Torque and Work - AP Physics C Mechanics

1. A beam of negligible mass is pivoted at its left end and lies initially horizontal. A weight $W_1=30\,\text{N}$ hangs at $r_1=0.40\,\text{m}$ from the pivot, and a second weight $W_2=20\,\text{N}$ hangs at $r_2=0.70\,\text{m}$ from the pivot, both on the same side. The weights act vertically downward, and the beam rotates in a vertical plane. At the instant shown, the beam remains horizontal so each weight produces a moment arm equal to its distance from the pivot. Ignore friction at the pivot and any other forces. The net torque about the pivot is the sum of the individual torques from the two weights. Using the given information, compute the net torque magnitude about the pivot at this instant.
   Answer: $\tau=26\,\text{N\,m}$ This question tests AP Physics C: Mechanics, specifically the concepts of torque and work in rotating systems. Torque is the rotational equivalent of force, calculated as τ = rFsinθ, where r is the distance from the pivot, F is the force, and θ is the angle between force and lever arm. Work in rotation is given by W = τθ, where θ is the angular displacement. In this scenario, two weights hang from a horizontal beam, creating torques that must be summed. Choice A is correct because τ₁ = r₁W₁ = (0.40 m)(30 N) = 12 N·m and τ₂ = r₂W₂ = (0.70 m)(20 N) = 14 N·m, giving a net torque of 12 + 14 = 26 N·m (both rotate the beam in the same direction). Choice C incorrectly calculates only half the total torque. To help students, emphasize that torques add algebraically when they act about the same axis. Practice with multiple forces at different positions, ensuring students understand sign conventions for clockwise vs. counterclockwise torques.
2. A disc of radius $r=0.25\,\text{m}$ rotates about its center on a frictionless axle. A constant tangential force $F=16\,\text{N}$ is applied at the rim so that the force stays perpendicular to the radius. The disc rotates through an angular displacement of $\Delta\theta=\pi\,\text{rad}$. Ignore any losses so all work is due to the applied torque. The torque magnitude is constant and equals $\tau=rF$. The rotational work done is $W=\tau\Delta\theta$. Using the given information, how much work is done by the force during this rotation?
   Answer: $W=4\pi\,\text{J}$ This question tests AP Physics C: Mechanics, specifically the concepts of torque and work in rotating systems. Torque is the rotational equivalent of force, calculated as τ = rFsinθ, where r is the distance from the pivot, F is the force, and θ is the angle between force and lever arm. Work in rotation is given by W = τθ, where θ is the angular displacement. In this scenario, a tangential force of 16 N at radius 0.25 m rotates a disc through π rad. Choice A is correct because the torque is τ = rF = (0.25 m)(16 N) = 4.0 N·m, and the work is W = τΔθ = (4.0 N·m)(π rad) = 4π J. Choice D incorrectly gives units of N·m instead of J, confusing torque with work. To help students, practice problems using π in angular measurements, ensuring comfort with symbolic answers. Emphasize dimensional analysis: torque (N·m) × angle (rad) = work (J).
3. A flywheel with rotational inertia $I$ is rotating with initial angular velocity $\omega_0$. A net work $W$ is done on the flywheel, bringing it to rest. Which of the following expresses the work $W$ done?
   Answer: $W = -\frac{1}{2}I\omega_0^2$ According to the work-energy theorem for rotation, the net work done is equal to the change in rotational kinetic energy: $W = \Delta K_{rot} = K_f - K_i$. Since the flywheel is brought to rest, its final kinetic energy $K_f$ is zero. The initial kinetic energy is $K_i = \frac{1}{2}I\omega_0^2$. Therefore, $W = 0 - \frac{1}{2}I\omega_0^2 = -\frac{1}{2}I\omega_0^2$.
4. A flywheel experiences a constant torque due to a motor. The motor applies a tangential force $F=80\,\text{N}$ at a radius $r=0.15\,\text{m}$ from the axis, and the force remains perpendicular to the radius. The flywheel starts from rest and rotates without friction. The motor continues to apply the same torque until the flywheel has rotated through $\Delta\theta=10\,\text{rad}$. Treat the torque as constant during this interval. The work done by a constant torque over an angular displacement is $W=\tau\Delta\theta$. Using the given information, determine the work done by the motor during the $10\,\text{rad}$ rotation.
   Answer: $W=120\,\text{J}$ This question tests AP Physics C: Mechanics, specifically the concepts of torque and work in rotating systems. Torque is the rotational equivalent of force, calculated as τ = rFsinθ, where r is the distance from the pivot, F is the force, and θ is the angle between force and lever arm. Work in rotation is given by W = τθ, where θ is the angular displacement. In this scenario, a tangential force of 80 N at radius 0.15 m rotates a flywheel through 10 rad. Choice A is correct because the torque is τ = rF = (0.15 m)(80 N) = 12 N·m (tangential means sin90° = 1), and the work is W = τΔθ = (12 N·m)(10 rad) = 120 J. Choice C incorrectly calculates only half the work, possibly from an arithmetic error. To help students, practice problems with various angular displacements, emphasizing that work accumulates over the entire rotation. Reinforce unit consistency: rad × N·m = J.
5. A rigid body with rotational inertia $I$ is initially at rest. A net work $W$ is done on the body, causing it to rotate. What is the final angular velocity $\omega$ of the body?
   Answer: $\sqrt{2W/I}$ The rotational work-energy theorem states that the net work done on a rigid body is equal to its change in rotational kinetic energy, $W_{net} = \Delta K_{rot}$. Since the body starts from rest, $\Delta K_{rot} = K_{rot, final} = \frac{1}{2}I\omega^2$. Setting $W = \frac{1}{2}I\omega^2$ and solving for $\omega$ gives $\omega = \sqrt{2W/I}$.
6. A pulley of radius $r=0.20\,\text{m}$ is turned by pulling on a light rope wrapped around its rim. A constant tangential force $F=50\,\text{N}$ is applied to the rope so the force stays perpendicular to the radius. The axle is frictionless, and the pulley rotates in its plane. The force is applied while the pulley turns through an angle $\Delta\theta=2.5\,\text{rad}$. Assume the rope does not slip on the rim. The torque from the pull is constant, so rotational work is $W=\tau\Delta\theta$. Using the given information, calculate the total work done by the applied force on the pulley during this rotation.
   Answer: $W=25\,\text{J}$ This question tests AP Physics C: Mechanics, specifically the concepts of torque and work in rotating systems. Torque is the rotational equivalent of force, calculated as τ = rFsinθ, where r is the distance from the pivot, F is the force, and θ is the angle between force and lever arm. Work in rotation is given by W = τθ, where θ is the angular displacement. In this scenario, a tangential force of 50 N is applied to a pulley of radius 0.20 m through an angle of 2.5 rad. Choice A is correct because the torque is τ = rF = (0.20 m)(50 N) = 10 N·m (tangential force means sin90° = 1), and the work is W = τΔθ = (10 N·m)(2.5 rad) = 25 J. Choice D incorrectly gives units of N·m instead of J, showing confusion between torque and work units. To help students, stress that work always has units of energy (joules), while torque has units of N·m. Practice dimensional analysis to verify that τΔθ gives units of energy.
7. A uniform thin rod of mass $M$ and length $L$ is pivoted at one end. It is released from rest in a horizontal position and swings downward. What is the work done by the gravitational force on the rod as it swings to its lowest point, the vertical position?
   Answer: $\frac{1}{2}MgL$ The work done by gravity depends on the vertical displacement of the center of mass. For a uniform rod, the center of mass is at its geometric center, a distance of $L/2$ from the pivot. As the rod swings from horizontal to vertical, its center of mass drops by a vertical distance of $h = L/2$. The work done by gravity is $W_g = Mgh = Mg(L/2)$.
8. A string is wrapped around a solid cylinder of radius $R$. A person pulls the string with a constant force $F$, causing the cylinder to rotate about its central axis. The force is always tangential to the cylinder's surface. What is the work done by the force $F$ when a length $L$ of string has been unwound?
   Answer: $FL$ The work done by a constant force is the product of the force and the distance over which it is applied. The force $F$ is applied over the length $L$ of the string that is unwound. Thus, $W = F \cdot L$. Alternatively, the torque is $\tau = FR$ and the angle is $\Delta\theta = L/R$. The work is $W = \tau\Delta\theta = (FR)(L/R) = FL$.
9. The instantaneous power delivered by a motor to a rotating shaft is given by $P(t) = C t^{1/2}$, where $C$ is a constant. What is the work done by the motor during the time interval from $t=0$ to $t=T$?
   Answer: $\frac{2}{3}CT^{3/2}$ Power is the rate at which work is done, $P = dW/dt$. To find the total work done, we must integrate the power function with respect to time: $W = \int_{0}^{T} P(t) dt = \int_{0}^{T} C t^{1/2} dt = C [\frac{t^{3/2}}{3/2}]_{0}^{T} = \frac{2}{3}CT^{3/2}$.
10. A uniform sphere of mass $M$ and radius $R$ rolls without slipping down an incline of angle $\theta$. At the same time, a tangential force $F$ parallel to the incline acts on the sphere at its topmost point. After the sphere has moved a distance $s$ down the incline, what is the total work done by force $F$?
    Answer: $2Fs$ The key insight is that when the sphere rolls distance $s$ down the incline, it rotates through angle $\phi = s/R$. The force $F$ is applied at the top of the sphere, which moves through a more complex path. As the sphere rolls, the point of application of force $F$ moves distance $s$ down the incline (translation) plus an additional distance $R\phi = R(s/R) = s$ due to the rotation of the sphere. Therefore, the total displacement of the point where force is applied is $s + s = 2s$. The work done is $W = F \cdot 2s = 2Fs$. Choice A ignores the rotational component. Choice B incorrectly applies a cosine factor. Choice D includes an unnecessary angular term.
11. A grinding wheel with a rotational inertia of $2.0 \, \text{kg} \cdot \text{m}^2$ slows uniformly from an angular speed of $30 \, \text{rad/s}$ to $10 \, \text{rad/s}$. What is the magnitude of the work done by friction on the wheel?
    Answer: $800 \, \text{J}$ The work done by the net torque (which is the frictional torque here) equals the change in rotational kinetic energy. $W = \Delta K_{rot} = \frac{1}{2}I(\omega_f^2 - \omega_i^2)$. $W = \frac{1}{2}(2.0)((10)^2 - (30)^2) = 1.0(100 - 900) = -800 \, \text{J}$. The magnitude of the work done is $800 \, \text{J}$.
12. A robotic arm rotates about a fixed pivot in a plane. A force of magnitude $F=30\,\text{N}$ is applied at a point $r=0.40\,\text{m}$ from the pivot. At the instant considered, the angle between the radius vector and the force is $\phi=60^\circ$. The pivot is frictionless, and only this applied force produces torque about the pivot. Use the standard torque magnitude formula $\tau=rF\sin\phi$. Assume all quantities are constant at this instant and treat the problem as two-dimensional. Ignore any gravitational torque on the arm. Using the given information, determine the torque produced by the mechanical arm.
    Answer: $\tau=10.4\,\text{N}\cdot\text{m}$ This question tests AP Physics C: Mechanics, specifically the concepts of torque and work in rotating systems. Torque is the rotational equivalent of force, calculated as τ = rFsinφ, where r is the distance from the pivot, F is the force, and φ is the angle between force and lever arm. Work in rotation is given by W = τθ, where θ is the angular displacement. In this scenario, a force of 30 N is applied at 0.40 m from the pivot at an angle of 60° to the radius vector. Choice B is correct because it accurately uses τ = rFsinφ = (0.40)(30)(sin 60°) = (0.40)(30)(0.866) = 10.4 N·m. Choice C is incorrect because it uses sin 30° instead of sin 60°, giving only 6.0 N·m. To help students, emphasize understanding which angle to use in the torque formula - it's the angle between the force vector and the position vector. Practice with various force orientations and ensure students can identify the correct angle for the sine function.
13. A balancing beam of negligible mass is pivoted at one end and initially held horizontal. A single weight of magnitude $W=50\,\text{N}$ hangs from the beam at a distance $r=0.60\,\text{m}$ from the pivot. The weight pulls vertically downward, and the beam rotates in a vertical plane. Consider the instant when the beam is horizontal so the angle between $\vec r$ and the weight is $90^\circ$. Ignore friction at the pivot and assume the weight remains attached at the same point. The torque magnitude from the weight is $\tau=rW\sin 90^\circ$. Using the given information, what is the torque about the pivot due to the hanging weight at this instant?
    Answer: $\tau=30\,\text{N\,m}$ This question tests AP Physics C: Mechanics, specifically the concepts of torque and work in rotating systems. Torque is the rotational equivalent of force, calculated as τ = rFsinθ, where r is the distance from the pivot, F is the force, and θ is the angle between force and lever arm. Work in rotation is given by W = τθ, where θ is the angular displacement. In this scenario, a 50 N weight hangs 0.60 m from a pivot on a horizontal beam, creating maximum torque. Choice A is correct because τ = rWsin(90°) = (0.60 m)(50 N)(1) = 30 N·m, since the weight acts vertically and the beam is horizontal. Choice D incorrectly labels the torque with units of joules instead of N·m, confusing torque with work. To help students, emphasize that hanging weights on horizontal beams always produce maximum torque (sin90° = 1). Practice identifying when forces are perpendicular to position vectors for simplified calculations.
14. What is the net work done on the disk during the first 4.0 seconds?
    Answer: $160 \, \text{J}$ The final angular velocity at $t=4.0$ s is $\omega_f = 8.0 \, \text{rad/s}$. The disk starts from rest, $\omega_i = 0$. The net work done is equal to the change in rotational kinetic energy. $W_{net} = \Delta K_{rot} = \frac{1}{2}I\omega_f^2 = \frac{1}{2}(5.0 \, \text{kg} \cdot \text{m}^2)(8.0 \, \text{rad/s})^2 = 2.5 \times 64 = 160 \, \text{J}$.
15. A pulley of radius $r=0.10\,\text{m}$ is rotated by a constant tangential pull on a rope. The applied force has magnitude $F=60\,\text{N}$ and stays perpendicular to the radius as the pulley turns. The axle is frictionless, and the pulley rotates in its plane. The force is applied while the pulley rotates through $\Delta\theta=4.0\,\text{rad}$. Assume no slipping and no dissipative losses. The torque is constant with magnitude $\tau=rF$. The work done by the torque is $W=\tau\Delta\theta$. Using the given information, calculate the total work done by the applied force.
    Answer: $W=24\,\text{J}$ This question tests AP Physics C: Mechanics, specifically the concepts of torque and work in rotating systems. Torque is the rotational equivalent of force, calculated as τ = rFsinθ, where r is the distance from the pivot, F is the force, and θ is the angle between force and lever arm. Work in rotation is given by W = τθ, where θ is the angular displacement. In this scenario, a tangential force of 60 N at radius 0.10 m rotates a pulley through 4.0 rad. Choice A is correct because the torque is τ = rF = (0.10 m)(60 N) = 6.0 N·m, and the work is W = τΔθ = (6.0 N·m)(4.0 rad) = 24 J. Choice B incorrectly multiplies by an extra factor of 10, possibly from a decimal error. To help students, emphasize careful unit tracking and decimal placement. Practice with small radii to ensure students don't automatically assume meters when given centimeters.
16. A motor applies a constant torque of $40 \, \text{N} \cdot \text{m}$ to a large turntable, causing it to accelerate from rest. How much work is done by the motor in the first 5 revolutions?
    Answer: $400\pi \, \text{J}$ First, convert the angular displacement from revolutions to radians: $\Delta\theta = 5 \, \text{rev} \times \frac{2\pi \, \text{rad}}{1 \, \text{rev}} = 10\pi \, \text{rad}$. Then, calculate the work done by the constant torque: $W = \tau \Delta\theta = (40 \, \text{N} \cdot \text{m})(10\pi \, \text{rad}) = 400\pi \, \text{J}$.
17. A constant force of magnitude $F$ is applied at a distance $R$ from the axis of a rotating disk. The force makes a constant angle of $60^{\circ}$ with the radial line from the axis to the point of application. What is the work done by this force when the disk rotates through an angle $\Delta\theta$?
    Answer: $FR \Delta\theta \sin(60^{\circ})$ The torque is given by $\tau = rF\sin\phi$, where $\phi$ is the angle between the position vector $\vec{r}$ and the force vector $\vec{F}$. Here, $r=R$ and $\phi=60^{\circ}$. So, $\tau = FR\sin(60^{\circ})$. The work done by this constant torque over an angular displacement $\Delta\theta$ is $W = \tau\Delta\theta = FR\Delta\theta\sin(60^{\circ})$.
18. The net torque on an engine component is described by $\tau(\theta) = A e^{-\theta/\theta_0}$, where $A$ and $\theta_0$ are positive constants. What is the work done by this torque as the component rotates from $\theta=0$ to a very large angle ($\theta \rightarrow \infty$)?
    Answer: $A \theta_0$ Work is the integral of torque with respect to angular position: $W = \int_0^\infty A e^{-\theta/\theta_0} d\theta$. The integral of $e^{ax}$ is $\frac{1}{a}e^{ax}$. Here, $a = -1/\theta_0$. So, $W = A [-\theta_0 e^{-\theta/\theta_0}]_0^\infty = -A\theta_0 [e^{-\infty} - e^0] = -A\theta_0 [0 - 1] = A\theta_0$.
19. A robotic arm rotates about a fixed pivot in a horizontal plane. A force of magnitude $F=40\,\text{N}$ is applied at a point $r=0.30\,\text{m}$ from the pivot. The force makes an angle $\phi=60^\circ$ with the arm (the radius vector from the pivot to the point of application). The applied force remains constant in magnitude and direction relative to the arm at that instant. Frictional effects are negligible, and only the torque from this force is considered. Use the standard torque magnitude relation $\tau=rF\sin\phi$. Using the given information, determine the torque produced about the pivot by the applied force.
    Answer: $\tau=10.4\,\text{N\,m}$ This question tests AP Physics C: Mechanics, specifically the concepts of torque and work in rotating systems. Torque is the rotational equivalent of force, calculated as τ = rFsinθ, where r is the distance from the pivot, F is the force, and θ is the angle between force and lever arm. Work in rotation is given by W = τθ, where θ is the angular displacement. In this scenario, a 40 N force is applied at 0.30 m from the pivot at a 60° angle to the arm, requiring calculation of torque. Choice B is correct because τ = rFsinφ = (0.30 m)(40 N)sin(60°) = (0.30)(40)(0.866) = 10.4 N·m. Choice A incorrectly calculates τ = rF without considering the angle, yielding 12 N·m. To help students, emphasize the importance of identifying the angle between force and position vectors. Use visual aids showing force decomposition and practice with various angles to build intuition about when torque is maximized (90°) or zero (0° or 180°).
20. A spinning flywheel is slowing down due to a constant frictional torque. Which of the following correctly describes the work done by the frictional torque and the work done by the net torque on the flywheel during one revolution?
    Answer: Work by frictional torque is negative; work by net torque is negative. The frictional torque opposes the angular velocity, so the angle between the torque vector and the infinitesimal angular displacement vector is $\pi$ radians. Thus, the work done by friction is negative. Since friction is the only torque mentioned that causes it to slow down, it is the net torque. Therefore, the net work done is also negative, consistent with the decrease in rotational kinetic energy.

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