm ≠ Rω. Kinetic friction then acts and dissipates mechanical energy until the motion matches (or the object stops).

6.6 Motion of Orbiting Satellites

For a satellite of small mass orbiting a massive body, total mechanical energy is conserved (gravity is the only force, and it is conservative) and angular momentum about the central body is conserved (gravity exerts no torque about its center). In a circular orbit K = GMm/(2r) and E = −K; on an ellipse, K and U trade and the satellite is fastest at its closest point. Escape needs E = 0, v = √(2GM/r).

Model: a satellite of much smaller mass than the central body; U_G = −GMm/r, zero at infinite separation.
Circular orbit: E, U, K and L are all constant. v = √(GM/r), K = GMm/(2r), E = −K.
Elliptical orbit: E and L constant; K and U trade. Fastest at the closest point: r_Av_A = r_Bv_B where v ⊥ r.
A bound orbit has E < 0. Escape speed sets E = 0: v = √(2GM/r).
Name the reason: no torque about the center (L), only a conservative force inside the system (E).

the total mechanical energy E = K + U of the system is constant it exerts no torque about that point, so the satellite’s angular momentum L about it is constant U becomes more negative and K grows: it speeds up (r v sin θ stays fixed) K = GMm/(2r), U = −2K and E = −K: a bound orbit has E < 0 the object can just escape: v_esc = √(2GM/r)

elliptical orbit: An oval orbit with the central body off center. The satellite moves fastest at its closest point and slowest at its farthest; the total energy and the angular momentum about the central body stay constant while K and U trade.
total orbital energy: The total mechanical energy of a satellite and the body it orbits, E = K + U with U_G = −GMm/r (zero at infinite separation). It is negative for a bound orbit; for a circular orbit E = −K = −GMm/(2r).
escape speed: The smallest speed that lets an object reach infinite separation from a body: the speed at which the total energy is zero, v = √(2GM/r).