Torque and Fixed-Axis Rotational Dynamics

Torque and Fixed-Axis Rotational Dynamics

Rotational kinematics describes how angular position, angular velocity, and angular acceleration change.

Rotational dynamics asks what causes those changes.

The rotational analogue of force is torque. For a force F applied at position r relative to a chosen origin,

(1)

Its magnitude is

(2)

where ϕ is the angle between r and F.

For a rigid body constrained to rotate about a fixed inertial axis, the component of the net external torque along that axis satisfies

(3)

This equation is the fixed-axis rotational analogue of

(4)

The analogy is powerful:

Fnet	↔τnet,	(5)
m	↔I,	(6)
a	↔α.	(7)

But the rotational equation carries an important condition: the axis must be specified, and the relevant torque and moment of inertia must refer to that same axis.

1 Torque as a vector

Torque is defined by the cross product

(8)

Its direction follows the right-hand rule.

Its magnitude is

(9)

Only the component of force perpendicular to r contributes:

(10)

Thus

(11)

Equivalently, define the perpendicular lever arm or moment arm

(12)

Then

(13)

2 Line of action and lever arm

The line of action of a force is the infinite straight line passing through the force vector.

The lever arm is the perpendicular distance from the rotation axis to that line.

This interpretation is often more useful than measuring the distance to the point of application.

If the line of action passes through the axis,

(14)

and therefore

(15)

A large force can produce zero torque if its line of action passes through the axis.

3 Sign convention for planar fixed-axis rotation

For rotation in the xy plane about the z axis, torque has only the relevant axial component

(16)

A common convention is:

• counterclockwise torque is positive,
• clockwise torque is negative.

This is equivalent to the right-hand-rule convention:

• positive τ_z points along +e_z,
• negative τ_z points along −e_z.

The signed torque should be used consistently with signed angular acceleration.

4 Net torque

Several forces can act simultaneously on a rigid body.

The net torque about the chosen origin is

(17)

For fixed-axis planar motion,

(18)

Torques must be summed algebraically, including signs.

5 Torque and angular momentum

The fundamental rotational dynamics law is

(19)

about an inertial origin.

For a rigid body rotating about a fixed z axis, the axial angular momentum is

(20)

Because the body is rigid and the axis is fixed relative to the body geometry,

(21)

Therefore

τext,z	=	(22)
	=	(23)
	= Iz.	(24)

Hence

(25)

6 Why the scalar equation is axial

The full vector relation is

(26)

It is generally not correct to replace this by

(27)

for arbitrary three-dimensional rigid body motion.

For a fixed axis, however, the axial component satisfies

(28)

Other components of external torque can be balanced by constraint forces and moments at bearings or supports.

This distinction becomes essential when the inertia tensor is introduced later in M05.

7 Particle derivation of the fixed-axis equation

The same result can be seen particle by particle.

For mass element m_i at perpendicular radius r_i⊥, tangential acceleration is

(29)

The tangential net force component is

(30)

Its axial torque contribution is

(31)

Summing gives

τz	= ∑ imiri⊥2α	(32)
	= α.	(33)

Since

(34)

we recover

(35)

8 Internal forces and external torque

For the whole rigid body, internal forces occur between particles of the body.

When internal forces satisfy Newton’s third law along the line joining each interacting pair, their torques cancel in the total system torque.

The rotational equation for the body therefore uses the net external torque:

(36)

This is analogous to translational dynamics:

(37)

9 Moment of inertia is the rotational inertia

For a given torque,

(38)

Thus a larger moment of inertia produces a smaller angular acceleration:

(39)

for fixed torque magnitude.

Mass distribution matters as strongly as total mass.

A ring and disk with equal mass and radius respond differently to the same torque because

Iring	= MR2,	(40)
Idisk	= MR2.	(41)

The disk therefore has twice the angular acceleration of the ring under the same axial torque.

10 Example 1: force applied to a disk

A uniform disk has

M	= 4.0 kg,	(42)
R	= 0.30 m.	(43)

A tangential force

(44)

acts at the rim.

The torque magnitude is

(45)

The disk moment of inertia is

I	= MR2	(46)
	= (4.0)(0.30)2	(47)
	= 0.18 kg m2.	(48)

Therefore

α	=	(49)
	=	(50)
	= 10 rad∕s2.	(51)

Thus

(52)

11 Example 2: same force at a different angle

Suppose the same

(53)

force is applied at the rim, but the angle between r and F is

(54)

The torque is

τ	= RF sin 30∘	(55)
	= (0.30)(6.0)(0.5)	(56)
	= 0.90 N m.	(57)

Thus

(58)

Only the perpendicular component of the applied force contributes to the torque.

12 Gravity acting on an extended rigid body

In a uniform gravitational field, the total gravitational force on an extended body can be treated as acting at the center of mass for the purpose of computing net torque.

If the center of mass position relative to a pivot is

(59)

then

(60)

The torque magnitude is

(61)

Here ϕ is the angle between r_CM and g.

13 Example 3: uniform rod pivoted at one end

A uniform rod of length L and mass M is horizontal and pivoted at one end.

Gravity acts at the center of mass,

(62)

The gravitational torque magnitude is

(63)

The moment of inertia about the pivot is

(64)

Therefore the initial angular acceleration magnitude is

|α|	=	(65)
	=	(66)
	= .	(67)

Thus

(68)

The mass cancels.

14 Pivot and bearing forces

A fixed pivot or bearing can exert forces on a rotating body.

If the torque origin is chosen at the pivot and the pivot force acts through that point, then its torque about the pivot is zero:

(69)

This is one reason pivot points are often convenient torque origins.

However, a real bearing can also exert a reaction couple or Friction torque.

A force passing through the axis has zero moment arm, but a pure applied couple need not.

15 Constraint reactions and full force balance

The equation

(70)

does not replace translational force balance.

A body rotating about a fixed axis generally has a moving center of mass unless the center of mass lies on the axis.

Therefore support forces may be required to produce the center of mass acceleration.

A complete rigid body problem may require both:

(71)

and

(72)

16 Distributed forces

If force is distributed continuously over a body, the differential torque is

(73)

The total torque is

(74)

For a tangential force per unit length f(s) acting along a rim or rod,

(75)

and the torque can be integrated directly.

A single resultant force can replace a distributed load for torque calculations only if its line of action is chosen to preserve the same net moment.

17 Couples

A couple consists of two equal and opposite parallel forces whose lines of action are separated.

The net force is zero:

(76)

But the net torque is nonzero.

If the perpendicular separation between the force lines is d,

(77)

The torque of a pure couple is independent of the choice of origin.

This makes a couple fundamentally different from a single force.

18 Tension and pulley torque

Consider a Pulley of radius R and moment of inertia I.

If tensions T₁ and T₂ act tangentially on opposite sides, the net pulley torque is

(78)

Thus

(79)

If a string does not slip on the pulley,

(80)

This couples translational dynamics of the masses to rotational dynamics of the pulley.

19 Example 4: two masses and a massive pulley

Masses m₁ and m₂ hang from a light string over a pulley of radius R and moment of inertia I.

Assume

(81)

and no slipping.

For m₁ moving upward,

(82)

For m₂ moving downward,

(83)

For the pulley,

(84)

Since

(85)

(86)

Add the two mass equations:

(87)

Substitute the pulley relation:

(88)

Therefore

(89)

The pulley inertia reduces the acceleration compared with an ideal massless pulley.

20 Why the two tensions differ

If the pulley were massless,

(90)

Then the pulley equation gives

(91)

For a massive pulley with nonzero angular acceleration,

(92)

so

(93)

The tension difference is exactly what produces the pulley’s rotational acceleration.

21 Constant net torque

If the moment of inertia is constant and the net axial torque is constant,

(94)

then

(95)

is constant.

The constant-angular-acceleration equations from M05-01 apply:

ω	= ω0 + αt,	(96)
𝜃 − 𝜃0	= ω0t + αt2,	(97)
ω2	= ω 02 + 2α(𝜃 − 𝜃 0).	(98)

Thus a constant net torque produces the rotational analogue of constant linear acceleration.

22 Example 5: constant motor torque

A rotor has

(99)

and is driven from rest by constant torque

(100)

Its angular acceleration is

(101)

After

(102)

(103)

Its angular displacement is

(104)

23 Braking torque

A braking torque acts opposite the angular velocity.

For constant moment of inertia,

(105)

If the torque opposes positive rotation,

(106)

and

(107)

For constant braking torque, the stopping time is

(108)

The stopping angle follows from

(109)

Hence

(110)

24 Example 6: braking a flywheel

A flywheel has

(111)

and initial angular speed

(112)

A constant braking torque of magnitude

(113)

is applied.

Then

(114)

The stopping time is

(115)

The stopping angle is

Δ𝜃	=	(116)
	=	(117)
	= 100 rad.	(118)

Thus

(119)

25 Rotational work

For a small angular displacement

(120)

a torque does work

(121)

for fixed-axis motion.

Therefore

(122)

For constant torque,

(123)

26 Work-energy check

Start from

(124)

Use

(125)

Then

(126)

Integrate:

W	= ∫ ωiωf Iω dω	(127)
	= Iωf2 −Iωi2.	(128)

Thus

(129)

This agrees with the rotational work-energy relation developed in M03.

27 Rotational power

Power is

(130)

Since

(131)

P	= τz	(132)
	= τzω.	(133)

Therefore

(134)

for fixed-axis motion.

In vector form for an applied torque and angular velocity,

(135)

28 Angular impulse

From

(136)

integrate over a time interval:

(137)

Thus the angular impulse is

(138)

For constant torque,

(139)

For a fixed rigid body axis,

(140)

Therefore

(141)

when torque is constant.

29 Variable torque

If torque depends on time,

(142)

then

(143)

For constant I,

(144)

If torque depends on angular position,

(145)

the work-energy form is often more convenient:

(146)

30 Example 7: linearly increasing torque

Suppose

(147)

with constant

(148)

For constant I,

(149)

Integrating from rest,

ω(t)	= ∫ 0tt′dt′	(150)
	= t2.	(151)

Then

𝜃(t) − 𝜃0	= ∫ 0tω(t′) dt′	(152)
	= ∫ 0t(t′)2 dt′	(153)
	= t3.	(154)

31 Static rotational equilibrium

If a body is in rotational equilibrium about a fixed axis,

(155)

Therefore

(156)

If the center of mass is also not accelerating,

(157)

Both translational and rotational equilibrium conditions are required for complete static equilibrium.

32 Dynamic rotational equilibrium

A body can have

(158)

while still rotating.

Then

(159)

and

(160)

Zero net torque means constant angular velocity, not necessarily zero angular velocity.

33 Torque about different origins

Torque depends on the chosen origin:

(161)

Changing the torque origin changes r and generally changes the torque.

Therefore every torque calculation should state or make clear:

(162)

A pure couple is a special exception: its net torque is independent of origin.

34 Axis reactions

A bearing constraining a rigid body to a fixed axis can supply reaction forces and possibly reaction moments.

The equation

(163)

uses the component of net external torque along the allowed rotation axis.

Constraint reactions perpendicular to the allowed rotational degree of freedom can enforce the fixed-axis condition.

35 When τ = Iα is not enough

The scalar relation

(164)

is appropriate when:

• the body is rigid,
• the rotation axis is fixed in an inertial frame,
• I_z is constant,
• torque and angular acceleration are referred to the same axis.

More general situations require additional mechanics.

Examples include:

• changing mass distributions,
• free three-dimensional rotation,
• precession,
• moving rotation axes,
• deformable bodies.

For these problems the full vector angular-momentum equation remains fundamental:

(165)

36 Connection to rolling

A rolling body combines translation of the center of mass with rotation.

The rotational equation remains useful:

(166)

for planar rigid body rolling under the appropriate conditions, while translation obeys

(167)

The no-slip kinematic condition then couples the two motions.

That is the subject of M05-07.

37 Common mistakes

1. Using the full force magnitude in τ = rF when the force is not perpendicular to r.
2. Measuring the lever arm to the force application point instead of perpendicular to the force line of action.
3. Forgetting the sign of clockwise versus counterclockwise torque.
4. Mixing torques calculated about different origins.
5. Using a moment of inertia about an axis different from the torque axis.
6. Writing τ = Iα as a universally valid three-dimensional vector law.
7. Forgetting that only the axial component gives τ_z = I_zα in the fixed-axis scalar formulation.
8. Including pivot force torque about the pivot when its line of action passes through the pivot.
9. Assuming a real bearing can never exert a friction torque or reaction couple.
10. Using T₁ = T₂ for a massive accelerating pulley.
11. Forgetting the no-slip relation a = αR in pulley problems.
12. Assuming zero net torque means zero angular velocity.
13. Ignoring translational force balance when the center of mass accelerates.
14. Forgetting that a couple can have zero net force but nonzero torque.
15. Using constant-angular-acceleration equations when torque or moment of inertia varies.

38 Practice exercises

1. A force of magnitude F is applied at distance r from an axis at angle ϕ to the radius vector. Derive the torque magnitude.
2. Explain geometrically why a force whose line of action passes through the axis produces zero torque.
3. Derive τ_z = I_zα from L_z = I_zω for a fixed rigid body axis.
4. Derive τ_z = I_zα by summing particle tangential equations.
5. A disk of mass M and radius R is acted on by a tangential rim force F. Find α.
6. Repeat the previous problem when the force makes angle 30^∘ with the radius vector.
7. A uniform rod of length L is horizontal and pivoted at one end. Derive its initial angular acceleration under gravity.
8. A rotor has I = 1.5 kg m² and a constant net torque of 6 N m. Starting from rest, find ω after 4 s.
9. Derive the acceleration of two hanging masses connected by a string over a massive pulley.
10. Explain why the tensions on the two sides of a massive accelerating pulley are unequal.
11. A flywheel with moment of inertia I and initial speed ω₀ is stopped by a constant opposing torque τ_b. Derive the stopping time and stopping angle.
12. Show that integrating τ_z = Iα over angle produces the rotational work-energy theorem.
13. Show that integrating torque over time produces angular impulse.
14. A torque varies as τ(t) = kt. Starting from rest, derive ω(t) and 𝜃(t).
15. Explain why ∑ τ_z = 0 does not imply ω = 0.

39 Summary

Torque is

(168)

Its magnitude can be written

(169)

The fundamental angular-momentum law is

(170)

For a rigid body rotating about a fixed inertial axis,

(171)

Rotational work and power are

W = ∫ τz d𝜃,		(172)
P = τzω.		(173)

Angular impulse is

(174)

Together with

(175)

these equations provide the dynamics foundation for rigid body rolling, developed next in M05-07.

References

References