Rotational Equilibrium and Newton's First Law - AP Physics 1

A turntable rotates at constant angular velocity while a student gently presses a finger on its edge. The motor adjusts so the rotation rate stays constant; angular acceleration is zero. What is the net torque about the turntable's axis?
Answer: It is zero. This question assesses understanding of rotational equilibrium and Newton's first law in rotation. The turntable rotates at constant angular velocity, so angular acceleration is zero. Newton's first law for rotation states that zero angular acceleration implies zero net torque about the axis. The motor's torque balances the frictional torque from the finger, maintaining equilibrium. Choice A is incorrect because friction's torque is canceled by the motor, not making net torque nonzero. To solve similar problems, always check if angular acceleration is zero, which implies net torque is zero regardless of motion.
A uniform rod is held horizontally by two hands applying upward forces at its ends. The rod is motionless and does not rotate; angular acceleration is zero. What can be concluded about the net torque on the rod about its center of mass?
Answer: The net torque is zero. This question assesses understanding of rotational equilibrium and Newton's first law in rotation. The rod is motionless with zero angular acceleration about its center of mass. Newton's first law for rotation means that zero angular acceleration requires zero net torque. The upward forces at the ends balance the weight and any torques they produce. Choice C is incorrect because multiple forces at different points can still result in zero net torque if balanced. To solve similar problems, always check if angular acceleration is zero, which implies net torque is zero regardless of motion.
A pulley with a rope wrapped around it is mounted on a fixed axle. Two students pull on opposite ends of the rope with different forces, yet the pulley rotates at constant angular velocity. The angular acceleration is measured to be $\alpha=0$ . What can be concluded about the net torque on the pulley about its axle?
Answer: The net torque must be zero. This question assesses understanding of rotational equilibrium and Newton's first law in rotational dynamics. In rotational terms, Newton's first law states that net torque is required to change angular velocity. Since the pulley rotates at constant velocity with α=0, the net torque about the axle must be zero. This comes from the equation τ_net = Iα, where zero α means zero net torque despite different forces. Choice A is incorrect because different forces can be balanced by other torques, like from the axle, to yield net zero. For pulleys or wheels, leverage the α=0 condition to conclude torque balance without needing force magnitudes.
A uniform ladder rests against a frictionless wall and rough floor without slipping. The ladder is stationary, and the angular acceleration is explicitly $\alpha=0$ . A student argues that because the wall is frictionless, the ladder must have a nonzero net torque. Which statement is supported about the ladder's net torque?
Answer: The net torque is zero. This question assesses understanding of rotational equilibrium and Newton's first law in rotational dynamics. Newton's first law in rotational form requires zero net torque for equilibrium. With the ladder stationary and α=0, the net torque is zero. This is because τ_net = Iα, and α=0 implies no net torque overall. Choice A is incorrect because the horizontal wall force can be part of a balanced torque system with floor friction and gravity. For static systems like ladders, select an axis and apply zero net torque directly from the α=0 condition.
A solid disk rotates on a vertical axle. A motor applies a torque, while a brake pad applies an opposing torque. The disk's angular speed remains constant for $5\,\text{s}$ , and the angular acceleration is $\alpha=0$ throughout that time. What is the net torque on the disk during the $5\,\text{s}$ interval?
Answer: Zero, because the torques must balance. This question assesses understanding of rotational equilibrium and Newton's first law in rotational dynamics. Newton's first law in rotation states that constant angular velocity persists without net torque. Since the disk's angular speed is constant with α=0, the net torque must be zero. This is because τ_net = Iα, and α=0 implies τ_net=0, meaning motor and brake torques balance. Choice A is incorrect because applied torques can exist but must cancel out for no net effect. In systems with opposing torques, verify equilibrium by confirming α=0 to conclude net torque is zero.
A turntable rotates at constant angular speed while a small motor provides torque and bearing friction provides an opposite torque. The angular acceleration is $\alpha=0$ . What is the net torque?
Answer: Zero, because the torques must sum to zero. This problem demonstrates rotational equilibrium in a friction-compensated system. With constant angular speed and α = 0, Newton's second law for rotation requires Στ = Iα = 0, giving zero net torque. The motor torque exactly balances the bearing friction torque, maintaining steady rotation. This is analogous to driving a car at constant speed where engine force balances air resistance. Choice D incorrectly applies the formula Iω, which represents angular momentum, not torque. The key insight is: constant angular velocity requires torque balance, not torque absence.
A uniform meterstick is balanced horizontally on a pivot located at the $40\,\text{cm}$ mark. A $1.0\,\text{N}$ weight hangs from the $10\,\text{cm}$ mark, and an unknown weight hangs from the $90\,\text{cm}$ mark. The stick remains at rest, and the angular acceleration is $\alpha=0$ . Which statement must be true about the net torque on the stick about the pivot?
Answer: The net torque about the pivot is zero. This question assesses understanding of rotational equilibrium and Newton's first law in rotational dynamics. According to Newton's first law for rotation, an object maintains its angular velocity without a net torque. With the meterstick at rest and α=0, the net torque about the pivot must be zero. This follows from τ_net = Iα, so zero angular acceleration means no net torque. Choice A is incorrect because even if one weight is farther, the magnitudes are such that torques balance for equilibrium. To solve similar problems, calculate individual torques and ensure their sum is zero when α=0 is given.
A uniform horizontal beam is hinged to a wall at its left end and supported at its right end by a vertical cable. The beam remains at rest while a $200\,\text{N}$ load hangs from its midpoint. The hinge can exert both horizontal and vertical forces. The system is not slipping or rotating, and the angular acceleration of the beam is explicitly $\alpha=0$ . Which statement about the net torque on the beam is correct?
Answer: The net torque about any axis is zero because $\alpha=0$ . This question assesses understanding of rotational equilibrium and Newton's first law in rotational dynamics. Newton's first law for rotation states that an object at rest or in constant angular motion will remain so unless acted upon by a net external torque. Given that the beam is at rest with angular acceleration α=0, the net torque on the beam must be zero about any axis. This holds because torque is related to angular acceleration by τ_net = Iα, so if α=0, then τ_net=0. Choice A is incorrect because while the load creates a clockwise torque, other forces like the cable and hinge provide balancing torques to make the net zero. When analyzing systems in equilibrium, always confirm α=0 to infer that both net force and net torque are zero for complete stability.
A door is held open and does not rotate while two people push on it at different distances from the hinge. The door's angular acceleration is $\alpha=0$ . What must be true?
Answer: The net torque about the hinge is zero. This scenario involves rotational equilibrium of a stationary door. With the door held open and α = 0, Newton's first law for rotation demands zero net torque about the hinge. The two push forces create torques about the hinge that must sum to zero, meaning they produce equal and opposite torques despite potentially different magnitudes and distances. Choice D incorrectly claims stationary objects experience no torques, confusing the absence of rotation with the absence of individual torques. The transferable principle is: rotational equilibrium requires torque balance, which can be achieved through various combinations of forces and lever arms.
A uniform disk on a low-friction axle rotates clockwise at constant angular velocity. Two tangential forces act at different radii, yet the rotation rate stays constant; angular acceleration is zero. Which statement about the disk's net torque about the axle is correct?
Answer: The net torque is zero. This question assesses understanding of rotational equilibrium and Newton's first law in rotation. Newton's first law for rotation states that an object will maintain its angular velocity unless acted upon by a net torque. Since the disk rotates at constant angular velocity, its angular acceleration is zero. Therefore, the net torque about the axle must be zero, as net torque equals moment of inertia times angular acceleration. Choice A is incorrect because constant rotation does not require nonzero net torque; only changing rotation does. To solve similar problems, always check if angular acceleration is zero, which implies net torque is zero regardless of motion.
A uniform rod lies on a rough horizontal table. A student pushes at one end while friction acts along the rod. The rod stays at rest and the angular acceleration is $\alpha=0$ . What must be true?
Answer: The net torque on the rod is zero. This scenario demonstrates rotational equilibrium in a static friction situation. With the rod at rest and α = 0, Newton's first law for rotation demands that the net torque be zero. The push force creates a torque about the rod's center, while friction forces distributed along the rod create opposing torques that exactly balance it. This prevents both translation and rotation of the rod. Choice D incorrectly suggests friction cannot exert torque, when in fact distributed friction forces can create significant torques about a pivot point. The transferable principle is: zero angular acceleration requires all torques to sum to zero, regardless of force types.
A uniform disk on a frictionless axle spins at constant angular velocity. Two tangential forces act at different radii, yet the disk's angular acceleration is explicitly $\alpha=0$ . What can be inferred about the net torque on the disk?
Answer: The net torque about the axle is zero. This question tests understanding of rotational equilibrium and Newton's first law for rotation. When an object has zero angular acceleration ( $\alpha = 0$ ), Newton's second law for rotation states that the net torque must be zero: $\Sigma \tau = I \alpha = 0$ . The disk maintains constant angular velocity, which means it's in rotational equilibrium despite the presence of individual forces creating torques. The key insight is that these individual torques must cancel out to produce zero net torque. Choice C incorrectly assumes that rotation requires nonzero net torque, confusing constant rotation with changing rotation. The transferable strategy is: zero angular acceleration always means zero net torque, regardless of whether the object is rotating or at rest.
A uniform ladder rests against a frictionless wall and a rough floor. It remains stationary without slipping, and angular acceleration is zero. Considering torques about the ladder's center, what is the net torque on the ladder?
Answer: The net torque is zero. This question assesses understanding of rotational equilibrium and Newton's first law in rotation. The ladder is stationary with zero angular acceleration about its center. Newton's first law for rotation requires zero net torque for zero angular acceleration. Forces from the wall, floor, and gravity produce torques that balance overall. Choice A is incorrect because the wall's force contributes to torque balance, not necessarily making net torque nonzero. To solve similar problems, always check if angular acceleration is zero, which implies net torque is zero regardless of motion.
A bicycle wheel is held on an axle while a hand applies a steady torque, and a brake pad provides an opposing torque. The wheel spins at constant angular velocity and $\alpha=0$ . What is true?
Answer: The net torque on the wheel is zero. This scenario illustrates rotational equilibrium during constant angular velocity motion. With the wheel spinning steadily and α = 0, Newton's second law for rotation gives Στ = Iα = 0, meaning zero net torque. The hand's applied torque exactly balances the brake pad's opposing torque, maintaining constant angular speed. This is the rotational analog of pushing an object at constant velocity against friction. Choice A incorrectly concludes the applied torque itself must be zero, rather than recognizing that opposing torques cancel. The key principle is: steady rotation requires balanced torques, not zero individual torques.
A uniform sign is attached to a wall by a hinge at its left edge and a cable from its right edge up to the wall, forming a triangle. The sign remains at rest, and the angular acceleration is explicitly $\alpha=0$ . A student claims the hinge force and cable tension must create equal and opposite forces. Which inference about the net torque on the sign is supported?
Answer: The net torque on the sign is zero. This question assesses understanding of rotational equilibrium and Newton's first law in rotational dynamics. The rotational version of Newton's first law requires zero net torque for constant angular motion or rest. With the sign at rest and α=0, the net torque on the sign is zero. This arises from τ_net = Iα, where zero acceleration confirms zero net torque from all forces. Choice C is incorrect because an angled cable can still contribute to balanced torques without causing net torque. For suspended objects, use the condition of α=0 to directly infer torque equilibrium independent of force directions.
A sign is suspended by two cables from a ceiling and remains motionless. Taking torques about the sign's center, the angular acceleration is $\alpha=0$ . What can be concluded about net torque?
Answer: The net torque on the sign is zero. This problem involves rotational equilibrium of a suspended sign. With the sign motionless and α = 0, Newton's first law for rotation requires zero net torque about any point, including the sign's center. The torques from the two cable tensions and the sign's weight must sum to zero when calculated about the center. This ensures the sign doesn't start rotating. Choice D incorrectly claims weight produces zero torque about any point - torque depends on the perpendicular distance from the line of action to the pivot. The transferable strategy is: for static equilibrium, verify that net torque equals zero about your chosen reference point.
A ceiling fan spins steadily at constant angular velocity. Air drag provides a resistive torque while the motor provides a driving torque, and the fan's angular acceleration is zero. What is the net torque on the fan about its rotation axis?
Answer: It is zero. This question assesses understanding of rotational equilibrium and Newton's first law in rotation. The fan spins at constant angular velocity, indicating zero angular acceleration. Newton's first law for rotation implies that zero angular acceleration means zero net torque about the axis. The motor torque balances the drag torque, producing no net effect. Choice A is incorrect because while the motor provides torque, it is balanced by drag in steady state. To solve similar problems, always check if angular acceleration is zero, which implies net torque is zero regardless of motion.
A door is held open at rest by two forces applied at different points: one student pushes near the knob while another pushes near the hinge. The door does not rotate, and the angular acceleration is $\alpha=0$ . What must be true about the net torque on the door about the hinge axis?
Answer: It is zero. This question assesses understanding of rotational equilibrium and Newton's first law in rotational dynamics. The first law for rotation dictates zero net torque for no change in angular motion. Since the door is at rest with α=0, the net torque about the hinge must be zero. This follows τ_net = Iα, so α=0 ensures τ_net=0 from the balancing pushes. Choice A is incorrect because forces at different radii can still produce equal and opposite torques if magnitudes differ appropriately. When multiple forces act, compute torques relative to the axis and set their sum to zero based on α=0.
A bicycle wheel spins at constant angular velocity on a stand. A brake pad lightly touches the rim while a motor applies a steady driving torque, and the wheel's angular acceleration is zero. What is the net torque on the wheel about its axle?
Answer: It is zero. This question assesses understanding of rotational equilibrium and Newton's first law in rotation. The wheel spins at constant angular velocity, so its angular acceleration is zero. According to Newton's first law for rotation, zero angular acceleration requires zero net torque about the axle. The driving torque from the motor balances the braking torque from friction, resulting in no net torque. Choice D is incorrect because multiple forces can act while still producing zero net torque in equilibrium. To solve similar problems, always check if angular acceleration is zero, which implies net torque is zero regardless of motion.
A ceiling fan rotates at constant angular speed on its vertical shaft. Air resistance exerts a torque opposite the rotation, but the fan's speed does not change, and $\alpha=0$ . Which statement about the net torque on the fan is correct during this steady rotation?
Answer: It is zero because the motor torque balances drag. This question assesses understanding of rotational equilibrium and Newton's first law in rotational dynamics. Newton's first law applied to rotation means constant angular speed requires no net torque. With the fan at constant speed and α=0, the net torque is zero as motor torque balances drag. This is evident from τ_net = Iα, confirming zero net torque when α=0. Choice A is incorrect because drag torque exists but is countered exactly by the motor for equilibrium. In powered rotating systems, use steady-state conditions like α=0 to determine that opposing torques sum to zero.

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