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Kurs: Fizyka > Rozdział 7

Lekcja 2: Momenty, moment siły i moment pędu

Moment bezwładności

Learn how the distribution of mass can affect the difficulty of causing angular acceleration.

What is rotational inertia?

Rotational inertia is a property of any object which can be rotated. It is a scalar value which tells us how difficult it is to change the rotational velocity of the object around a given rotational axis.

Rotational inertia plays a similar role in rotational mechanics to mass in linear mechanics. Indeed, the rotational inertia of an object depends on its mass. It also depends on the distribution of that mass relative to the axis of rotation.

When a mass moves further from the axis of rotation it becomes increasingly more difficult to change the rotational velocity of the system. Intuitively, this is because the mass is now carrying more momentum with it around the circle (due to the higher speed) and because the momentum vector is changing more quickly. Both of these effects depend on the distance from the axis.

Rotational inertia is given the symbol

I

. For a single body such as the tennis ball of mass

m

(shown in Figure 1), rotating at radius

r

from the axis of rotation the rotational inertia is

I = m r^{2}

and consequently rotational inertia has SI units of

kg \cdot m^{2}

Rotational inertia is also commonly known as moment of inertia. It is also sometimes called the second moment of mass; the 'second' here refers to the fact that it depends on the length of the moment arm squared.

How does rotational inertia relate to Newton's 2ⁿᵈ law?

Rotational inertia takes the place of mass in the rotational version of Newton's 2ⁿᵈ law.

Consider a mass

m

attached to one end of a massless rod. The other end of the rod is hinged so that the system can rotate about the central hinge point as shown in Figure 2.

We now start rotating the system by applying a tangential force

F_{T}

to the mass. From Newton’s 2ⁿᵈ law,

F_{T} = m a_{T}

this can also be written as

F_{T} = m (r α)

Newton's 2ⁿᵈ law relates force to acceleration. In rotational mechanics torque

τ

takes the place of force. Multiplying both sides by the radius gives the expression we want.

\begin{aligned} F_{T} r & = m (r α) r \\ τ & = m r^{2} α \\ τ & = I α \end{aligned}

This expression can now be used to find the behavior of a mass in response to a known torque.

Zadanie 1a:

A motor capable of producing a constant torque of $100 Nm$ ‍ and a maximum rotation speed of $150 rad / s$ ‍ is connected to a flywheel with rotational inertia $0.1 kg m^{2}$ ‍. What angular acceleration will the flywheel experience as the motor is switched on?

Exercise 1b:

How long will the flywheel take to reach a steady speed if starting from rest?

How can we calculate rotational inertia in general?

Often mechanical systems are made of many masses connected together, or complex shapes.

It is possible to calculate the total rotational inertia for any shape about any axis by summing the rotational inertia of each mass.

\begin{aligned} I & = m_{1} r_{1}^{2} + m_{2} r_{2}^{2} + \dots \\ = Σ m_{i} r_{i}^{2} \end{aligned}

Zadanie 2a:

Consider the object shown in figure 3(a). What is its rotational inertia?

Zadanie 2b:

Consider the alternate case of Figure 3(b) of the same system rotating about a different axis. What would you expect the rotational inertia to be in this case?

How can we find the rotational inertia of complex shapes?

For more complicated shapes, it is generally necessary to use calculus to find the rotational inertia. However, for many common geometric shapes it is possible to find tables of equations for the rotational inertia in textbooks or other sources. These typically give the moment of inertia for a shape rotated about its centroid (which often corresponds with the shapes center of mass).

For example, the rotational inertia of a solid cylinder with radius

r

rotated about a central axis is

I = \frac{1}{2} m r^{2}

and for a hollow cylinder with inner and outer radii

r_{i}

and

r_{o}

respectively,

I = \frac{m (r_{i}^{2} + r_{o}^{2})}{2}

Expressions for other simple shapes are shown in Figure 4.

Complex shapes can often be represented as combinations of simple shapes for which there exists a known equation for rotational inertia. We can then combine these rotational inertia to find that of the composite object.

The problem that we will likely run into when combining simple shapes is that the equations tell us the rotational inertia as found about the centroid of the shape and this does not necessarily correspond to the axis of rotation of our composite shape. We can account for this using the parallel axis theorem.

The parallel axis theorem allows us to find the moment of inertia of an object about a point

o

as long as we known the moment of inertia of the shape around its centroid

c

, mass

m

and distance

d

between points

o

and

c

I_{o} = I_{c} + m d^{2}

Exercise 3:

If the shape shown in Figure 5 is made by welding three $10 mm$ ‍ thick metal discs (each with mass $50 kg$ ‍) to a metal ring with mass $100 kg$ ‍. If rotated about a central axis (out of the page), what is the rotational inertia of the object?
Rozwiązanie:
We begin by finding the rotational inertia of the four components separately.
Using the previously given equation for a hollow cylinder, we can find the rotational inertia $I_{b}$ ‍ of the big disc. This centroid of this component already coincides with the rotational axis of the final part so no correction is necessary.
$\begin{aligned} I_{b} & = \frac{1}{2} m (r_{i}^{2} + r_{o}^{2}) \\ = \frac{1}{2} (100 kg) \cdot ({0.75}^{2} + 1^{2}) m^{2} \\ ≃ 78.125 kg \cdot m^{2} \end{aligned}$ ‍
Using the same procedure for the smaller discs, for now centered on their own rotational axes:
$\begin{aligned} I_{s} & = \frac{1}{2} m r_{c}^{2} \\ = \frac{1}{2} \cdot (50 \cdot kg) (0.30 m)^{2} \\ = 2.25 kg \cdot m^{2} \end{aligned}$ ‍
Because each of these three discs are rotating about a point distance $d$ ‍ away from their respective centroids we can use the parallel axis theorem to find the effective rotational inertia $I_{s}^{'}$ ‍.
$d = \frac{1}{2} (1 + 0.75) = 0.875 m$ ‍
$\begin{aligned} I_{s}^{'} & = I_{s} + m d^{2} \\ = (2.25 kg \cdot m^{2}) + (50 kg) \cdot (0.875 m)^{2} \\ ≃ 40.5 kg \cdot m^{2} \end{aligned}$ ‍
As each of the three small discs are at the same radius from the rotational axis of the part, we can treat them as one part with three times the rotational inertia. Finally, the rotational inertial of the object is:
$\begin{aligned} I & ≃ 78 + 3 \cdot 40.5 \\ ≃ 200 kg \cdot m^{2} \end{aligned}$ ‍

Where else does rotational inertia come up in physics?

Rotational inertia is important in almost all physics problems that involve mass in rotational motion. It is used to calculate angular momentum and allows us to explain (via conservation of angular momentum) how rotational motion changes when the distribution of mass changes. It also is needed to find the energy which is stored as rotational kinetic energy in a spinning flywheel.

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