Rotational Equilibrium and Newton's First Law - AP Physics C Mechanics

A seesaw pivots at its center and stays level while two students sit on opposite sides. A $300\,\text{N}$ student sits $1.2\,\text{m}$ to the left of the pivot, and a second student sits to the right with an unknown weight $W$ at a distance $0.80\,\text{m}$ . The forces act vertically downward at their seats, creating torques with lever arms measured from the pivot. The support force at the pivot produces no torque about the pivot. The seesaw is motionless, so Newton's First Law in rotational form requires $\sum \tau_{\text{pivot}}=0$ . Considering the forces acting on the seesaw, calculate the force needed at a specific point to maintain equilibrium.
Answer: $W=450\,\text{N}$ , since $300(1.2)=W(0.80)$ . This question tests understanding of rotational equilibrium and Newton's First Law in rotational form in AP Physics C: Mechanics. Rotational equilibrium occurs when the sum of torques acting on an object is zero, meaning the object is not accelerating rotationally. Newton's First Law in rotational form states that an object at rest will remain so unless acted upon by a net external torque. In this scenario, we have a seesaw with two students sitting on opposite sides of a central pivot, where we need to find the unknown weight to maintain balance. Choice B is correct because applying torque balance about the pivot: clockwise torque = counterclockwise torque, so 300 N × 1.2 m = W × 0.80 m, giving W = 360/0.80 = 450 N. Choice A is incorrect because it reverses the lever arms in the calculation, using 300(0.80) = W(1.2) instead of the correct relationship. To help students: Draw a clear diagram showing the pivot, forces, and lever arms. Emphasize that torque = force × perpendicular distance from pivot, and practice setting up the torque balance equation systematically with proper signs for clockwise vs counterclockwise torques.
A uniform disk of mass $M$ and radius $R$ is pivoted at a point on its rim. A horizontal force $F$ is applied at the opposite point on the rim. For the disk to remain in rotational equilibrium, what must be the relationship between $F$ and the other parameters?
Answer: $F = \frac{Mg}{2}$ The disk is pivoted at a point on its rim, so the center of mass is at distance $R$ from the pivot. The gravitational torque about the pivot is $\tau_g = MgR$ (taking clockwise as positive). The applied force $F$ acts at the opposite point on the rim, which is at distance $2R$ from the pivot, creating torque $\tau_F = F(2R)$ in the opposite direction. For equilibrium: $MgR = F(2R)$ , so $F = \frac{Mg}{2}$ . Choice B assumes the lever arm for $F$ is $R$ . Choice C uses incorrect geometry. Choice D assumes $F$ acts at distance $R/2$ .
The rotational analog of Newton's first law of motion implies that an object's state of rotation will not change unless...
Answer: ...a net external torque is applied to it. Newton's first law for rotation states that a rigid body will remain at rest or rotate with a constant angular velocity unless acted upon by a net external torque. Torque is the rotational analog of force, and a net torque causes a change in the state of rotation (i.e., an angular acceleration).
At which mark must a second mass $m_2 = 3M$ be placed to keep the meter stick in static equilibrium?
Answer: the 70 cm mark For static equilibrium, the net torque about the pivot must be zero. Let the pivot be the origin (50 cm). The torque from $m_1$ is counter-clockwise: $\tau_1 = m_1 g r_1 = (2M)g(50-20\text{ cm})$ . The torque from $m_2$ is clockwise: $\tau_2 = -m_2 g r_2 = -(3M)g(x-50\text{ cm})$ . Setting the sum of torques to zero: $(2M)g(30) - (3M)g(x-50) = 0$ . This simplifies to $60 = 3(x-50)$ , so $20 = x-50$ , and $x = 70$ cm.
Which of the following statements best describes the net torque on the rotor?
Answer: The net torque is zero because the angular velocity is constant. According to Newton's first law for rotation, an object will maintain a constant angular velocity if and only if the net external torque acting on it is zero. The engine provides a forward torque, which is exactly balanced by the resistive torque from air drag, resulting in a net torque of zero and thus a constant angular velocity.
For the rod to be in rotational equilibrium, what must be the relationship between the applied force $F$ and the weight $W$ ?
Answer: $F L \sin\theta = W \frac{L}{2} \cos\theta$ To be in rotational equilibrium, the net torque about the hinge must be zero. The weight $W$ acts at the center of the rod ( $L/2$ ) and its lever arm is $(L/2)\cos\theta$ , creating a clockwise torque. The force $F$ acts at the end ( $L$ ) and its lever arm is $L\sin\theta$ , creating a counter-clockwise torque. Setting the magnitudes of these torques equal: $F(L\sin\theta) = W(\frac{L}{2}\cos\theta)$ .
Consider a wheel free to rotate about a fixed axle. For the wheel to be in a state of dynamic rotational equilibrium, which condition must be satisfied?
Answer: The net torque on the wheel must be zero, and its angular velocity must be constant and non-zero. Dynamic rotational equilibrium is a state where the object is rotating but its angular velocity is constant (and not zero). For the angular velocity to be constant, the angular acceleration must be zero. According to Newton's laws for rotation, zero angular acceleration implies that the net torque acting on the wheel is zero.
A uniform disk of radius $R$ and mass $M$ is mounted on a frictionless axle through its center. A string is wound around the disk's edge and a mass $m$ hangs from the string. The system is released from rest but remains stationary. Which condition must be satisfied?
Answer: The string tension must be zero and the mass $m$ must be supported by another force If the system remains stationary after release, both translational and rotational equilibrium must be maintained. For the hanging mass: $\sum F = 0$ requires the weight $mg$ to be balanced by another upward force (not string tension, since that would create torque on the disk). For the disk: $\sum \tau = 0$ requires no net torque, so string tension must be zero. Choice A describes motion, not equilibrium. Choice B incorrectly assumes non-zero string tension. Choice C is unnecessarily restrictive.
Where must the pivot be placed for the bar to be in static equilibrium?
Answer: At a distance of $L/4$ from End A. For the bar to be in static equilibrium under its own weight, the pivot must provide a normal force that exactly opposes the gravitational force, and there must be zero net torque. To achieve zero net torque from gravity, the pivot must be placed directly under the center of mass. This way, the lever arm for the gravitational force is zero, resulting in zero torque.
When analyzing a rigid body in static equilibrium, the net torque is calculated to be zero. This conclusion is valid if the torques are summed about which of the following points?
Answer: Any point, whether it is on or off the body. A key principle of static equilibrium is that if the net force on an object is zero and the net torque about one point is zero, then the net torque about any other point is also zero. Therefore, for convenience, one can choose any point as the axis of rotation when applying the condition $\sum \tau = 0$ .
Which of the following motions is impossible for the body?
Answer: The body is moving with a constant non-zero linear acceleration. If the vector sum of external forces is zero ( $\sum F = 0$ ), the body is in translational equilibrium. This means its linear acceleration must be zero ( $\sum F = ma$ ). Therefore, it is impossible for the body to have a constant non-zero linear acceleration. It can be at rest, move with constant velocity, or rotate, as long as the net torque condition is also met for that state of rotational motion.
A 1.8 m uniform beam is hinged at the left end and supports a 180 N sign at the right end. The beam's weight is 90 N acting at its center. A cable attaches to the beam 1.8 m from the hinge and pulls at an angle $\theta$ above the beam, providing an upward component that creates counterclockwise torque. The beam remains horizontal, so clockwise and counterclockwise torques must balance. With zero angular acceleration, Newton's First Law in rotational form requires $\sum \tau_{\text{hinge}}=0$ . Based on the system described, which change increases the counterclockwise torque from the cable without changing the cable tension magnitude?
Answer: Increase $\theta$ so $T\sin\theta$ increases. This question tests understanding of rotational equilibrium and Newton's First Law in rotational form in AP Physics C: Mechanics. Rotational equilibrium requires the net torque about any point to be zero, with torque depending on both the force magnitude and its perpendicular distance from the pivot. In this scenario, the cable creates counterclockwise torque through its vertical component T sin(θ), which must balance the clockwise torques from the beam and sign weights. Choice A is correct because increasing θ increases sin(θ), thereby increasing the vertical component of the tension and its torque without changing the tension magnitude T. Choice B is incorrect because cos(θ) relates to the horizontal component which doesn't contribute to torque about a horizontal axis, choice C would increase clockwise torque making balance harder, and choice D incorrectly references moment of inertia which doesn't affect static torque. To help students: Use vector decomposition to identify which force components create torque, remember that torque depends on the perpendicular distance between the force line and pivot, and practice varying one parameter while holding others constant to see the effect.
Which of the following statements must be true for the object to be in rotational equilibrium?
Answer: The object's center of mass must be located directly below the pivot point. For the object to be in rotational equilibrium (hang at rest), the net torque about the pivot must be zero. The pivot force exerts no torque about the pivot point itself. The only other force is gravity, which acts at the center of mass. For the torque due to gravity to be zero, its lever arm must be zero. This occurs when the center of mass is on the vertical line passing through the pivot point, i.e., directly above or below it. For stable equilibrium, it must be directly below.
A uniform beam of mass $M$ and length $L$ is supported by two vertical cables at its ends. The beam carries a point load $W$ at distance $a$ from the left end. If the tension in the left cable is $T_L$ and in the right cable is $T_R$ , which relationship must hold for rotational equilibrium about the left end?
Answer: $T_R L = Wa + Mg\frac{L}{2}$ Taking torques about the left end (clockwise positive): The weight $W$ creates torque $Wa$ , the beam's weight $Mg$ creates torque $Mg(L/2)$ , and the right cable tension creates torque $-T_R L$ . For equilibrium: $Wa + Mg\frac{L}{2} - T_R L = 0$ , so $T_R L = Wa + Mg\frac{L}{2}$ . Choice B incorrectly subtracts the beam weight. Choice C uses $(L-a)$ which would be correct if taking torques about the right end. Choice D combines both errors.
A horizontal beam of length $4.0\,\text{m}$ is held in static equilibrium by two vertical support cables at its ends. A $300\,\text{N}$ load hangs $1.0\,\text{m}$ from the left end, and the beam's own weight is negligible. The upward tensions $T_L$ and $T_R$ act at the left and right ends, respectively, and the load's weight acts downward at its attachment point. Taking torques about the left end, the lever arm for $T_R$ is $4.0\,\text{m}$ and for the load is $1.0\,\text{m}$ . The beam does not rotate, so Newton's First Law in rotational form implies $\sum \tau_{\text{left}}=0$ . Considering the forces acting on the object, calculate the force needed at a specific point to maintain equilibrium.
Answer: $T_R=75\,\text{N}$ upward, from $T_R(4.0)=300(1.0)$ . This question tests understanding of rotational equilibrium and Newton's First Law in rotational form in AP Physics C: Mechanics. Rotational equilibrium occurs when the sum of torques acting on an object is zero, meaning the object is not accelerating rotationally. Newton's First Law in rotational form states that an object at rest will remain so unless acted upon by a net external torque. In this scenario, we have a horizontal beam with a load closer to the left support, requiring us to find the right support tension using torque balance. Choice A is correct because taking torques about the left end: clockwise torque from load = counterclockwise torque from right tension, so 300 N × 1.0 m = TR × 4.0 m, giving TR = 300/4.0 = 75 N upward. Choice C is incorrect because it assumes equal sharing of the load, ignoring that the load is closer to the left support, which must carry more of the weight. To help students: Demonstrate that the support closer to the load carries more weight. Practice choosing convenient pivot points (like one support) to eliminate unknown forces from the torque equation.
A rigid body can be in translational equilibrium without being in rotational equilibrium. Which of the following situations is an example of this?
Answer: A pair of equal and opposite, non-collinear forces acting on a stationary rod. A pair of equal and opposite forces (a couple) results in a zero net force, so the body is in translational equilibrium (its center of mass will not accelerate). However, because the forces are non-collinear, they produce a net torque, causing an angular acceleration. Therefore, the body is not in rotational equilibrium.
A seesaw is in rotational equilibrium about a frictionless pivot, remaining horizontal. A 500 N person sits 1.0 m to the left of the pivot, and a 250 N person sits 2.0 m to the right. Both forces act downward, and their lever arms are measured from the pivot along the plank. The plank's weight acts at the pivot and produces negligible torque. The torques balance so that clockwise torque equals counterclockwise torque, consistent with Newton's First Law in rotational form ( $\sum \tau=0$ ). Based on the system described, which condition must be met for the system to remain in rotational equilibrium?
Answer: $\sum \tau_{\text{pivot}}=0$ and $\sum F=0$ This question tests understanding of rotational equilibrium and Newton's First Law in rotational form in AP Physics C: Mechanics. Rotational equilibrium requires that both the net torque and net force on an object equal zero, ensuring no linear or angular acceleration. In this scenario, the seesaw experiences torques from two people: 500 N × 1.0 m = 500 N·m counterclockwise and 250 N × 2.0 m = 500 N·m clockwise, which balance perfectly. Choice A is correct because for complete static equilibrium, both conditions must be satisfied: Στ = 0 (no angular acceleration) and ΣF = 0 (no linear acceleration). Choice B is incorrect because torque balance is essential for rotational equilibrium, choice C incorrectly applies the rotational dynamics equation to a static situation, and choice D contradicts the equilibrium condition. To help students: Emphasize that static equilibrium requires two separate conditions to be met simultaneously, practice identifying all forces and their points of application, and reinforce that torque depends on both force magnitude and lever arm distance.
A uniform beam of mass $M$ and length $L$ is pivoted at its center. Two forces are applied: a force $F_1 = 20\text{ N}$ downward at a distance $L/3$ from the pivot, and a force $F_2$ upward at a distance $L/4$ from the pivot on the opposite side. If the beam remains in rotational equilibrium, what is the magnitude of $F_2$ ?
Answer: $26.7\text{ N}$ For rotational equilibrium, the net torque about the pivot must be zero. Taking counterclockwise as positive: $\tau_{net} = F_2 \cdot \frac{L}{4} - F_1 \cdot \frac{L}{3} = 0$ . Substituting values: $F_2 \cdot \frac{L}{4} = 20 \cdot \frac{L}{3}$ . Solving: $F_2 = \frac{20 \cdot 4}{3} = \frac{80}{3} = 26.7\text{ N}$ . Choice A uses $F_2 = F_1 \cdot \frac{3}{4}$ incorrectly. Choice C assumes equal lever arms. Choice D multiplies instead of dividing by the ratio.
A uniform $3.0\,\text{m}$ beam supports a hanging sign and is pinned to a wall at the left end, remaining horizontal. The beam's weight is $120\,\text{N}$ acting at its center, and the sign's weight is $200\,\text{N}$ acting at the right end. A cable attaches to the right end and makes a $40^\circ$ angle above the beam, exerting tension $T$ along its force vector. Taking torques about the wall pin, the pin's reaction forces create no torque, while the weights create clockwise torque balanced by the cable's counterclockwise torque. The system is static, so Newton's First Law in rotational form requires $\sum \tau_{\text{pin}}=0$ . Considering the forces acting on the object, calculate the force needed at a specific point to maintain equilibrium.
Answer: $T=\dfrac{(120)(1.5)+(200)(3.0)}{3.0\sin 40^\circ}=402\,\text{N}$ . This question tests understanding of rotational equilibrium and Newton's First Law in rotational form in AP Physics C: Mechanics. Rotational equilibrium occurs when the sum of torques acting on an object is zero, meaning the object is not accelerating rotationally. Newton's First Law in rotational form states that an object at rest will remain so unless acted upon by a net external torque. In this scenario, we have a beam pinned to a wall with its weight and a sign's weight creating clockwise torques that must be balanced by the cable's tension. Choice A is correct because the vertical component of tension (T sin 40°) creates the counterclockwise torque: T sin 40° × 3.0 m = 120 N × 1.5 m + 200 N × 3.0 m, giving T = 780/(3.0 sin 40°) = 402 N. Choice B is incorrect because it uses cos 40° instead of sin 40° for the vertical component of tension that creates the torque. To help students: Draw force diagrams showing tension components and emphasize that only the perpendicular component of a force contributes to torque. Practice decomposing angled forces and identifying which component creates torque about the chosen pivot.
A balanced mobile uses a light horizontal bar pivoted at its center, staying at rest. A 2.0 N weight hangs 0.40 m left of the pivot, and a 1.0 N weight hangs 0.10 m right of the pivot. A third weight of 1.5 N is moved along the right side to reestablish equilibrium. Each weight exerts a downward force vector, and each produces torque equal to $rF$ about the pivot. The system is in rotational equilibrium, so the sum of clockwise and counterclockwise torques is zero, consistent with Newton's First Law in rotational form. Considering the forces acting on the object, where should the 1.5 N weight be placed (distance from the pivot on the right) to maintain equilibrium?
Answer: $0.47\ \text{m}$ to the right of the pivot This question tests understanding of rotational equilibrium and Newton's First Law in rotational form in AP Physics C: Mechanics. Rotational equilibrium occurs when the net torque about the pivot is zero, meaning clockwise and counterclockwise torques balance perfectly. In this scenario, the 2.0 N weight creates counterclockwise torque (2.0 × 0.40 = 0.80 N·m), while the 1.0 N weight creates clockwise torque (1.0 × 0.10 = 0.10 N·m), leaving a net counterclockwise torque of 0.70 N·m that must be balanced by the 1.5 N weight. Choice A is correct because setting up the balance equation: 0.80 = 0.10 + 1.5 × d, where d is the distance right of the pivot, gives 1.5d = 0.70, so d = 0.70/1.5 = 0.467 m ≈ 0.47 m. Choice C (0.07 m) might result from arithmetic errors, while choice D incorrectly places the weight on the left side. To help students: Create a torque inventory table listing each force, its lever arm, and resulting torque with proper signs, always verify that the sum of all torques equals zero in the final configuration, and use dimensional analysis to ensure the answer has correct units.

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