AP Physics 1 Flashcards: Rolling

Study Rolling in AP Physics 1 with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

AP Physics 1

Rolling

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QUESTION
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What is the instantaneous speed of the contact point on a rolling wheel (relative to ground)?

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ANSWER

vcontact=0v_{contact}=0. Contact point is instantaneous center of rotation for pure rolling.

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Flashcard 1: What is the instantaneous speed of the contact point on a rolling wheel (relative to ground)?

Answer: vcontact=0v_{contact}=0. Contact point is instantaneous center of rotation for pure rolling.

Flashcard 2: Find the rolling distance given angular displacement θ\theta and radius rr.

Answer: s=rθs = r \theta. Rolling distance equals radius times angular displacement.

Flashcard 3: Calculate the moment of inertia for a ring rolling about its central axis.

Answer: I=mr2I = m r^2. For a thin ring, all mass is concentrated at radius rr.

Flashcard 4: What is the formula for potential energy of a rolling object at height hh?

Answer: PE=mghPE = mgh. Gravitational potential energy depends on mass, gravity, and height.

Flashcard 5: What is the static friction magnitude for rolling down an incline at angle θ\theta (using β=IcmmR2\beta=\frac{I_{cm}}{mR^2})?

Answer: fs=β1+βmgsinθf_s=\frac{\beta}{1+\beta}mg\sin\theta. Friction provides torque to maintain rolling constraint on incline.

Flashcard 6: What is the moment of inertia of a solid disk or solid cylinder about its center?

Answer: Icm=12mR2I_{cm}=\frac{1}{2}mR^2. Mass distributed uniformly throughout solid disk or cylinder.

Flashcard 7: Calculate the moment of inertia for a ring rolling about its central axis.

Answer: I=mr2I = m r^2. For a thin ring, all mass is concentrated at radius rr.

Flashcard 8: What is the direction of static friction on a driven wheel that is powered to accelerate forward without slipping?

Answer: Static friction points backward. Friction opposes slipping tendency; powered wheel would slip forward.

Flashcard 9: What is the direction of static friction on a freely rolling wheel pulled forward at its axle (no applied torque)?

Answer: Static friction points backward. Friction provides torque to accelerate rotation when pulled at axle.

Flashcard 10: State the formula for the total mechanical energy of a rolling object.

Answer: E=KEtrans+KErotE = KE_{\text{trans}} + KE_{\text{rot}}. Total energy is the sum of translational and rotational kinetic energies.

Flashcard 11: What is the formula for the work done by torque on a rolling object?

Answer: W=τθW = \tau \theta. Work equals torque times angular displacement.

Flashcard 12: What is the moment of inertia for a solid sphere rolling about its central axis?

Answer: I=25mr2I = \frac{2}{5} m r^2. For a solid sphere, II is two-fifths of mr2mr^2.

Flashcard 13: What is the moment of inertia of a hollow cylinder rolling about its central axis?

Answer: I=mr2I = m r^2. For a thin-walled cylinder, all mass is at radius rr.

Flashcard 14: Find ω\omega for rolling without slipping if vcm=6m/sv_{cm}=6\,\text{m/s} and R=0.50mR=0.50\,\text{m}.

Answer: ω=12rad/s\omega=12\,\text{rad/s}. Apply ω=vcmR=60.5=12\omega = \frac{v_{cm}}{R} = \frac{6}{0.5} = 12 rad/s.

Flashcard 15: What is the linear acceleration of an object rolling without slipping down an incline at angle θ\theta?

Answer: a=gsinθ1+IcmmR2a=\frac{g\sin\theta}{1+\frac{I_{cm}}{mR^2}}. Derived from Newton's second law with rolling constraint.

Flashcard 16: What is the formula for the centripetal force on a rolling object?

Answer: Fc=mv2rF_c = \frac{mv^2}{r}. Centripetal force equals mass times velocity squared over radius.

Flashcard 17: What is the formula for potential energy of a rolling object at height hh?

Answer: PE=mghPE = mgh. Gravitational potential energy depends on mass, gravity, and height.

Flashcard 18: What is the speed of a point on the rim at the top of a rolling wheel (relative to ground)?

Answer: vtop=2vcmv_{top}=2v_{cm}. Top point moves at vcmv_{cm} plus rim speed ωR=vcm\omega R = v_{cm}.

Flashcard 19: What is the parallel-axis theorem for moment of inertia?

Answer: I=Icm+md2I=I_{cm}+md^2. Relates moment of inertia about any axis to that about center of mass.

Flashcard 20: Find acma_{cm} for rolling without slipping if α=8rad/s2\alpha=8\,\text{rad/s}^2 and R=0.25mR=0.25\,\text{m}.

Answer: acm=2.0m/s2a_{cm}=2.0\,\text{m/s}^2. Apply acm=αR=8×0.25=2.0a_{cm} = \alpha R = 8 \times 0.25 = 2.0 m/s².

Flashcard 21: What is the formula for the moment of inertia of a solid cylinder rolling about its central axis?

Answer: I=12mr2I = \frac{1}{2} m r^2. For a solid cylinder, moment of inertia equals half mass times radius squared.

Flashcard 22: What is the moment of inertia for a solid sphere rolling about its central axis?

Answer: I=25mr2I = \frac{2}{5} m r^2. For a solid sphere, II is two-fifths of mr2mr^2.

Flashcard 23: What is the rolling-without-slipping condition relating vcmv_{cm} and ω\omega for radius RR?

Answer: vcm=ωRv_{cm}=\omega R. Center of mass velocity equals angular velocity times radius for pure rolling.

Flashcard 24: What is the condition for rolling without slipping?

Answer: v=rθv = r \theta. For no slipping, contact point velocity equals zero.

Flashcard 25: State the formula for the total mechanical energy of a rolling object.

Answer: E=KEtrans+KErotE = KE_{\text{trans}} + KE_{\text{rot}}. Total energy is the sum of translational and rotational kinetic energies.

Flashcard 26: What is the moment of inertia for a thin spherical shell rolling about its central axis?

Answer: I=23mr2I = \frac{2}{3} m r^2. For a hollow sphere, II is two-thirds of mr2mr^2.

Flashcard 27: Find the rolling distance given angular displacement θ\theta and radius rr.

Answer: s=rθs = r \theta. Rolling distance equals radius times angular displacement.

Flashcard 28: What is the expression for the angular displacement of a rolling object?

Answer: θ=sr\theta = \frac{s}{r}. Angular displacement equals arc length divided by radius.

Flashcard 29: What is the moment of inertia of a thin spherical shell about its center?

Answer: Icm=23mR2I_{cm}=\frac{2}{3}mR^2. All mass concentrated at surface of hollow sphere.

Flashcard 30: What is the moment of inertia of a hollow cylinder rolling about its central axis?

Answer: I=mr2I = m r^2. For a thin-walled cylinder, all mass is at radius rr.

Flashcard 31: What is the minimum coefficient μs\mu_s needed to roll without slipping down an incline (using β=IcmmR2\beta=\frac{I_{cm}}{mR^2})?

Answer: μsβ1+βtanθ\mu_s\ge\frac{\beta}{1+\beta}\tan\theta. Ensures friction force doesn't exceed maximum static friction.

Flashcard 32: What is the expression for the angular displacement of a rolling object?

Answer: θ=sr\theta = \frac{s}{r}. Angular displacement equals arc length divided by radius.

Flashcard 33: What is the equation for the translational kinetic energy of a rolling object?

Answer: KEtrans=12mv2KE_{\text{trans}} = \frac{1}{2} mv^2. Translational KE depends on mass and linear velocity squared.

Flashcard 34: What is the condition for rolling without slipping?

Answer: v=rθv = r \theta. For no slipping, contact point velocity equals zero.

Flashcard 35: What is the moment of inertia for a thin spherical shell rolling about its central axis?

Answer: I=23mr2I = \frac{2}{3} m r^2. For a hollow sphere, II is two-thirds of mr2mr^2.

Flashcard 36: What is the rolling-without-slipping condition relating acma_{cm} and α\alpha for radius RR?

Answer: acm=αRa_{cm}=\alpha R. Center of mass acceleration equals angular acceleration times radius for pure rolling.

Flashcard 37: What is the equation for the translational kinetic energy of a rolling object?

Answer: KEtrans=12mv2KE_{\text{trans}} = \frac{1}{2} mv^2. Translational KE depends on mass and linear velocity squared.

Flashcard 38: What is the moment of inertia of a solid sphere about its center?

Answer: Icm=25mR2I_{cm}=\frac{2}{5}mR^2. Mass distributed uniformly throughout solid sphere.

Flashcard 39: What is the formula for the work done by torque on a rolling object?

Answer: W=τθW = \tau \theta. Work equals torque times angular displacement.

Flashcard 40: What is the direction of static friction on a freely rolling wheel going down an incline without slipping?

Answer: Static friction points up the incline. Friction prevents wheel from sliding down faster than it rolls.

Flashcard 41: What is the formula for the centripetal force on a rolling object?

Answer: Fc=mv2rF_c = \frac{mv^2}{r}. Centripetal force equals mass times velocity squared over radius.

Flashcard 42: Identify the correct aa down a ramp: disk vs hoop, same mm and RR, rolling without slipping.

Answer: Disk has larger aa than hoop. Disk has β=0.5\beta=0.5 vs hoop's β=1\beta=1, giving larger aa.

Flashcard 43: Identify which reaches the bottom first (no slipping): hoop or solid sphere, same RR and same ramp.

Answer: Solid sphere. Sphere has smaller β=IcmmR2\beta=\frac{I_{cm}}{mR^2}, so larger acceleration.

Flashcard 44: What is the formula for the moment of inertia of a solid cylinder rolling about its central axis?

Answer: I=12mr2I = \frac{1}{2} m r^2. For a solid cylinder, moment of inertia equals half mass times radius squared.

Flashcard 45: What is the moment of inertia of a hoop (thin-walled cylinder) about its center?

Answer: Icm=mR2I_{cm}=mR^2. All mass concentrated at distance RR from center.

Flashcard 46: What is the total kinetic energy of a rigid object rolling without slipping?

Answer: K=12mvcm2+12Icmω2K=\frac{1}{2}mv_{cm}^2+\frac{1}{2}I_{cm}\omega^2. Sum of translational and rotational kinetic energies.