Rotational Work, Power, and Energy Transfer

(Topic)

Rotational Work, Power, and Energy Transfer

M03-09 established the rotational kinetic energy

1 Krot = -Iω2 2

(1)

for a rigid body rotating about a fixed axis.

The next question is how rotational kinetic energy changes.

A force transfers energy when its point of application moves. In rotational motion, the same physical idea can be expressed through torque and angular displacement:

|-------------| dW = τ ⋅ d𝜃.| ---------------

(2)

For fixed-axis rotation,

|------------| -dW--=--τ d𝜃.|

(3)

The corresponding instantaneous power is

|----------| P--=-τ-⋅ ω.-

(4)

For a fixed axis,

|--------| |P = τ ω.| ----------

(5)

These relations describe energy transfer in shafts, motors, flywheels, Pulleys, brakes, turbines, gears, and rotating machinery.

1 From force work to torque work

Consider a force F applied to a rigid body at position r measured from a fixed rotation axis.

The differential work is

dW = F ⋅ dr.

(6)

For a small rigid rotation,

dr = d𝜃 × r.

(7)

Substitute:

= F ⋅ (d𝜃 × r).

(8)

Using the scalar triple-product identity,

F ⋅ (d𝜃 × r) = (r × F) ⋅ d𝜃.

(9)

Since

τ = r × F,

(10)

we obtain

|-------------| dW--=--τ-⋅ d𝜃.-

(11)

Figure 1. The ordinary work F ⋅ dr produced by a force on a rotating body can be written as torque times angular displacement.

2 Sign of rotational work

For fixed-axis rotation,

dW = τ d𝜃.

(12)

Therefore:

dW > 0 when torque and angular displacement have the same sign,
dW = 0 when the torque has no component along the rotation axis,
dW < 0 when torque opposes the angular displacement.

A motor driving a shaft usually does positive rotational work.

A brake opposing rotation usually does negative rotational work.

3 Finite rotational work

For a torque that may vary with angular position,

|------------------| | ∫ 𝜃f | |W = τ(𝜃)d𝜃. | --------𝜃i----------|

(13)

If torque is constant,

|-----------------------| W = τ(𝜃 − 𝜃) = τΔ 𝜃. | ---------f----i----------

(14)

Angular displacement must be expressed in radians because the differential relation

ds = r d𝜃

(15)

assumes radian measure.

4 Torque angle graphs

The integral

∫ W = τ d𝜃

(16)

has a geometric interpretation.

On a graph of torque versus angular displacement:

area above the angular axis contributes positive work,
area below the angular axis contributes negative work,
total signed area equals rotational work.

Figure 2. The signed area under a torque-versus-angle curve equals rotational work.

The units are

(N m )(rad ).

(17)

Radians are dimensionless in SI, so the result has units

N m = J.

(18)

5 Torque is not energy

Torque and energy can both be expressed using the dimensional combination N m, but they are different physical quantities.

Torque is a rotational moment:

τ = r × F.

(19)

Work and energy are scalars:

∫ W = τ ⋅ d𝜃.

(20)

A torque of

10 N m

(21)

is not the same physical quantity as

10 J.

(22)

The numerical dimensions coincide, but the physical meanings and transformation properties do not.

6 Rotational work-energy theorem

For a rigid body rotating about a fixed axis,

τnet = Iα.

(23)

The net rotational work is

∫ 𝜃f Wnet = τnetd𝜃. 𝜃i

(24)

Use

τnet = I α

(25)

and

α = ω dω-. d𝜃

(26)

Then

W_net	= ∫ Iωd𝜃	(27)
	= ∫ _{ω_i}^ω_fIω dω.	(28)

For constant moment of inertia,

|--------1-------1-----| |Wnet = -Iω2f − -Iω2i.| ---------2-------2-----|

(29)

Therefore

|--------------| -Wnet-=-ΔKrot.--

(30)

7 Rotational power

Power is the time rate of work:

P = dW--. dt

(31)

Using

dW = τ ⋅ d𝜃,

(32)

we obtain

d𝜃 P = τ ⋅dt-.

(33)

Therefore

|----------| P--=-τ-⋅ ω.-

(34)

For a fixed axis,

|--------| -P-=-τ-ω.-

(35)

Figure 3. Torque acting through angular velocity transfers energy at the rate P = τω. For a fixed moment of inertia, this is also dK_rot∕dt.

8 Power and rotational kinetic energy

For fixed I,

1 2 Krot = -Iω . 2

(36)

Differentiate:

	= Iω	(37)
	= Iωα.	(38)

Since

τnet = Iα,

(39)

we obtain

|--------------| |dKrot-= τ ω.| --dt------net---

(40)

Thus

|--------------| |Pnet = dKrot-.| ----------dt---|

(41)

9 Constant torque

If a constant net torque τ₀ acts through angular displacement Δ𝜃,

Wnet = τ0Δ 𝜃.

(42)

Then

1 1 τ0Δ 𝜃 = -Iω2f − -Iω2i. 2 2

(43)

Therefore

|------------------| | 2 2 2τ0Δ𝜃- | |ωf = ω i + I .| -------------------

(44)

10 Example 1: flywheel accelerated by constant torque

A flywheel has

2 I = 12 kg m .

(45)

A constant net torque

τ = 18 N m

(46)

acts while the wheel turns through

Δ 𝜃 = 30 rad.

(47)

It starts from rest.

The work is

W = (18)(30) = 540 J.

(48)

Set this equal to final rotational kinetic energy:

1 2 540 = -(12)ωf. 2

(49)

Thus

ω2f = 90,

(50)

and

|----------------| -ωf-=-9.49-rad-∕s.|

(51)

11 Variable torque example

Suppose

( 𝜃) τ(𝜃) = τ0 1 − -- 𝜃0

(52)

for

0 ≤ 𝜃 ≤ 𝜃0.

(53)

The work is

W	= ∫ ₀^𝜃₀τ₀d𝜃	(54)
	= τ₀₀^𝜃₀	(55)
	= τ₀𝜃₀.	(56)

Geometrically, this is the area of a triangle under the torque angle graph.

12 Torsional springs

A torsional spring produces a restoring torque proportional to angular displacement:

|----------| -τs =-− κ-𝜃,

(57)

where κ is the torsional spring constant.

The potential energy satisfies

τs = − dUs-. d 𝜃

(58)

Thus

dUs − κ𝜃 = − ----, d𝜃

(59)

dU ---s= κ𝜃. d 𝜃

(60)

Integrating,

1 Us = -κ𝜃2 + C. 2

(61)

Choosing

Us(0) = 0

(62)

gives

|-----------| | 1- 2 | Us = 2κ𝜃 . | -------------

(63)

Figure 4. A torsional spring stores rotational potential energy U_s = 1
2 κ𝜃² and exerts the restoring torque τ_s = −κ𝜃.

13 Example 2: torsional spring release

A rotor with moment of inertia I is attached to a torsional spring of constant κ.

The rotor is released from rest at angular displacement 𝜃₀.

Initially,

Ki = 0

(64)

and

U = 1κ𝜃2. i 2 0

(65)

𝜃 = 0,

(66)

the spring potential energy is zero and the rotational kinetic energy is maximum:

1Iω2 = 1-κ𝜃2. 2 max 2 0

(67)

Therefore

|--------------| | ∘ -- | |ωmax = 𝜃0 κ.| -------------I--

(68)

14 Flywheel energy storage

A flywheel stores energy as rotational kinetic energy:

|----------------| | 1- 2 | |Eflywheel = 2 Iω .| ------------------

(69)

The usable energy between two angular speeds is

|------------------------| | 1- ( 2 2 ) | |ΔE = 2 I ω high − ωlow .| --------------------------

(70)

Because the energy depends on ω², doubling angular speed multiplies stored kinetic energy by four for the same I.

Figure 5. Flywheel energy grows quadratically with angular speed. Energy extracted as the wheel slows is the difference between two values of 1
2 Iω².

15 Example 3: usable flywheel energy

A flywheel has

I = 20 kg m2.

(71)

It operates between

ω_high	= 300 rad∕s,	(72)
ω_low	= 100 rad∕s.	(73)

The usable energy is

ΔE	= (20)	(74)
	= 10(90000 − 10000)	(75)
	= 800000 J.	(76)

Thus

|--------------| |ΔE = 800kJ. | ---------------

(77)

16 Motor torque and power

For a motor shaft,

P = τ ω.

(78)

If torque is approximately constant,

P ∝ ω.

(79)

If the motor is operating in an approximately constant-power regime,

|--------| | P | |τ = --.| ------ω--

(80)

Therefore available torque decreases as angular speed increases.

Figure 6. At constant torque, power increases linearly with angular speed. At constant power, torque decreases as 1∕ω.

17 Example 4: shaft torque from power

A motor delivers

P = 24 kW

(81)

ω = 160 rad∕s.

(82)

Then

τ	=	(83)
	=	(84)
	= 150 N m.	(85)

Therefore

|------------| τ-=--150N-m.--

(86)

18 Rotational braking

A brake applies a torque opposite the angular velocity.

Thus

τ b ⋅ ω < 0.

(87)

The braking power is negative:

|--------| |Pb < 0. | ---------

(88)

If a constant braking torque of magnitude τ_b acts through stopping angle Δ𝜃,

1- 2 − τbΔ𝜃 = − 2Iω 0.

(89)

Therefore

|----------| | 2 | Δ 𝜃 = Iω0.| -------2τb--

(90)

The stopping angle grows as ω₀².

19 Regenerative rotational braking

Negative rotational work does not have to become thermal energy.

In regenerative braking, a rotating system can drive an electrical machine as a generator.

Then rotational kinetic energy is transferred into electrical energy:

Krot − → Eelectrical.

(91)

The torque on the rotor still opposes the motion, so

Protor < 0.

(92)

The destination of the removed mechanical energy depends on the physical mechanism.

20 Rotational damping

A common damping model is a torque proportional to angular velocity:

|----------| |τd = − bω,| ------------

(93)

where b > 0 is a rotational damping coefficient.

The damping power is

P_d	= τ_dω	(94)
	= −bω².	(95)

Therefore

|------------| |P = − bω2. | --d----------

(96)

The damping torque always removes rotational mechanical energy when ω≠0.

Figure 7. Viscous rotational damping produces negative power P_d = −bω², continuously removing rotational mechanical energy.

21 Free decay under viscous rotational damping

Consider a rotor with no applied torque except

τd = − bω.

(97)

The equation of motion is

dω- Idt = − bω.

(98)

Separate variables:

dω b ---= − -dt. ω I

(99)

Integrating,

|----------−-bt∕I-| -ω(t)-=-ω0e-----.|

(100)

The rotational kinetic energy is

|--------------−2bt∕I-| -Krot(t) =-K0e------.|

(101)

Energy decays twice as fast in the exponent as angular speed because kinetic energy depends on ω².

22 Belts, pulleys, and tangential power transfer

A belt applies tangential force F_t at pulley radius R.

The torque is

τ = F R. t

(102)

If the pulley edge moves at tangential speed

v = R ω,

(103)

then the linear power transmitted by the belt is

P = Ftv.

(104)

Substitute

v = R ω :

(105)

P = F R ω = τ ω. t

(106)

Therefore

|----------| -Ftv-=-τω.-|

(107)

Figure 8. A tangential belt force transfers the same power whether described linearly as F_tv or rotationally as τω.

23 Ideal mechanical transmission

In an ideal lossless rotational transmission,

Pin = Pout.

(108)

Thus

|τinωin =-τoutωout.| ------------------

(109)

A decrease in angular speed can therefore accompany an increase in torque.

For a real transmission with efficiency η,

------------- |P = ηP .| --out-----in-|

(110)

Hence

|------------------| -τoutωout-=-ητinωin.|

(111)

24 Example 5: ideal speed reduction

An ideal transmission reduces angular speed from

ω = 300 rad∕s in

(112)

ωout = 60rad ∕s.

(113)

τin = 20N m,

(114)

power conservation gives

τinωin = τoutωout.

(115)

Thus

|------------------------| | 300 | |τout = 20 ----= 100 N m. | ----------60--------------

(116)

25 Combined translational and rotational power

For a rigid body in planar motion,

K = 1M V2 + 1I ω2. 2 CM 2 CM

(117)

Differentiate:

dK-- dt = M ACM ⋅ VCM + ICM α ω.

(118)

Using the translational and rotational equations of motion,

|----------------------------| |dK--= F ⋅ V + τextω.| -dt------ext,net---CM-----CM----

(119)

Thus rigid-body power naturally separates into translational and rotational parts.

26 Contact forces and energy transfer

The work of a contact force depends on the velocity of the point where the force acts.

For a force F_c applied at contact point C,

|-------------| Pc-=--Fc ⋅-vC.|

(120)

For ideal rolling on a fixed surface,

vC = 0,

(121)

so static friction can produce zero power.

For a moving belt or moving surface,

vC ⁄= 0,

(122)

so frictional contact can transfer mechanical power.

This point-of-application view is the safest way to reason about work by contact forces.

27 Common mistakes

Treating torque itself as energy because both can use N m.
Using degrees rather than radians in W = ∫ τ d𝜃.
Using W = τΔ𝜃 when torque varies with angle.
Forgetting the sign of braking or damping torque.
Using P = τω without considering the angle between torque and angular velocity in three dimensions.
Forgetting that flywheel energy scales as ω².
Assuming constant torque means constant power.
Assuming constant power means constant torque.
Forgetting that viscous damping gives P_d = −bω².
Assuming all negative rotational work becomes heat.
Using a force displacement different from the displacement of the actual force application point.
Forgetting transmission efficiency in real power-transfer systems.

28 Practice exercises

A constant torque of 15 N m acts through 8 rad. Find the work.
A torque varies as τ(𝜃) = 4𝜃 in SI units. Find the work from 𝜃 = 0 to 𝜃 = 3 rad.
Derive the rotational work-energy theorem from τ = Iα.
A shaft rotates at 200 rad∕s while transmitting torque 75 N m. Find the power.
A flywheel with I = 10 kg m² increases speed from 50 to 150 rad∕s. Find the increase in stored rotational kinetic energy.
A constant braking torque of magnitude 40 N m stops a flywheel with I = 5 kg m² from 20 rad∕s. Find the stopping angle.
A torsional spring has κ = 12 N m∕rad and is twisted by 0.50 rad. Find the stored energy.
A rotor with I = 3 kg m² is released from 𝜃₀ = 0.40 rad on a torsional spring with κ = 75 N m∕rad. Find the maximum angular speed.
Show that a belt force satisfies F_tv = τω for a pulley of radius R.
An ideal transmission has ω_in∕ω_out = 4. If the input torque is 25 N m, find the output torque.
Repeat the preceding problem for a transmission efficiency of 0.85.
A viscous rotational damper has τ_d = −bω. Derive the damping power.
Solve Iω = −bω and obtain the exponential decay of angular speed.
Explain why the energy of a viscously damped rotor decays with e^−2bt∕I.
Give an example where a contact force does zero work and another where a contact force transfers nonzero power.

29 Summary

Rotational work is

|-----∫--------| W = τ ⋅ d𝜃. ----------------

(123)

For fixed-axis rotation,

|-----∫------| | | W = τ d𝜃.| --------------

(124)

The rotational work-energy theorem is

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

(125)

Rotational power is

|----------| P--=-τ-⋅ ω.-

(126)

A torsional spring stores

|-----------| | 1- 2 | Us = 2κ𝜃 . | -------------

(127)

A flywheel stores

|-------------| Krot = 1Iω2. | --------2------

(128)

Viscous rotational damping has

---------------------------- | 2 | -τd =-− bω,----Pd-=--−-bω--.-

(129)

For ideal power transmission,

|----------------| -τinωin =-τoutωout.-

(130)

This completes the main work-energy sequence and provides the bridge into momentum methods.

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] K. R. Symon, Mechanics, 3rd ed., Addison-Wesley, 1971.

[4] H. D. Young and R. A. Freedman, University Physics with Modern Physics, 15th ed., Pearson, 2020.

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

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Rotational Work, Power, and Energy Transfer