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AP Physics 1 Flashcards: Torque

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

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What this deck covers

This deck focuses on Torque, giving you a quick way to review the definitions, rules, and examples that matter most for AP Physics 1.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

AP Physics 1 Flashcards: Torque

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QUESTION

What is the torque if F=0F=0F=0?

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ANSWER

Zero. No force means no rotational effect can be produced.

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Flashcard 1: What is the torque if F=0F=0F=0?

Answer: Zero. No force means no rotational effect can be produced.

Flashcard 2: What does FFF represent in the torque formula?

Answer: Magnitude of the force applied. The force magnitude directly affects the rotational strength.

Flashcard 3: What effect does increasing θ\thetaθ to 90∘90^\circ90∘ have on torque?

Answer: Torque increases. Approaching 90∘90^\circ90∘ increases sin⁡(θ)\sin(\theta)sin(θ) toward its maximum value.

Flashcard 4: What happens to torque if the angle is decreased from 90∘90^\circ90∘?

Answer: Torque decreases. Smaller angles reduce sin⁡(θ)\sin(\theta)sin(θ), decreasing rotational effect.

Flashcard 5: State the effect on torque if the force is perpendicular to the lever arm.

Answer: Torque is maximized. Perpendicular force gives sin⁡(90°)=1\sin(90°) = 1sin(90°)=1, the maximum torque.

Flashcard 6: What does τ\tauτ represent in the torque formula?

Answer: Torque. Greek letter tau represents the rotational moment or turning effect.

Flashcard 7: State the relationship between torque and rotational equilibrium.

Answer: Net torque must be zero. No net torque means no angular acceleration occurs.

Flashcard 8: Calculate torque: r=1r=1r=1 m, F=10F=10F=10 N, θ=45∘\theta=45^\circθ=45∘.

Answer: τ=7.07\tau = 7.07τ=7.07 Nm. τ=1×10×sin⁡(45°)=10×0.707=7.07\tau = 1 \times 10 \times \sin(45°) = 10 \times 0.707 = 7.07τ=1×10×sin(45°)=10×0.707=7.07 Nm.

Flashcard 9: Which quantity is vector: force, mass, or torque?

Answer: Torque. Torque has both magnitude and direction, making it a vector.

Flashcard 10: What is the formula for torque?

Answer: τ=r⋅F⋅sin⁡(θ)\tau = r \cdot F \cdot \sin(\theta)τ=r⋅F⋅sin(θ). Fundamental formula where all three factors determine rotational effect.

Flashcard 11: What is the effect of negative torque on rotation?

Answer: Causes clockwise rotation. Negative torque produces rotation opposite to positive direction.

Flashcard 12: Find the lever arm if torque is 202020 Nm and force is 555 N.

Answer: r=4r = 4r=4 m. From τ=rFsin⁡(θ)\tau = rF\sin(\theta)τ=rFsin(θ), assuming θ=90°\theta = 90°θ=90°: r=τ/F=20/5=4r = \tau/F = 20/5 = 4r=τ/F=20/5=4 m.

Flashcard 13: What does rrr represent in the torque formula?

Answer: Distance from pivot point to force application. The lever arm length determines how much turning advantage you have.

Flashcard 14: Which direction is positive torque conventionally?

Answer: Counterclockwise. Standard physics convention for positive rotational direction.

Flashcard 15: Identify the symbol for torque in equations.

Answer: τ\tauτ. Greek letter tau is the standard physics symbol for torque.

Flashcard 16: Find torque: r=3r=3r=3 m, F=15F=15F=15 N, θ=90∘\theta=90^\circθ=90∘.

Answer: τ=45\tau = 45τ=45 Nm. τ=3×15×sin⁡(90°)=3×15×1=45\tau = 3 \times 15 \times \sin(90°) = 3 \times 15 \times 1 = 45τ=3×15×sin(90°)=3×15×1=45 Nm.

Flashcard 17: Calculate the torque if r=2r=2r=2 m, F=10F=10F=10 N, θ=0∘\theta=0^\circθ=0∘.

Answer: τ=0\tau = 0τ=0 Nm. sin⁡(0°)=0\sin(0°) = 0sin(0°)=0, so torque equals zero regardless of rrr and FFF.

Flashcard 18: Calculate torque for r=3r=3r=3 m, F=10F=10F=10 N, θ=60∘\theta=60^\circθ=60∘.

Answer: τ=25.98\tau = 25.98τ=25.98 Nm. τ=3×10×sin⁡(60°)=30×0.866=25.98\tau = 3 \times 10 \times \sin(60°) = 30 \times 0.866 = 25.98τ=3×10×sin(60°)=30×0.866=25.98 Nm.

Flashcard 19: Calculate torque for r=1.5r=1.5r=1.5 m, F=20F=20F=20 N, θ=45∘\theta=45^\circθ=45∘.

Answer: τ=21.21\tau = 21.21τ=21.21 Nm. τ=1.5×20×sin⁡(45°)=30×0.707=21.21\tau = 1.5 \times 20 \times \sin(45°) = 30 \times 0.707 = 21.21τ=1.5×20×sin(45°)=30×0.707=21.21 Nm.

Flashcard 20: Find the torque for r=2r=2r=2 m, F=5F=5F=5 N, θ=30∘\theta=30^\circθ=30∘.

Answer: τ=5\tau = 5τ=5 Nm. τ=2×5×sin⁡(30°)=2×5×0.5=5\tau = 2 \times 5 \times \sin(30°) = 2 \times 5 \times 0.5 = 5τ=2×5×sin(30°)=2×5×0.5=5 Nm.

Flashcard 21: When is the torque zero?

Answer: When θ=0∘\theta = 0^\circθ=0∘ or sin⁡(θ)=0\sin(\theta) = 0sin(θ)=0. Force parallel to lever arm produces no rotational effect.

Flashcard 22: State the torque if F=8F=8F=8 N, r=0.5r=0.5r=0.5 m, and θ=90∘\theta=90^\circθ=90∘.

Answer: τ=4\tau = 4τ=4 Nm. τ=8×0.5×sin⁡(90°)=8×0.5×1=4\tau = 8 \times 0.5 \times \sin(90°) = 8 \times 0.5 \times 1 = 4τ=8×0.5×sin(90°)=8×0.5×1=4 Nm.

Flashcard 23: State the effect on torque if θ\thetaθ changes from 90∘90^\circ90∘ to 0∘0^\circ0∘.

Answer: Torque decreases to zero. Maximum torque at 90°90°90° reduces to zero at 0°0°0°.

Flashcard 24: Choose the effect of increasing rrr on torque.

Answer: Torque increases. Torque is directly proportional to lever arm distance.

Flashcard 25: State the effect of doubling rrr on torque if FFF and θ\thetaθ are constant.

Answer: Torque doubles. Torque is directly proportional to lever arm length.

Flashcard 26: State the right-hand rule in torque context.

Answer: Curl fingers in rotation direction; thumb points in torque direction. Standard method to determine torque vector direction in 3D.

Flashcard 27: Identify the direction of torque for negative angle.

Answer: Clockwise. Negative angles conventionally indicate clockwise rotation direction.

Flashcard 28: Identify the unit of torque in the SI system.

Answer: Newton-meter (Nm). Force times distance gives units of energy per radian.

Flashcard 29: What happens to torque if the force increases?

Answer: Torque increases. Torque is directly proportional to applied force magnitude.

Flashcard 30: Find the net torque: τ1=5\tau_1=5τ1​=5 Nm, τ2=−3\tau_2=-3τ2​=−3 Nm.

Answer: τnet=2\tau_{net} = 2τnet​=2 Nm. Vector addition: 5+(−3)=25 + (-3) = 25+(−3)=2 Nm in positive direction.