OBJECTIVE:  I will be able to use the Moment Of Inertia (M.O.I.) to describe how an object will rotate about a fixed axis after today's class.

WORDS FOR TODAY:

• centrifugal acceleration (inaccurate)
• centripetal acceleration
• centripetal force (accurate but can be misleading)
• tangential velocity (flying off the disk (meters/second))
• angular speed (internal rotation (radians/second)
• angular acceleration

• moment of inertia (a means to determine rotation about a fixed axis)

FORMULAE:

There are a veritable FLOOD of new terms to get comfortable with in this unit... so let's start NOW:

Term	Formula	SI units	Description	Notes
radian	θ	radians	2π (in radians) = 360o	1 radian = 57.3 degrees
period	T	seconds	the period = time for one full rotation
angular speed	ω or ∆θ/dt or dθ/dt	radians/sec	velocity at any radial distance "r" of a rotating object	angles ALWAYS described in radians
tangential velocity	v = ωr	m/s	linear velocity at any radial distance "r"
centripetal acceleration	ac = v2/r or rω2	m/s2	acceleration of an object following a circular path	Be careful -- radians (by definition) are unitless
Arc length	s		measured in meters
angular acceleration	α	d2θ/dt2 or dω/dt	at = rα
Linear Torque	dFsinθ or RFsinθ	nm	d = displacement through which the force acts	Torque IS NOT WORK
Rotational Torque	RF	nm	R = radius through which the force acts	Torque IS NOT WORK
Moment of Inertia	various	kgm2	see notes
Moment of Inertia	∑τ = Fr = Iα
Moment of Inertia	I =∫r2dm	kgm

BUT WAIT!!! THERE'S MORE!

WORK O' THE DAY:

Welcome to the wonderful world of the 'moment of inertia'

Anytime an object is rotated, a certain force must be applied to that object in order to induce that object to rotate. However, force isn't the only variable to consider as we must also consider how far away from the pivot point the force is applied and the mass of the object we are trying to rotate (*whew*)

In mathematical terms, we often relate Moment Of Inertia (I'm capitalizing for emphasis) to torque as follows:

∑τ = Fr = Iα

or we can qualitative summarize that as:

"The sum of the torques acting on a rotating object is equal to the force acting on that object times the radius over with that force is acting"

which is also equal to:

"The moment of inertia of that object times the angular acceleration that object experiences"

YIKES.....

Let's take a look at the following chart (as shown on page 303)

Notice there is NO fixed point that is the Moment Of Inertia so in that regard it is nothing at all like the center of mass.

Instead, think of Moment Of Inertia as a way to help us understand HOW a rotating object will behave when sent in motion about a fixed axis.

Let's see if we can de-mystify that a bit with a (hopefully) somewhat more simple example.

Consider the case of a thin rod. We CHOOSE to explore what it will take to make that rod rotate about a particular point as shown below:

Let's say we want to determine the amount of force it will take to get a 1.00 meter long, thin rod with mass of 1.00 kg, spinning about a fixed central point at 2 radians/sec²

The first thing we do is to LOOK UP Moment Of Inertia from our handy dandy table:

For a long thin rod rotating about a central point, the Moment Of Inertia is found to be:

(1/12)ML²

as you might expect, M = mass (in kgs), and L = Length (in meters)

Take a moment to recall our working definition of Moment Of Inertia:

Moment of inertia helps us understand the relationship between mass, length and angular acceleration for an object rotating about a fixed point.

As always, we begin with our defining formula:

∑τ = Fr

∑τ = Iα

So Fr = Iα

(oddly enough, I prefer to write these all down at once as:

∑τ = Fr = Iα

So... let's revisit our original investigation: "How much force must we apply to this long thin rod to get it rotating about a fixed central point at 2.00 radians/sec?"

Now let's get to work:

Fr = Iα

Rewriting that we get:

F = Iα/r

Now substituting for the appropriate MOI:

F = (1/12)(ML2)α/r

Now substituting for the appropriate MOI:

F = (1/12)(1.00 kg)(1.00 m²)(α) / r

Now substituting for the appropriate α and r:

F = (1/12)(1.00 kg)(1.00 m²)(2.00 rad/sec²) / (.5m)

F = .332 N (I hope!)

In otherwords, if we want to get a long thin rod of mass 1.00 kg and length 1.00 m to rotate about a fixed, central point at 2.00 rad/sec, we must apply a force of .332 to the end of the rod in a clockwise direction!

YIKES!

To reiterate, MOI let's us evaulate how an object will rotate about a fixed point (or axis).

Now let's see how this works if we pivot about an end point of that same rod... do you suppose it will take more or less force to do that?

Why is r now L and not .5 L as before?

Let's mush on in a bit more abbreviate fashion... As before we want to know how much force it will take to make this long thin rod pivot about a fixed point (at this point 0 cm) so let's start with our guiding formula:

Fr = Iα

Rewriting that we get:

F = Iα/r

Now substituting for the appropriate MOI:

F = (1/3)(ML²)α/r

Now substituting for the appropriate remaining terms:

F = (1/3)(1.00 kg)(1.00 m²)(2.00 rad/sec²) / (1.00m)

F = .67 N (I hope!)

So twice as much for is required...which kinda makes sense, vous avez?

(please check my math!)

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So...by way of final analysis. MOI let's us knit together the relationship between:

1) pivot position

2) radial arm length

3) force

4) angular acceleration

NOTES:

• MOI has standard international units of kgm²
• MOI does NOT dictate a specific point in space, it is all about the relationships between the 4 items listed above.

1. Step through example 10.4 (recall that the mass of the rod acts through a center of mass with radius = 1/2 L)
2. ignore examples 10.5 - 10.6
3. skim through section 10.6

Review this:

Start with equatio 10.20:

I =∫r²dm

"The moment of inertial of an object can be found by integrating the radius of the object with respect to the mass of the object"

Look familiar? Let's take a go for a long, thin rod (yes we already know the answer but let's take a look at the actual calculation):

1) recollect from our previous unit that when dealing with dm considerations that we MUST get mass in terms of length, width and height.

2) The good news is that for a 'long, thin rod' we have only one dimension, length

so let's begin with the defining formula:

I =∫r²dm

• Now let's get dm in terms of x. Remember that we must work in terms of linear density λ
• We know that λ represents mass/length
• λ multiplied by x (length) gives us mass so.... dm = λdx
• rewrite our equation with linear density values (substitute x for r and λxdx for dm):

I =∫x²λdx

in this case, the long thin rod is of uniform density so we can pull λ out of the integral:

I = λ∫x²dx

solve and evaulate over the length of the rod to each side:

I = λx³ from +L/2 to - L/2

I = λ(L³/8)/3 - (-λL³/8)/3)

I = (2λL³/8)/3

I = 1/12λL³

However, for a long, UNIFORM rod, the linear density is simple M/L so drop that in and VOILA!

I = 1/12ML²

NOTE: If the rod was NOT of uniform density and we had the actual value for λ in terms of x, then we could simply plug and chuck with that value instead of washing it out at the end!

You can take a look at 10.8 if you're a glutton for mathematical punishment, but I won't ask you to (HAH!)

If time permits... the INFAMOUS PARALELL AXIS THEOREM!!!

 

 

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HOMEWORK: Problem 27

ACK!!! NOT 28) (if you didn't do it last night, don't worry though it's pretty simple-- beware of the distractor)

HINT: Remember that what I'm calling 'linear' torque is dfsinθ and rotational torque is simply RF

29, 31 & 34