Rotational Equilibrium and Newton's First Law
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AP Physics 1 › Rotational Equilibrium and Newton's First Law
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?
The net torque is zero only if the net force on the stick is zero.
The net torque is nonzero because the stick has weight.
The net torque about the pivot is zero.
The net torque is clockwise because the $90,\text{cm}$ weight is farther from the pivot.
Explanation
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 rod lies horizontally on two supports, one near each end. A $50,\text{N}$ weight is hung somewhere along the rod, and the rod remains at rest without tipping. The angular acceleration is $\alpha=0$. Without calculating forces, what must be true about the net torque on the rod about its center of mass?
It is nonzero unless the $50,\text{N}$ weight is at the center.
It equals the rod’s weight times half its length.
It is nonzero because the supports exert upward forces.
It is zero.
Explanation
This question assesses understanding of rotational equilibrium and Newton's first law in rotational dynamics. Newton's first law for rotation ensures no change in angular velocity without net torque. Given the rod is at rest with α=0, the net torque about its center of mass must be zero. This is supported by τ_net = Iα, so α=0 directly implies τ_net=0. Choice B is incorrect because upward support forces create torques that balance with the weight's torque. Always choose a convenient axis like the center of mass and apply the zero net torque condition when α=0.
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?
It equals the torque from the larger force only.
It is zero only if the net force on the door is zero.
It is zero.
It is nonzero because two forces act at different radii.
Explanation
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 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?
Each push produces zero torque because the door is not moving.
The net torque about the hinge is zero.
The larger force must be applied farther from the hinge.
The net force on the door is zero, so torque cannot exist.
Explanation
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 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?
Equal to $I\omega$ because the turntable is rotating.
Nonzero, because constant speed requires a constant net torque.
Zero, because the torques must sum to zero.
Zero only if the net force on the turntable is also zero.
Explanation
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 ceiling fan rotates steadily at constant angular speed while air resistance exerts a drag torque. The fan’s angular acceleration is $\alpha=0$. What is the net torque on the fan?
It is nonzero because the fan is rotating.
It must point upward along the rotation axis.
It is zero because the motor torque balances the drag torque.
It equals the drag torque because drag always dominates.
Explanation
This question addresses rotational equilibrium in a steady-state rotating system. Despite the fan's continuous rotation at constant angular speed, α = 0 means the net torque must be zero according to Στ = Iα. The motor provides a driving torque while air resistance creates an equal and opposite drag torque, resulting in zero net torque. This is analogous to an object moving at constant velocity under balanced forces. Choice C incorrectly assumes rotation requires net torque, confusing motion with change in motion. The key insight is: constant angular velocity (even if nonzero) means zero net torque, just as constant linear velocity means zero net force.
A Ferris wheel rotates at constant angular velocity. The motor provides a driving torque while resistive torques oppose motion. The wheel’s angular acceleration is $\alpha=0$. What follows?
The net torque on the wheel is zero.
The net torque must point in the direction of rotation.
The motor torque must be zero since the wheel is not speeding up.
Rotational equilibrium requires the net force on the wheel to be zero.
Explanation
This question addresses rotational equilibrium in a large rotating system. Despite the Ferris wheel's continuous rotation at constant angular velocity, α = 0 means the net torque must be zero according to Στ = Iα. The motor's driving torque exactly balances all resistive torques from friction and air resistance. This maintains steady rotation without angular acceleration. Choice C incorrectly assumes net torque must align with rotation direction, confusing torque (which changes rotation) with the rotation itself. The key strategy is: constant angular velocity always indicates zero net torque, regardless of the rotation speed.
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?
The net torque on the wheel is zero.
The applied torque must be zero because $\alpha=0$.
The brake pad cannot exert a torque since it is a contact force.
The net force on the wheel must be zero, so torques are irrelevant.
Explanation
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 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?
The weight of the sign produces zero torque about any point.
Each cable must have the same tension.
The net torque on the sign is zero.
The net force must be nonzero to prevent rotation.
Explanation
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 uniform beam is supported at two points and remains at rest under its own weight plus a hanging load. The angular acceleration is $\alpha=0$. Which inference is valid?
The support forces must be equal in magnitude.
The net torque on the beam is zero.
The hanging load exerts no torque because it is vertical.
If the net force is zero, the net torque must be nonzero.
Explanation
This question tests understanding of rotational equilibrium in beam statics. Since the beam remains at rest with α = 0, the net torque must be zero according to Newton's first law for rotation. All torques from the support forces, beam weight, and hanging load must sum to zero about any chosen pivot point. This is a fundamental requirement for static equilibrium of extended objects. Choice C incorrectly suggests vertical forces exert no torque, ignoring that torque depends on the perpendicular distance to the pivot point. The transferable strategy is: in static equilibrium, both net force and net torque must equal zero.