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Flashcards · AP® Physics 1: Algebra-Based · Unit 6

Unit 6: Energy and Momentum of Rotating Systems flashcards

37 cards on topics 6.1 to 6.6. Pick a topic or study the whole unit.

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6.1 Rotational Kinetic Energy

What is rotational kinetic energy?
K = ½Iω², with ω in rad/s and I in kg·m². It is the sum of ½mv² for every piece (v = rω). It is a scalar: the sense of rotation makes no difference.
A wheel with I = 0.40 kg·m² spins at 5.0 rad/s. What is its kinetic energy?
K = ½Iω² = ½ × 0.40 × 5.0² = 5.0 J. Halving ω would make it one quarter.
How do you find the kinetic energy of an object that moves and spins?
K_total = ½Mv_cm² + ½I_cm ω². About a fixed pivot use ½I_pivot ω² alone. Never mix the two descriptions.
Can an object have kinetic energy if its center of mass is at rest?
Yes. A flywheel spinning on a fixed axle has rotational kinetic energy even though its center of mass is still and its total momentum is zero.
Which stores more energy at the same ω: a hoop or a disc of equal mass and radius?
The hoop (I = MR²) stores twice the energy of the disc (I = ½MR²), because K = ½Iω² and its mass is farther from the axis.
A 2.0 kg ball moves at 3.0 m/s and spins at 30 rad/s with I_cm = 0.010 kg·m². What is its total kinetic energy?
Translational ½ × 2.0 × 3.0² = 9.0 J plus rotational ½ × 0.010 × 30² = 4.5 J, total 13.5 J.

6.2 Torque and Work

How much work does a constant torque do?
W = τΔθ, with Δθ in radians and W in J. It comes from W = Fs with arc length s = rΔθ.
A constant torque of 2.0 N·m acts through 3.0 revolutions. How much work does it do?
Δθ = 3.0 × 2π ≈ 18.8 rad, so W = 2.0 × 18.8 ≈ 37.7 J. Always convert revolutions to radians.
How does the net work done by torques relate to rotation?
W_net = ½Iω² − ½Iω₀²: the net work equals the change in rotational kinetic energy.
What sign is the work done by axle friction on a spinning wheel?
Negative. A torque opposing the rotation takes energy out; a torque in the same sense as the rotation does positive work.
How do you find the work done by a torque that changes with angle?
Find the area under the τ–θ graph. A torque rising steadily from 0 to 4.0 N·m over 3.0 rad does ½ × 3.0 × 4.0 = 6.0 J.
How does a stopping angle depend on initial angular speed against a constant friction torque?
It grows with ω₀², because τΔθ = ½Iω₀². Doubling ω₀ multiplies the stopping angle by 4.

6.3 Angular Momentum and Angular Impulse

What is the angular momentum of a rigid body spinning about a fixed axis?
L = Iω, in kg·m²/s, measured about the stated axis.
Does an object moving in a straight line have angular momentum?
Yes, about any point off its path: L = rmv sin θ = mvd, where d is the perpendicular distance from the point to the line of motion. About a point on the path, L = 0.
What is the rotational impulse–momentum theorem?
Net angular impulse equals change in angular momentum: τ_net Δt = ΔL = L − L₀. Angular impulse has units N·m·s = kg·m²/s.
How do you read torque and angular impulse from graphs?
Slope of an L–t graph = net torque. Area under a τ–t graph = angular impulse (area below the axis is negative).
A constant net torque of 0.50 N·m acts for 6.0 s. What is the change in angular momentum?
ΔL = τΔt = 0.50 × 6.0 = 3.0 kg·m²/s, with the same sign as the torque.
Can a small torque produce a bigger change in angular momentum than a large one?
Yes, if it acts for longer. Compare angular impulses τΔt, not torques.

6.4 Conservation of Angular Momentum

When is angular momentum conserved?
When the net external torque on the system is zero. Internal torques give equal and opposite angular impulses, so they cannot change the total.
A spinning skater pulls in her arms, halving her rotational inertia. What happens to ω and K?
L = Iω is constant, so ω doubles. K = L²/(2I) also doubles; the extra energy comes from the work she does pulling in.
A disk (I = 0.20 kg·m²) at 6.0 rad/s has a disk (I = 0.10 kg·m²) at rest dropped onto it. What is the common angular speed?
I₁ω₁ = (I₁ + I₂)ω, so ω = 0.20 × 6.0 / 0.30 = 4.0 rad/s. Do not average the angular speeds.
Is kinetic energy conserved when angular momentum is?
Not necessarily. Objects sticking together lose kinetic energy; pulling mass inwards gains it. Track K with K = L²/(2I).
What graph tests conservation of angular momentum from data?
Plot ω against 1/I. A straight line through the origin shows L is constant (slope = L).
Why can you use angular momentum but not linear momentum for an object hitting a pivoted rod?
The pivot exerts an external force, so linear momentum changes. That force acts at the axis, so its torque about the pivot is zero and L about the pivot is conserved.

6.5 Rolling

What is the condition for rolling without slipping?
v_cm = rω (and Δx_cm = rΔθ, a_cm = rα). The contact point is momentarily at rest and the top point moves at 2v_cm.
What is the kinetic energy of a rolling object?
K_total = ½Mv_cm² + ½I_cm ω². With I_cm = βMr² and no slipping, K = ½(1 + β)Mv_cm².
Why is mechanical energy conserved for an object rolling without slipping?
The contact point is at rest, so static friction does no work. Its torque only shares gravity's energy between translation and rotation.
How fast is an object moving after rolling from rest down a height h?
v = √(2gh / (1 + β)), independent of mass and radius. A hoop (β = 1) reaches √(gh); a frictionless sliding block reaches the faster √(2gh).
Which wins a race down a ramp: a solid sphere, a disc or a hoop?
The solid sphere (β = ⅖), then the disc (½), then the hoop (1). Lower β puts more energy into translation. Mass and radius do not matter.
What fraction of a rolling disc's kinetic energy is rotational?
One third. With β = ½, rotational ½ × ½Mv² is ⅓ of the total ½(1 + ½)Mv².
What does kinetic friction do when a wheel rolls while slipping?
The contact point slides, so v_cm ≠ rω. Kinetic friction opposes the sliding, changes v_cm and ω until rolling begins, and dissipates energy meanwhile.

6.6 Motion of Orbiting Satellites

Why are a satellite's angular momentum and mechanical energy constant?
Gravity points at the central body's centre, so it exerts no torque about it: L is constant. Gravity is the only force, so E = ½mv² − GMm/r is constant.
What are the energies of a satellite in a circular orbit?
K = GMm/(2r), U_g = −GMm/r and E = −GMm/(2r). So E = −K and U_g = 2E. Every bound orbit has negative total energy.
Where is a satellite in an elliptical orbit fastest?
At the closest point. E and L are constant, so as U_g falls K rises. At the closest and farthest points L = mvr, so v ∝ 1/r there.
What is escape speed?
v_esc = √(2GM/r): the speed that makes E = 0, so the object only approaches zero speed as r grows without limit. It is √2 × circular orbital speed and does not depend on the object's mass.
What is the orbital speed of a satellite in a circular orbit?
v² = GM/r, with r from the centre of the central body. A higher circular orbit has less kinetic energy but more (less negative) total energy.
A probe moves at 6.0 km/s at 1.0 × 10⁷ m from a planet's centre. How fast is it at its farthest point, 3.0 × 10⁷ m away?
L = mvr is the same at both points, so v = 6.0 × (1.0 / 3.0) = 2.0 km/s.

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