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246

Week 5: Torque and Rotation in One Dimension

5.3.2: Table of Useful Moments of Inertia

Finally, here is a table of a few useful moments of inertia of simple uniform objects. In each case I indicate the value about an axis through the symmetric center of mass of the object, because we can use the parallel axis theorem and the perpendicular axis theorem to find the moments of inertia around at least some alternative axes.

Shape

Icm

Rod from −L/2 to L/2

1

M L2

 

 

 

 

12

 

 

 

 

Ring

M R2

 

 

 

 

Disk

 

1

M R2

2

 

 

 

 

 

 

Sphere

 

2

M R2

5

 

 

 

 

 

 

Spherical Shell

 

2

M R2

3

 

 

 

 

 

Generic “Round” Mass of Mass M and radius R

βM R2

 

 

 

 

 

 

Table 3: A few useful moments of inertia of symmetric objects around an axis of symmetry through their center of mass. You should probably know all of the moments in this table and should be able to evaluate the first three by direct integration.

5.4: Torque as a Cross Product

This section will be rather abbreviated this week; next week we will cover it in gory detail as a vector relation. For the moment, however, we need to make a number of observations that will help us solve problems. First, we know that the one-dimensional torque produced by any single force acting on a rigid object a distance r from a pivot axis is just:

τ = rF = rF sin(φ)

(487)

where F is just the component of the force perpendicular to the (shortest) vector ~r from the pivot axis to the point of application. This is really just one component of the total torque, mind you, but it is the one we have learned so far and are covering this week.

~

First, let’s make an important observation. Provided that ~r and F lie in a plane (so that the one dimension is the right dimension) the magnitude of the torque is the magnitude of the cross

~

product of ~r and F :

~

(488)

τ = |~r × F | = rF sin(φ) = rF = r F

I’ve used the fact that I can move the sin(φ) around to write this in terms of:

r = r sin(φ)

(489)

which is the component of ~r perpendicular to F , also known as the moment arm of the torque. This is a very useful form of the torque in many problems. It it equally well expressible in terms of the familiar:

F = F sin(φ),

(490)

Week 5: Torque and Rotation in One Dimension

247

~

the component of F perpendicular to ~r. This form, too, is often useful. In fact, both forms may be useful (to evaluate di erent parts of the total torque) in a single problem!

If we let the vector torque be defined by:

~

(491)

= ~r × F

marvelous things will happen. Next week we will learn about them, and will learn about how to evaluate this a variety of ways. For now let’s just learn one.

~

The vector torque has a magnitude |~r × F | = rF sin(φ) and points in the direction given by the right hand rule.

The right hand rule, in turn, is the following:

~ ~

The direction of the vector cross product A × B is in the direction the thumb of your right hand points when you begin with the fingers of your right hand lined up with the

~ ~

vector A and then curl them naturally through the angle φ < π into the direction of B.

That is, if you imagine “grasping” the axis in the direction of the torque with your right hand, your

~

fingers will curl around in the direction from ~r to F through the smaller of the two angles in between them (the one less than π).

You will get lots of practice with this rule, but be sure to practice with your right right hand, not your wrong right hand. Countless students (and physics professors and TAs!) have been embarrassed be being caught out evaluating the direction of cross products with their left hand111. Don’t be one of them!

The direction of the torque matters, even in one dimension. There is no better problem to demonstrate this than the following one, determine what direction a spool of rope resting on a table will roll when one pulls on the rope.

Example 5.4.1: Rolling the Spool

I’m not going to quite finish this one for you, as there are a lot of things one can ask and it is a homework problem. But I do want you to get a good start.

The spool in figure 67 is wrapped many times around with string. It is sitting on a level, rough

~ ~

table so that for weak forces F it will freely roll without slipping (although for a large enough F of course it will slip or even rise up o of the table altogether).

The question is, which direction will it roll (or will it not roll until it slides) for each of the three directions in which the string is pulled.

The answer to this question depends on the direction of the total torque, and the relevant pivot is the point that does not move when it rolls, where the (unknown!) force of static friction acts. If we choose the pivot to be the point where the spool touches the table, then gravity, the normal

~

force and static friction all exert no torque! The only source of torque is F .

So, what is the direction of the torque for each of the three forces drawn, and will a torque in that direction make the mass roll to the left, the right, or slide (or not move)?

Think about it.

111Let he or she who is without sin cast the first stone, I always say. As long as it is cast with the right hand...