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[parent] example of Rotational Work, Power, and Energy Transfer

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GRE Physics Companion: Rotational Work, Power, and Energy Transfer

The essential rotational energy relations are

|-----∫------|
W  =    τ d𝜃,|
--------------
(1)

|--------|
|P = τ ω,|
----------
(2)

and

|----------(------)--|
|            1-  2   |
|Wnet = Δ    2Iω    .|
---------------------
(3)

PIC

Figure 1. A compact strategy for rotational work and Power problems. Use torque-angle area for work, torque times angular speed for power, and rotational kinetic energy for state changes.

1 High-value GRE facts

  1. Constant torque gives W = τΔ𝜃.
  2. Variable torque requires W = ∫ τ d𝜃.
  3. The area under a τ versus 𝜃 graph is work.
  4. rotational power is P = τω for a fixed axis.
  5. Torque and energy may share N m dimensions but are different quantities.
  6. Flywheel energy scales as ω2.
  7. Constant power implies τ = P∕ω.
  8. A torsional spring stores 1
2κ𝜃2.
  9. Viscous rotational damping gives P = −bω2.
  10. An ideal transmission conserves mechanical power.

Part I: Original GRE-style problems

Problem 1: constant torque work

A constant torque of 6 N m acts through 10 rad. The work is

  1. 0.6 J
  2. 6 J
  3. 16 J
  4. 60 J
  5. 600 J

Problem 2: rotational power

A shaft rotates at 80 rad∕s while transmitting torque 25 N m. The power is

  1. 0.32 kW
  2. 2.0 kW
  3. 3.2 kW
  4. 20 kW
  5. 200 kW

Problem 3: torque-angle graph

Torque increases linearly from zero at 𝜃 = 0 to τ0 at 𝜃 = 𝜃0. The work is

  1. 0
  2. τ0∕𝜃0
  3. 1
2τ0𝜃0
  4. τ0𝜃0
  5. 2τ0𝜃0

Problem 4: flywheel speed scaling

A flywheel’s angular speed doubles while its moment of inertia remains constant. Its rotational kinetic energy is multiplied by

  1. 1∕2
  2. 2
  3. 4
  4. 8
  5. 16

Problem 5: torsional spring energy

A torsional spring has constant κ and is twisted through angle 𝜃. Its stored energy is

  1. κ𝜃
  2. κ𝜃2
  3. 1
2κ𝜃
  4. 1
2κ𝜃2
  5. 2κ𝜃2

Problem 6: constant power motor

A motor delivers constant power P at angular speed ω. Its torque is

  1. Pω
  2. P∕ω
  3. ω∕P
  4. P∕ω2
  5. independent of ω

Problem 7: damping power

A rotational damper obeys τd = −bω. Its power is

  1. +bω
  2. −bω
  3. +bω2
  4. −bω2
  5. −bω3

Problem 8: ideal transmission

An ideal transmission reduces angular speed by a factor of 5. If input torque is τin, output torque is

  1. τin∕25
  2. τin∕5
  3. τin
  4. 5τin
  5. 25τin

Problem 9: braking angle

A rotor with moment of inertia I and initial angular speed ω0 is stopped by constant braking torque magnitude τb. The stopping angle is

  1. Iω0∕τb
  2. Iω02∕(2τ b)
  3. 2Iω02∕τ b
  4. τb∕(Iω02)
  5. I∕τb

Problem 10: belt power

A tangential belt force F acts at radius R of a Pulley rotating at ω. The transmitted power is

  1. Fω
  2. FR
  3. FRω
  4. Fω∕R
  5. FR∕ω

Problem 11: constant torque from rest

A rotor with I = 5 kg m2 starts from rest. Net torque 20 N m acts through 10 rad. The final angular speed is closest to

  1. 4.0 rad∕s
  2. 6.3 rad∕s
  3. 8.9 rad∕s
  4. 20 rad∕s
  5. 40 rad∕s

Problem 12: torque versus energy

Which statement is correct?

  1. Torque and energy are identical because both use N m.
  2. Torque is a rotational moment, while work is a scalar energy transfer.
  3. Torque is measured in joules.
  4. Work has a direction in space.
  5. Torque can be added directly to kinetic energy.

Part II: Complete worked solutions

Solution 1

W  =  τΔ 𝜃 = (6)(10) = 60J.
(4)

Answer: (D).

Solution 2

P  = τω =  (25 )(80 ) = 2000W  =  2.0 kW.
(5)

Answer: (B).

Solution 3

The graph is a triangle:

      1-
W  =  2τ0𝜃0.
(6)

Answer: (C).

Solution 4

Since K = 1
2Iω2, doubling ω multiplies K by 4. Answer: (C).

Solution 5

     1
U  = --κ𝜃2.
     2
(7)

Answer: (D).

Solution 6

From P = τω,

τ =  P-.
     ω
(8)

Answer: (B).

Solution 7

P  = τ ω = (− bω)ω =  − bω2.
 d    d
(9)

Answer: (D).

Solution 8

Ideal power conservation gives

τ ω  =  τ  ω   .
 in in    out  out
(10)

If ωout = ωin∕5, then

τ   =  5τ .
 out     in
(11)

Answer: (D).

Solution 9

τbΔ𝜃 =  1Iω20,
        2
(12)

so

       Iω20-
Δ 𝜃 =  2τb .
(13)

Answer: (B).

Solution 10

Since τ = FR,

P = τ ω = F R ω.
(14)

Answer: (C).

Solution 11

W  = (20)(10) = 200 J.
(15)

Then

      1
200 = --(5)ω2,
      2
(16)

so

    √ ---
ω =   80 = 8.94 rad∕s.
(17)

Answer: (C).

Solution 12

Torque is a moment of force. Work is the scalar integral of torque through angular displacement. Answer: (B).

2 GRE checklist

  1. Use radians in torque-angle work.
  2. Use area under a τ versus 𝜃 graph for variable torque.
  3. Use P = τω for fixed-axis power.
  4. Use Wnet = Δ(1
2Iω2) for rotational state changes.
  5. For flywheels, remember the ω2 dependence.
  6. For constant power, use τ = P∕ω.
  7. For damping, check that power is negative.
  8. For ideal transmissions, conserve power rather than torque.

References

References

[1]   J. R. Taylor, Classical Mechanics, University Science Books, 2005.

[2]   D. Kleppner and R. Kolenkow, An Introduction to Mechanics, 2nd ed., Cambridge University Press, 2014.

[3]   OpenStax, University Physics, Volume 1, Rice University, 2016.


"example of Rotational Work, Power, and Energy Transfer" is owned by bloftin.
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Keywords:  GRE physics, rotational work, rotational power, torque, angular displacement, flywheel, torsional spring, damping, motor torque, transmission

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Cross-references: displacement, kinetic energy, scalar, Pulley, force, magnitude, moment of inertia, mechanical power, rotational damping, torsional spring, dimensions, rotational power, graph, rotational kinetic energy, speed, work, Power, rotational work, relations, energy

This is version 1 of example of Rotational Work, Power, and Energy Transfer, born on 2026-10-03.
Object id is 1388, canonical name is ExampleOfRotationalWorkPowerAndEnergyTransfer.
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Physics Classification: 45.40.-f (Dynamics and kinematics of rigid bodies)
 45.20.Dd (Newtonian mechanics)
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