Energy Methods for Rigid Bodies

Energy Methods for Rigid Bodies

A rigid body is a system of particles whose mutual distances remain fixed. Because the particles cannot move independently, the particle-system energy decomposition developed in M03-08 takes a particularly useful form.

For planar rigid body motion,

(1)

The first term describes translation of the center of mass. The second describes rotation about the center of mass.

For pure rotation about a fixed axis,

(2)

These results make many rigid body problems much easier than treating every mass element separately.

1 Rigid body motion as a particle system

Consider a rigid body made of particles indexed by i.

The total kinetic energy is

(3)

From the particle-system result,

(4)

For a rigid body, the motion relative to the center of mass is constrained to be rotational.

In planar motion,

(5)

where r_i′ is measured from the center of mass.

Thus

(6)

2 Derivation of rotational kinetic energy

The relative kinetic energy is

(7)

For planar rotation about the center of mass,

(8)

where r_i⊥′ is the perpendicular distance to the rotation axis.

Therefore

Krel	= ∑ imiω2 2	(9)
	= ω2 ∑ imi 2.	(10)

Define the moment of inertia about the center-of-mass axis:

(11)

Hence

(12)

The total kinetic energy is therefore

(13)

3 Continuous rigid bodies

For a continuous mass distribution,

(14)

Thus

(15)

The moment of inertia plays the same role in rotational kinetic energy that mass plays in translational kinetic energy.

For common bodies about symmetry axes through the center of mass:

Ihoop	= MR2,	(16)
Isolid disk	= MR2,	(17)
Isolid sphere	= MR2,	(18)
Ithin spherical shell	= MR2.	(19)

4 Pure translation

If

(20)

then every point of the body has the same velocity.

The kinetic energy becomes

(21)

This is exactly the kinetic energy of a particle of mass M moving with the center of mass.

5 Pure rotation about a fixed axis

If the body rotates about a fixed axis and the center of mass has no translational motion,

(22)

for an axis through the center of mass.

Then

(23)

If the fixed axis does not pass through the center of mass, the body still has total kinetic energy

(24)

where I_O is the moment of inertia about the fixed axis O.

The parallel-axis theorem,

(25)

makes this equivalent to the center-of-mass decomposition.

6 Derivation using the parallel-axis theorem

Suppose a body rotates about a fixed axis O located a distance d from the center-of-mass axis.

The center of mass moves with speed

(26)

Thus

K	= M(ωd)2 + ICMω2	(27)
	= ω2.	(28)

Since

(29)

we obtain

(30)

7 Work done by a torque

For a force F acting at a point whose position from a chosen origin is r, the torque is

(31)

For an infinitesimal rigid rotation d𝜃, the displacement of the force application point due to rotation is

(32)

The corresponding work is

(33)

Use the scalar triple-product identity:

dW	= F ⋅ (d𝜃 × r)	(34)
	= (r × F) ⋅ d𝜃.	(35)

Therefore

(36)

For fixed-axis rotation,

(37)

Integrating,

(38)

For constant torque,

(39)

8 Rotational work-energy theorem

For rotation about a fixed axis,

(40)

Multiply by d𝜃:

(41)

Using

(42)

and

(43)

we obtain

Iαd𝜃	= Iω dt	(44)
	= Iω dω.	(45)

Integrating,

(46)

For constant I,

(47)

Thus

(48)

9 Power in rotational motion

Instantaneous power is

(49)

Using

(50)

we obtain

(51)

Therefore

(52)

for fixed-axis rotation.

More generally,

(53)

For a rigid body undergoing translation and rotation, external mechanical power can be written

(54)

when the torque is taken about the center of mass.

10 Rigid body work-energy theorem

For an ideal rigid body,

(55)

The work-energy theorem becomes

(56)

when internal rigid constraints do no net work and no additional internal-energy storage is modeled.

Thus

(57)

If conservative potentials are present,

(58)

11 Ideal rigid constraints and internal work

A perfect rigid body maintains fixed distance between every pair of material points.

Under the ideal rigid body model, constraint forces enforce the geometry without changing the internal separations.

Thus they do no net internal work associated with deformation.

Real materials are never perfectly rigid. They can store elastic energy, vibrate, heat, and dissipate energy. The rigid body model intentionally suppresses those internal degrees of freedom.

This is why the rigid body work-energy theorem can be written using only external work and rigid body kinetic energy.

12 Rolling without slipping

A rigid body rolling without slipping combines translation and rotation.

The no-slip kinematic condition is

(59)

The kinetic energy is

(60)

Using

(61)

we obtain

(62)

Factor:

(63)

13 Static friction in ideal rolling

For pure rolling on a fixed surface, the instantaneous contact point is momentarily at rest relative to the surface.

Therefore the static friction force can act while doing zero instantaneous work at the point of contact:

(64)

Static friction can still be essential dynamically because it produces torque and determines the rotational acceleration.

This is an important distinction:

(65)

This statement applies to ideal rolling on a fixed, nondeforming surface. Moving or deforming contacts require more careful energy accounting.

14 Rolling down a frictionless-energy incline

Consider a rigid body of mass M, radius R, and center-of-mass moment of inertia I_CM rolling without slipping from rest through a vertical drop h.

Static friction does no work in the ideal model, so mechanical energy is conserved:

(66)

Use

(67)

Then

(68)

Factor:

(69)

Therefore

(70)

Hence

(71)

15 Example 1: solid cylinder rolling downhill

For a solid cylinder,

(72)

Then

(73)

Therefore

(74)

Thus

(75)

The corresponding angular speed is

(76)

16 Example 2: solid sphere rolling downhill

For a solid sphere,

(77)

Therefore

(78)

Since

(79)

we obtain

(80)

Thus

(81)

17 Which rolling object reaches the bottom faster?

Write

(82)

Then

(83)

For objects released from the same height, a smaller β gives larger final speed.

Examples:

βhoop	= 1,	(84)
βshell	= ,	(85)
βdisk	= ,	(86)
βsphere	= .	(87)

Therefore the solid sphere has the greatest final speed among these ideal examples because the smallest fraction of its energy is stored in rotation.

18 Example 3: torque accelerating a flywheel

A flywheel has moment of inertia

(88)

A constant torque

(89)

acts through

(90)

The work is

(91)

Thus

(92)

If the flywheel starts from rest,

(93)

Therefore

(94)

so

(95)

19 Example 4: motor power and torque

A motor supplies

(96)

to a shaft rotating at

(97)

Using

(98)

the torque is

τ	=	(99)
	=	(100)
	= 125 N m.	(101)

Thus

(102)

20 Energy method for a rolling body with external work

Suppose a rolling body is acted on by a motor or applied force while changing height.

A general energy equation is

(103)

For rolling,

(104)

Thus

	MV i2 + ICMωi2 + U i + Wnc	(105)
	= MV f2 + ICMωf2 + U f.	(106)

If no slip holds,

(107)

can be used to reduce the number of unknowns.

21 Work of a force applied away from the center of mass

A force applied away from the center of mass can change both translation and rotation.

For a rigid body, the external power can be decomposed as

(108)

The first term changes center-of-mass kinetic energy.

The second changes rotational kinetic energy.

This provides a useful energy interpretation of why an off-center force can simultaneously accelerate the center of mass and spin the body.

22 Three-dimensional rotational kinetic energy

For general three-dimensional rigid body rotation,

(109)

where I_CM is the inertia tensor about the center of mass.

If the body rotates about a principal axis,

(110)

and the expression reduces to

(111)

The full tensor treatment belongs naturally in a later rotational-dynamics article, but the energy structure is already visible here.

23 Common mistakes

1. Using only MV _CM² for a rolling or rotating body.
2. Using only Iω² when the center of mass also translates.
3. Using the wrong moment of inertia axis.
4. Forgetting the parallel-axis theorem for rotation about an offset fixed axis.
5. Writing W = τΔ𝜃 when torque varies strongly with angle.
6. Confusing torque with work; torque has units of N m but is not energy.
7. Forgetting that rotational power is P = τω.
8. Assuming static friction must do negative work during pure rolling on a fixed surface.
9. Using V _CM = Rω when slipping occurs.
10. Treating all rolling objects as having the same final speed from the same height.
11. Forgetting that moment of inertia depends on mass distribution, not only total mass.

24 Practice exercises

1. A solid disk of mass M and radius R rotates about its symmetry axis with angular speed ω. Find its rotational kinetic energy.
2. A solid sphere rolls without slipping at speed v. Write its total kinetic energy entirely in terms of M and v.
3. Starting from the particle-system expression for kinetic energy, derive

   (112)

   for planar rigid body motion.

4. Derive the fixed-axis rotational work-energy theorem from τ = Iα.
5. A constant torque of 5 N m acts through 12 rad. Find the work done.
6. A shaft rotates at 80 rad∕s while transmitting 4.0 kW. Find the torque.
7. A hoop rolls without slipping from rest through vertical drop h. Find its final center-of-mass speed.
8. Repeat the previous problem for a solid disk.
9. Repeat the previous problem for a solid sphere.
10. Rank a hoop, spherical shell, disk, and solid sphere by final speed after rolling from the same height.
11. Explain why static friction can produce torque while doing zero work in ideal pure rolling.
12. Show that K = I_Oω² for rotation about a fixed offset axis using the parallel-axis theorem.
13. A wheel rolls without slipping while a horizontal external force acts at its axle. Write an energy equation relating applied work to translation and rotation.
14. For three-dimensional rotation, explain why K_rot = ω ⋅ Iω reduces to Iω² about a principal axis.
15. Give a physical example where an off-center force adds both translational and rotational kinetic energy.

25 Summary

For planar rigid body motion,

(113)

For fixed-axis rotation,

(114)

Rotational work is

(115)

For constant torque,

(116)

Rotational power is

(117)

For rigid body translation and rotation,

(118)

For rolling without slipping,

(119)

A rolling body’s kinetic energy is

(120)

These results extend the particle-system energy method directly into rigid body mechanics.

The next article, M03-10, develops rotational work, power, and energy transfer in greater depth.

References

References