AP Physics C Mechanics Quiz: Torque And Work
20 questions · exam conditions
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Torque And WorkQuestion 1 of 20

A gyroscope with its axis tilted precesses with a constant angular speed about a vertical axis. The gravitational force exerts a torque on the gyroscope, causing the precession. What is the work done by this gravitational torque during one complete precession cycle?

Zero, because the torque vector is always perpendicular to the angular displacement of precession.
Positive, because the torque is required to maintain the precession against dissipative forces.
Negative, because the potential energy of the gyroscope's center of mass does not change.
It cannot be determined without knowing the gyroscope's spin and precession speeds.
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AP Physics C Mechanics Quiz

AP Physics C Mechanics Quiz: Torque And Work

Practice Torque And Work in AP Physics C Mechanics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Torque And Work, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics C Mechanics.

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.

All questions

Question 1

A gyroscope with its axis tilted precesses with a constant angular speed about a vertical axis. The gravitational force exerts a torque on the gyroscope, causing the precession. What is the work done by this gravitational torque during one complete precession cycle?

  1. Zero, because the torque vector is always perpendicular to the angular displacement of precession. (correct answer)
  2. Positive, because the torque is required to maintain the precession against dissipative forces.
  3. Negative, because the potential energy of the gyroscope's center of mass does not change.
  4. It cannot be determined without knowing the gyroscope's spin and precession speeds.
Explanation: The work done by a torque is given by W=∫τ⃗⋅dθ⃗W = \int \vec{\tau} \cdot d\vec{\theta}. For a precessing gyroscope, the gravitational torque vector is horizontal. The angular displacement vector for the precession is along the vertical axis of precession. Since the torque vector is always perpendicular to the angular displacement vector, their dot product is zero, and the work done is zero. This torque changes the direction of the angular momentum but not its magnitude.

Question 2

A net torque given by τ=τ0sin⁡(θ)\tau = \tau_0 \sin(\theta) is applied to a rotor, where τ0\tau_0 is a positive constant. What is the work done by this torque as the rotor moves from an angular position of θ=0\theta = 0 to θ=π\theta = \pi radians?

  1. 00
  2. τ0\tau_0
  3. πτ0\pi \tau_0
  4. 2τ02\tau_0 (correct answer)
Explanation: The work done is the integral of the torque over the angular displacement: W=∫0πτ0sin⁡(θ)dθ=τ0[−cos⁡(θ)]0π=−τ0(cos⁡(π)−cos⁡(0))=−τ0(−1−1)=2τ0W = \int_{0}^{\pi} \tau_0 \sin(\theta) d\theta = \tau_0 [-\cos(\theta)]_{0}^{\pi} = -\tau_0 (\cos(\pi) - \cos(0)) = -\tau_0 (-1 - 1) = 2\tau_0.

Question 3

The net torque on a flywheel varies with its angular position θ\theta. The torque is initially 20 N⋅m20 \, \text{N} \cdot \text{m} at θ=0\theta=0 and decreases linearly to 0 N⋅m0 \, \text{N} \cdot \text{m} at θ=5.0 rad\theta=5.0 \, \text{rad}.

What is the total work done by the net torque on the flywheel as it rotates from θ=0\theta = 0 to θ=5.0 rad\theta = 5.0 \, \text{rad}?

  1. 25 J25 \, \text{J}
  2. 50 J50 \, \text{J} (correct answer)
  3. 75 J75 \, \text{J}
  4. 100 J100 \, \text{J}
Explanation: The work done is the area under the torque versus angular position graph. Since the torque decreases linearly, the graph is a triangle. The area of the triangle is W=12×base×height=12(5.0 rad)(20 N⋅m)=50 JW = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} (5.0 \, \text{rad})(20 \, \text{N} \cdot \text{m}) = 50 \, \text{J}.

Question 4

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?

  1. Work by frictional torque is negative; work by net torque is negative. (correct answer)
  2. Work by frictional torque is positive; work by net torque is negative.
  3. Work by frictional torque is negative; work by net torque is positive.
  4. Work by frictional torque is zero; work by net torque is zero.
Explanation: The frictional torque opposes the angular velocity, so the angle between the torque vector and the infinitesimal angular displacement vector is π\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.

Question 5

A pulley of radius r=0.10 mr=0.10\,\text{m} is rotated by a constant tangential pull on a rope. The applied force has magnitude F=60 NF=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 Δθ=4.0 rad\Delta\theta=4.0\,\text{rad}. Assume no slipping and no dissipative losses. The torque is constant with magnitude τ=rF\tau=rF. The work done by the torque is W=τΔθW=\tau\Delta\theta. Using the given information, calculate the total work done by the applied force.

  1. W=24 JW=24\,\text{J} (correct answer)
  2. W=240 JW=240\,\text{J}
  3. W=6.0 JW=6.0\,\text{J}
  4. W=24 N mW=24\,\text{N\,m}
Explanation: 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.

Question 6

A pulley of radius r=0.20 mr=0.20\,\text{m} is turned by pulling on a light rope wrapped around its rim. A constant tangential force F=50 NF=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 Δθ=2.5 rad\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=τΔθW=\tau\Delta\theta. Using the given information, calculate the total work done by the applied force on the pulley during this rotation.

  1. W=25 JW=25\,\text{J} (correct answer)
  2. W=5.0 JW=5.0\,\text{J}
  3. W=125 JW=125\,\text{J}
  4. W=2.0 N mW=2.0\,\text{N\,m}
Explanation: 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.

Question 7

A mechanical arm rotates about a pivot, and a force F=25 NF=25\,\text{N} is applied at a point r=0.40 mr=0.40\,\text{m} from the pivot. The force is applied at an angle ϕ=45∘\phi=45^\circ relative to the arm (the radius vector), in the plane of rotation. Assume the pivot is frictionless and the force magnitude remains constant at that instant. Only the torque from this applied force is considered. The torque magnitude is τ=rFsin⁡ϕ\tau=rF\sin\phi. Using the given information, determine the torque produced about the pivot.

  1. τ=7.1 N m\tau=7.1\,\text{N\,m} (correct answer)
  2. τ=10 N m\tau=10\,\text{N\,m}
  3. τ=14 N m\tau=14\,\text{N\,m}
  4. τ=0.28 N m\tau=0.28\,\text{N\,m}
Explanation: 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 25 N force is applied at 0.40 m from the pivot at a 45° angle. Choice A is correct because τ = rFsinφ = (0.40 m)(25 N)sin(45°) = (0.40)(25)(0.707) = 7.1 N·m. Choice B incorrectly calculates τ = rF without the angle factor, yielding 10 N·m. To help students, use the mnemonic that torque depends on the 'perpendicular' component of force. Practice with 45° angles specifically, as sin(45°) = cos(45°) = 0.707 is a common value students should memorize.

Question 8

A balancing beam of negligible mass is pivoted at one end and initially held horizontal. A single weight of magnitude W=50 NW=50\,\text{N} hangs from the beam at a distance r=0.60 mr=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 r⃗\vec r and the weight is 90∘90^\circ. Ignore friction at the pivot and assume the weight remains attached at the same point. The torque magnitude from the weight is τ=rWsin⁡90∘\tau=rW\sin 90^\circ. Using the given information, what is the torque about the pivot due to the hanging weight at this instant?

  1. τ=30 N m\tau=30\,\text{N\,m} (correct answer)
  2. τ=15 N m\tau=15\,\text{N\,m}
  3. τ=300 N m\tau=300\,\text{N\,m}
  4. τ=30 J\tau=30\,\text{J}
Explanation: 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.

Question 9

A uniform disc of radius r=0.40 mr=0.40\,\text{m} is mounted on a low-friction axle. A force of magnitude F=30 NF=30\,\text{N} is applied at the rim but not tangentially; instead, it makes an angle of 30∘30^\circ with the radius. The force lies in the plane of the disc and stays at this angle with respect to the radius at the point of application. Air resistance is negligible, and the disc remains rigid. The torque magnitude about the center is given by τ=rFsin⁡ϕ\tau=rF\sin\phi, where ϕ\phi is the angle between r⃗\vec r and F⃗\vec F. Using the given information, what is the torque exerted by the force about the disc's center?

  1. τ=12 N m\tau=12\,\text{N\,m}
  2. τ=10.4 N m\tau=10.4\,\text{N\,m}
  3. τ=6.0 N m\tau=6.0\,\text{N\,m} (correct answer)
  4. τ=24 N m\tau=24\,\text{N\,m}
Explanation: 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 30 N force is applied at the rim (0.40 m) at a 30° angle to the radius. Choice C is correct because τ = rFsinφ = (0.40 m)(30 N)sin(30°) = (0.40)(30)(0.5) = 6.0 N·m. Choice A incorrectly uses sin(90°) instead of sin(30°), yielding 12 N·m, which would be the torque if the force were tangential. To help students, use diagrams showing the force vector and its angle relative to the position vector. Emphasize that the angle in the torque formula is between the force and position vectors, not the force and some reference direction.

Question 10

A disc of radius r=0.25 mr=0.25\,\text{m} rotates about its center on a frictionless axle. A constant tangential force F=16 NF=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 Δθ=π rad\Delta\theta=\pi\,\text{rad}. Ignore any losses so all work is due to the applied torque. The torque magnitude is constant and equals τ=rF\tau=rF. The rotational work done is W=τΔθW=\tau\Delta\theta. Using the given information, how much work is done by the force during this rotation?

  1. W=4π JW=4\pi\,\text{J} (correct answer)
  2. W=2π JW=2\pi\,\text{J}
  3. W=8π JW=8\pi\,\text{J}
  4. W=4π N mW=4\pi\,\text{N\,m}
Explanation: 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).

Question 11

A robotic arm rotates about a fixed pivot in a horizontal plane. A force of magnitude F=40 NF=40\,\text{N} is applied at a point r=0.30 mr=0.30\,\text{m} from the pivot. The force makes an angle ϕ=60∘\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 τ=rFsin⁡ϕ\tau=rF\sin\phi. Using the given information, determine the torque produced about the pivot by the applied force.

  1. τ=12 N m\tau=12\,\text{N\,m}
  2. τ=10.4 N m\tau=10.4\,\text{N\,m} (correct answer)
  3. τ=24 N m\tau=24\,\text{N\,m}
  4. τ=6.0 J\tau=6.0\,\text{J}
Explanation: 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°).

Question 12

A flywheel experiences a constant torque due to a motor. The motor applies a tangential force F=80 NF=80\,\text{N} at a radius r=0.15 mr=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 Δθ=10 rad\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=τΔθW=\tau\Delta\theta. Using the given information, determine the work done by the motor during the 10 rad10\,\text{rad} rotation.

  1. W=120 JW=120\,\text{J} (correct answer)
  2. W=12 JW=12\,\text{J}
  3. W=60 JW=60\,\text{J}
  4. W=800 JW=800\,\text{J}
Explanation: 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.

Question 13

A beam of negligible mass is pivoted at its left end and lies initially horizontal. A weight W1=30 NW_1=30\,\text{N} hangs at r1=0.40 mr_1=0.40\,\text{m} from the pivot, and a second weight W2=20 NW_2=20\,\text{N} hangs at r2=0.70 mr_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.

  1. τ=26 N m\tau=26\,\text{N\,m} (correct answer)
  2. τ=2.6 N m\tau=2.6\,\text{N\,m}
  3. τ=13 N m\tau=13\,\text{N\,m}
  4. τ=52 N m\tau=52\,\text{N\,m}
Explanation: 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.

Question 14

A robotic arm rotates about a fixed pivot in a plane. A force of magnitude F=30 NF=30\,\text{N} is applied at a point r=0.40 mr=0.40\,\text{m} from the pivot. At the instant considered, the angle between the radius vector and the force is ϕ=60∘\phi=60^\circ. The pivot is frictionless, and only this applied force produces torque about the pivot. Use the standard torque magnitude formula τ=rFsin⁡ϕ\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.

  1. τ=12 N ⁣\cdot ⁣m\tau=12\,\text{N\!\cdot\!m}
  2. τ=10.4 N ⁣\cdot ⁣m\tau=10.4\,\text{N\!\cdot\!m} (correct answer)
  3. τ=6.0 N ⁣\cdot ⁣m\tau=6.0\,\text{N\!\cdot\!m}
  4. τ=20.8 N ⁣\cdot ⁣m\tau=20.8\,\text{N\!\cdot\!m}
Explanation: 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.

Question 15

A constant net torque of magnitude τ\tau is applied to a rigid body with rotational inertia II, causing it to rotate through an angular displacement Δθ\Delta\theta from rest. What is the work done by the torque on the rigid body?

  1. τΔθ\tau \Delta\theta (correct answer)
  2. τ/Δθ\tau / \Delta\theta
  3. τω\tau \omega, where ω\omega is the final angular velocity
  4. τα\tau \alpha, where α\alpha is the angular acceleration
Explanation: The work done by a constant torque τ\tau is defined as the product of the torque and the angular displacement Δθ\Delta\theta. The formula is W=τΔθW = \tau \Delta\theta. The rotational inertia II is not directly needed for this calculation.

Question 16

The torque exerted on a rotating disk is given by the function τ(θ)=kθ2\tau(\theta) = k\theta^2, where kk is a positive constant and θ\theta is the angular position in radians. What is the work done by this torque as the disk rotates from θ=0\theta = 0 to a final angular position θ=θf\theta = \theta_f?

  1. kθf2k\theta_f^2
  2. 2kθf2k\theta_f
  3. 12kθf2\frac{1}{2}k\theta_f^2
  4. 13kθf3\frac{1}{3}k\theta_f^3 (correct answer)
Explanation: Work done by a variable torque is calculated by integrating the torque with respect to the angular displacement. W=∫θiθfτ(θ)dθW = \int_{\theta_i}^{\theta_f} \tau(\theta) d\theta. In this case, W=∫0θfkθ2dθ=[13kθ3]0θf=13kθf3−0=13kθf3W = \int_{0}^{\theta_f} k\theta^2 d\theta = [\frac{1}{3}k\theta^3]_{0}^{\theta_f} = \frac{1}{3}k\theta_f^3 - 0 = \frac{1}{3}k\theta_f^3.

Question 17

A grinding wheel with a rotational inertia of 2.0 kg⋅m22.0 \, \text{kg} \cdot \text{m}^2 slows uniformly from an angular speed of 30 rad/s30 \, \text{rad/s} to 10 rad/s10 \, \text{rad/s}. What is the magnitude of the work done by friction on the wheel?

  1. 400 J400 \, \text{J}
  2. 800 J800 \, \text{J} (correct answer)
  3. 900 J900 \, \text{J}
  4. 1000 J1000 \, \text{J}
Explanation: The work done by the net torque (which is the frictional torque here) equals the change in rotational kinetic energy. W=ΔKrot=12I(ωf2−ωi2)W = \Delta K_{rot} = \frac{1}{2}I(\omega_f^2 - \omega_i^2). W=12(2.0)((10)2−(30)2)=1.0(100−900)=−800 JW = \frac{1}{2}(2.0)((10)^2 - (30)^2) = 1.0(100 - 900) = -800 \, \text{J}. The magnitude of the work done is 800 J800 \, \text{J}.

Question 18

A uniform thin rod of mass MM and length LL 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?

  1. 00
  2. 14MgL\frac{1}{4}MgL
  3. 12MgL\frac{1}{2}MgL (correct answer)
  4. MgLMgL
Explanation: 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/2L/2 from the pivot. As the rod swings from horizontal to vertical, its center of mass drops by a vertical distance of h=L/2h = L/2. The work done by gravity is Wg=Mgh=Mg(L/2)W_g = Mgh = Mg(L/2).

Question 19

A disk is free to rotate about a fixed axle. An applied torque of τapp=+15 N⋅m\tau_{app} = +15 \, \text{N} \cdot \text{m} acts in the direction of motion, while a frictional torque of τfr=−5 N⋅m\tau_{fr} = -5 \, \text{N} \cdot \text{m} opposes the motion. What is the net work done on the disk as it rotates through an angle of 2π2\pi radians?

  1. 10π J10\pi \, \text{J}
  2. 20π J20\pi \, \text{J} (correct answer)
  3. 30π J30\pi \, \text{J}
  4. 40π J40\pi \, \text{J}
Explanation: The net torque on the disk is the sum of the applied and frictional torques: τnet=τapp+τfr=15 N⋅m−5 N⋅m=10 N⋅m\tau_{net} = \tau_{app} + \tau_{fr} = 15 \, \text{N} \cdot \text{m} - 5 \, \text{N} \cdot \text{m} = 10 \, \text{N} \cdot \text{m}. The net work done is Wnet=τnetΔθ=(10 N⋅m)(2π rad)=20π JW_{net} = \tau_{net} \Delta\theta = (10 \, \text{N} \cdot \text{m})(2\pi \, \text{rad}) = 20\pi \, \text{J}.

Question 20

The instantaneous power delivered by a motor to a rotating shaft is given by P(t)=Ct1/2P(t) = C t^{1/2}, where CC is a constant. What is the work done by the motor during the time interval from t=0t=0 to t=Tt=T?

  1. CT1/2CT^{1/2}
  2. 12CT−1/2\frac{1}{2}CT^{-1/2}
  3. 23CT3/2\frac{2}{3}CT^{3/2} (correct answer)
  4. 32CT3/2\frac{3}{2}CT^{3/2}
Explanation: Power is the rate at which work is done, P=dW/dtP = dW/dt. To find the total work done, we must integrate the power function with respect to time: W=∫0TP(t)dt=∫0TCt1/2dt=C[t3/23/2]0T=23CT3/2W = \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}.