Week 5: Torque and Rotation in

One Dimension

Summary

•Rotations in One Dimension are rotations of a solid object about a single axis. Since we are free to choose any arbitrary coordinate system we wish in a problem, we can without loss of generality select a coordinate system where the z-axis represents the (positive or negative) direction or rotation, so that the rotating object rotates “in” the xy plane. Rotations of a rigid body in the xy plane can then be described by a single angle θ, measured by convention in the counterclockwise direction from the positive x-axis.

•Time-dependent Rotations can thus be described by:

a)The angular position as a function of time, θ(t).

b)The angular velocity as a function of time,

ω(t) = dθdt

c) The angular acceleration as a function of time,

α(t) = dω = d2θ dt dt2

Hopefully the analogy between these “one dimensional” angular coordinates and their one dimensional linear motion counterparts is obvious.

•Forces applied to a rigid object perpendicular to a line drawn from an axis of rotation exert a torque on the object. The torque is given by:

τ= rF sin(φ) = rF = r F

•The torque (as we shall see) is a vector quantity and by convention its direction is perpendicular

~

to the plane containing ~r and F in the direction given by the right hand rule. Although we won’t really work with this until next week, the “proper” definition of the torque is:

~

~τ = ~r × F

• Newton’s Second Law for Rotation in one dimension is:

τ = Iα

where I is the moment of inertia of the rigid body being rotated by the torqe about a given/specified axis of rotation. The direction of this (one dimensional) rotation is the righthanded direction of the axis – the direction your right handed thumb points if you grasp the axis with your fingers curling around the axis in the direction of the rotation or torque.

235

236	Week 5: Torque and Rotation in One Dimension

• The moment of inertia of a point particle of mass m located a (fixed) distance r from some axis of rotation is:

I= mr2

•The moment of inertia of a rigid collection of point particles is:

X

I = miri2

i

• the moment of inertia of a continuous solid rigid object is:

Z

I = r2dm

•The rotational kinetic energy of a rigid body (total kinetic energy of all of the chunks of mass that make it up) is:

Krot = 12 Iω2

• The work done by a torque as it rotates a rigid body through some angle dθ is:

dW = τ dθ

Hence the work-kinetic energy theorem becomes:

Z

W = τ dθ = Krot

•Consequently rotational work, rotational potential energy, and rotational kinetic energy call all be simply added in the appropriate places to our theory of work and energy. The total mechanical energy includes both the total translational kinetic energy of the rigid body treated as if it is a total mass located at its center of mass plus the kinetic energy of rotation around its center of mass:

Ktot = Kcm + Krot

This is a special case of the last theorem we proved last week.

•If we know the moment of inertia Icm of a rigid body about a given axis through its center of mass, the Parallel Axis Theorem permits us to find the moment of inertia of a rigid body of mass m around a new axis parallel to this axis and displaced from it by a distance rcm:

Inew = Icm + mrcm2

•For a distribution of mass with planar symmetry (mirror symmetry about the plane of rotation or distribution only in the plane of rotation), if we let z point in the direction of an axis of rotation perpendicular to this plane and x and y be perpendicular axes in the plane of rotation, then the Perpendicular Axis Theorem states that:

Iz = Ix + Iy

5.1: Rotational Coordinates in One Dimension

In the last week/chapter, you learned how a collection of particles can behave like a “particle” of the same total mass located at the center of mass as far as Newton’s Second Law is concerned. We also saw at least four examples of how problems involving systems of particles can be decomposed into two separate problems – one the motion of the center of mass, which generally obeys Newtonian

Week 5: Torque and Rotation in One Dimension

237

dynamics as if the whole system is “a particle”, and the other the motion in the center of mass system107.

This decomposition is useful (as we saw) even if the system has many particles in it and is fluid or non-interacting, but it is very useful in helping us to describe the motion of rigid bodies. This is because the most general motion of a rigid object is the translation of (the center of mass of) the object according to the total force acting on it and Newton’s Second Law (as demonstrated last week), plus the rotation of that body about its center of mass as unbalanced forces exert a torque on the object.

The first part we are very very familiar with at this point and we’ll take it for granted that you can solve for the motion of the center of mass of a rigid object given any reasonable net force. The second we are not familiar with at all, and we will now take the next two weeks to study it in detail and understand it, as rotation is just as important and common as translation when it comes to understanding the motion of nearly everything we see on a daily basis. Doors rotate about hinges, tires rotate about axles, electrons and protons “just rotate” because of their intrinsic spin, our fingers and toes and head and arms and legs rotate about their joints, our whole bodies rotate about their center of mass when we get up in the morning, when we do a twirl on ice skates, when we summersault on a trampoline, the entire Earth rotates around its axis while revolving around the sun which rotates on its axis while revolving around the Galactic center which... just goes to show that rotation really is ubiquitous, and pretending that it isn’t important or worthy of understanding is not an option, even for future physicians or non-rocket-scientist bio majors.

It will take two weeks (and maybe even longer, for some of you) because rotation is a wee bit complicated. For many of you, it will be the most di cult single topic we cover this semester, if only because rotation is best described by means of the Evil Cross Product108 . Just as we started our study of coordinate motion with motion in only one dimension, so we will start our study or rotation with “one dimensional rotation” of a rigid body, that is, the rotation of a rigid object through an angle θ about a single fixed axis109.

Eventually we want to be able to treat arbitrary rigid objects, ones that have their mass symmetrically but non-uniformly distributed (e.g. basketballs or ninja stars) or non-uniformly and not particularly symmetrically distributed (e.g. the human body, automobiles, blobs of putty of arbitrary shape). But at the moment even the rotation of a basketball on the tip of a player’s finger seems like too much for us to handle

We therefore start with the simplest possible example – a “rigid” system with all of its mass concentrated in a single point that rotates around some fixed axis. Consider a small “pointlike” ball of mass m on a rigid massless unstretchable rod, portrayed in figure 59. The rod itself is pivoted on a frictionless axle in the center so that the mass is constrained to move only on the dashed circle in the plane of the picture. The mass therefore maintains a constant distance from the pivot – r is a constant – but the angle θ can vary in time as external forces act on the system.

The very first things we need to do are to bring to mind the set of rotational coordinates that

107In particular, we solved elastic collisions in the center of mass frame (where they were easy) while the center of mass of the colliding system obeyed (trivial) Newtonian dynamics, we looked at the exploding rocket where the center of mass followed the parabolic/Newtonian trajectory, we saw that inelastic collisions turn all of the kinetic energy in the center of mass frame into heat, and we proved that in general the kinetic energy of a system in the lab is the sum of the kinetic energy of the system (treated as a particle moving at speed vCM ) plus the kinetic energy of all of the particles in the center of mass frame – this latter being the energy lost in a completely inelastic collision or conserved in an elastic one!

108Wikipedia: http://www.wikipedia.org/wiki/Cross Product. Something that is covered both in this Wikipedia

article and in the online Math Review supplement, so now is a really, really great time to pause in reading this chapter and skip o to refresh your memory of it. It is a memory, we hope, isn’t it? If not, then by all means skip o to learn it...

109The “direction” of a rotation is considered to be along the axis of its rotation in a right handed sense described later below. So a “one dimensional rotation” is the rotation of any object about a single axis – it does not imply that the object being rotated is in any sense one dimensional.

238	Week 5: Torque and Rotation in One Dimension

y

r

s v

θ

r x

Figure 59: A small ball of mass m rotates about a frictionless pivot, moving in a circle of radius r.

we have already introduced for doing kinematics of a rotating object. Since r is fixed, the position of the particle is uniquely determined by the positive angle θ(t), measured by convention as a counterclockwise rotation about the z-axis from the +x-axis as drawn in figure 59. We call θ the angular position of the particle.

We can easily relate r and θ to the real position of the particle. The distance the particle must move in the counterclockwise direction from the standard reference position at (x = r, y = 0) around the circular arc to an arbitrary position on the circle is s = rθ. s (the arc length) is a one dimensional coordinate that describes its motion on the arc of the circle itself, and if we know r and s (the latter measured from the +x-axis) we know exactly where the particle is in the x-y plane.

We recall that the tangential velocity of the particle on this circle is then

vt =	ds	=	d(rθ)	= r	dθ	= rω	(461)
	dt		dt		dt

where we remind you of the angular velocity ω = dθdt . Note that for a rigid body vr = drdt = 0, that is, the particle is constrained by the rigid rod or solidity of the body to move in circles of constant radius r about the center of rotation or pivot so its speed moving towards or away from the circle is zero.

Similarly, we can di erentiate one more time to find the tangential acceleration:

at =	dvt	= r	dω	= r	d2θ	= rα	(462)
	dt		dt		dt2

where α = dωdt = ddt22θ is the angular accleration of the particle.

Although the magnitude of vr = 0, we note well that the direction of ~vt is constantly changing and we know that ar = −v2/r = −rω2 which we derived in the first couple of weeks and by now have used repeatedly to solve many problems.

All of this can reasonably be put in a small table that lets us compare and contrast the one dimensional arc coordinates with the associated angular coordinates:

5.2: Newton’s Second Law for 1D Rotations

With these coordinates in hand, we can now consider the angular version of Newton’s Second Law

~

for a force F applied to this particle as portrayed in figure 60. This is an example of a “rigid” body

Week 5: Torque and Rotation in One Dimension										239
	Angular					Arc Length
	θ					s = rθ
		dθ				ds
	ω =					vt =				= rω
			dt					dt
	α =	dω				at =		dvt		= rα
								dt
			dt

Table 2: Coordinates used for angular/rotational kinemetics in one dimension. Note that θ is the rotation angle around a given fixed axis, in our picture above the z-axis, and that θ must be given in (dimensionless) radians for these relations to be true. Remember C = 2πr is the formula for the circumference of a circle and is a special case of the general relation s = rθ, but only when θ = 2π radians.

y

r F

Ft

φ

Τ Fr x

~

Figure 60: A force F is applied at some angle φ (relative to ~r) to the ball on the pivoted massless rod.

rotation, but because we aren’t yet ready to tackle extended objects all of the mass is concentrated in the ball at radius r. We’ll handle true, extended rigid objects shortly, once we understand a few basic things well.

Since the rod is rigid, and pivoted by an unmovable frictionless axle of some sort in the center, the tension in the rod opposes any motion along r. If the particle is moving around the circle at some speed vt (not shown), we expect that:

v2			2
Fr − T = F cos(φ) − T = −mar = −m	t	= −mrω		(463)
	r

(where r is an outward directed radius, note that the acceleration is in towards the center) as usual.

The rotational motion is what we are really interested in. Newton’s Law tangent to the circle is just:

Ft = F sin(φ) = mat = mrα

(464)

For reasons that will become clear in a bit, we will find it very useful to multiply this whole equation by r and redefine rFt to be a new quantity called the torque, given the symbol τ . We will also collect the factors of r and multiply them by the m to make a new quantity called the moment of intertia and give it the symbol I:

τ = rFt = rF sin(φ) = mr2α = Iα

(465)