AP Physics 1 Quiz: Rotational Equilibrium And Newtons First Law

What this quiz covers

This quiz focuses on Rotational Equilibrium And Newtons First Law, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. The net torque is clockwise because the $90\,\text{cm}$ weight is farther from the pivot.
2. The net torque about the pivot is zero. (correct answer)
3. The net torque is zero only if the net force on the stick is zero.
4. The net torque is nonzero because the stick has weight.

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.

Question 2

1. It is zero. (correct answer)
2. It is nonzero because the supports exert upward forces.
3. It is nonzero unless the $50\,\text{N}$ weight is at the center.
4. It equals the rod’s weight times half its length.

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.

Question 3

1. It is nonzero because two forces act at different radii.
2. It is zero. (correct answer)
3. It is zero only if the net force on the door is zero.
4. It equals the torque from the larger force only.

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.

Question 4

1. The larger force must be applied farther from the hinge.
2. The net torque about the hinge is zero. (correct answer)
3. The net force on the door is zero, so torque cannot exist.
4. Each push produces zero torque because the door is not moving.

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.

Question 5

1. Nonzero, because constant speed requires a constant net torque.
2. Zero, because the torques must sum to zero. (correct answer)
3. Zero only if the net force on the turntable is also zero.
4. Equal to $I\omega$ because the turntable is rotating.

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.

Question 6

1. It equals the drag torque because drag always dominates.
2. It is zero because the motor torque balances the drag torque. (correct answer)
3. It is nonzero because the fan is rotating.
4. It must point upward along the rotation axis.

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.

Question 7

1. The applied torque must be zero because $\alpha=0$.
2. The brake pad cannot exert a torque since it is a contact force.
3. The net torque on the wheel is zero. (correct answer)
4. 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.

Question 8

1. The net torque is nonzero because the forces are different.
2. The net torque must be zero. (correct answer)
3. The net torque equals the larger force times the pulley radius.
4. The net torque is zero only if the rope tensions are equal.

Explanation: 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.

Question 9

1. The net torque is zero. (correct answer)
2. The net torque equals the rod’s weight times half its length.
3. The net torque is nonzero because two forces act at different points.
4. The net torque is zero only if both upward forces are equal to the weight.

Explanation: 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.

Question 10

1. The net torque on the sign is zero. (correct answer)
2. Each cable must have the same tension.
3. The net force must be nonzero to prevent rotation.
4. The weight of the sign produces zero torque about any point.

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.

Question 11

1. The net torque on the wheel is zero. (correct answer)
2. The motor torque must be zero since the wheel is not speeding up.
3. The net torque must point in the direction of rotation.
4. 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.

Question 12

1. The net torque on the beam is zero. (correct answer)
2. The support forces must be equal in magnitude.
3. The hanging load exerts no torque because it is vertical.
4. 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.

Question 13

1. The net torque is nonzero because the load creates a clockwise torque.
2. The net torque about any axis is zero because $\alpha=0$. (correct answer)
3. The net torque is zero only if the net force is zero.
4. The net torque is zero only about the hinge, not about other points.

Explanation: 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.

Question 14

1. The net torque on the sign is zero. (correct answer)
2. The net torque is nonzero unless the hinge force equals the sign’s weight.
3. The net torque is nonzero because the cable is angled.
4. The net torque is zero only if the net force is not zero.

Explanation: 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.

Question 15

1. The net torque is nonzero because the wall exerts only a horizontal force.
2. The net torque is zero. (correct answer)
3. The net torque is zero only if the ladder’s center of mass is at the midpoint.
4. The net torque is zero only if the normal forces at wall and floor are equal.

Explanation: 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.

Question 16

1. It is nonzero because the push is applied far from the hinge.
2. It is zero. (correct answer)
3. It equals the push force regardless of where it is applied.
4. It must be nonzero unless the hinge forces sum to zero.

Explanation: This question assesses understanding of rotational equilibrium and Newton's first law in rotation. The door is at rest with zero angular acceleration about the hinge. Newton's first law for rotation states that zero angular acceleration implies zero net torque. The push and hinge forces produce torques that cancel each other. Choice A is incorrect because while the push is far from the hinge, the hinge's reaction torque balances it in equilibrium. To solve similar problems, always check if angular acceleration is zero, which implies net torque is zero regardless of motion.

Question 17

1. It is nonzero because both forces point downward.
2. It is zero. (correct answer)
3. It equals $2\,\text{N}\cdot 0.40\,\text{m}$ regardless of the other force.
4. It is zero only if the net force on the stick is zero.

Explanation: This question assesses understanding of rotational equilibrium and Newton's first law in rotation. The meterstick is at rest with zero angular acceleration about the pivot. Newton's first law for rotation requires that zero angular acceleration corresponds to zero net torque. The torques from the two downward forces balance each other out. Choice A is incorrect because forces in the same direction can still produce torques that cancel if at appropriate distances. To solve similar problems, always check if angular acceleration is zero, which implies net torque is zero regardless of motion.

Question 18

1. The net torque is nonzero because the wall exerts a force.
2. The net torque is zero. (correct answer)
3. The net torque must point toward the floor due to friction.
4. The net torque equals the normal force at the floor times the ladder length.

Explanation: 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.

Question 19

1. It equals the motor torque because the fan is spinning.
2. It equals the drag torque because drag is always present.
3. It is zero. (correct answer)
4. It is nonzero unless the net force on the fan is zero.

Explanation: 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.

Question 20

1. It is nonzero because drag provides a torque.
2. It is zero because the motor torque balances drag. (correct answer)
3. It is zero only if there is no drag torque.
4. It equals the fan’s angular momentum divided by time.

Explanation: 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.