Formal

1.

OBJECTIVE 1.

To understand the effect of moment of inertia rotational motion of rigid bodies.

2.

To determine experimentally the moment of inertia of some objects about an axis.

2.

THEORY

3.1

Moment of Inertia & Radius of Gyration Given that a body rotates about an axis AA' (Fig 1a), and that its mass is made up of small elements of mass ∆m 1 , ∆m 2 , ∆m 3 …∆m n ; its moment of inertia is given by r 2 m, and the resistance offered by the body is measured by the sum r 1 2 ∆ m 1 + r 2 2 ∆ m 2 + r 3 2 ∆ m 3 + ... r n 2 ∆ m n . By taking the number of elements over a domain Ω of a body, the moment of inertia can be expressed as :

I = ∫ r 2 dm

(1) Ω

We then consider the radius of gyration r of the body to be defined as the distance at which the entire mass of the body should be concentrated at if its moments of inertia with respect to the axis of rotation AA' is to remain unchanged. We can then express moment of inertia

I as:

I = k2m Figure 1a

3.2

(k is a constant)

(2)

Figure 1b

Parallel Axis Theorem The moment of inertia of a body with respect to any axis ZZ' (Fig.1b) can be expressed as the sum of its inertia about an axis GG' parallel to ZZ' through its centre of gravity (CG), and the square of the perpendicular distance d between the axes :

-1-

I Z = I G + md 2

(3)

where IZ is the moment of inertia of the body about axis ZZ' and I G is the moments of inertia about the axis GG' passing through the centre of gravity. This theorem is useful in calculating the moment of inertia of a complex shape which has been divided into a collection of simple ones.

3.3

THE ROTATING-MASS ASSEMBLY The moment of inertia of objects can be experimentally determined by using the Rotatingmass Assembly method. Figure 2 shows the experimental set-up of the experiment. The total moment of inertia of the system in rotation as shown can be expressed as a function of r , which is the distance of each mass M from the axis of rotation. Treating the masses attached on the horizontal bar as point masses, the moment of inertia I for the system can be expressed as :

where

I = 2 M r2 + Ih + Iv + Ip

(4)

I = 2 M r2 + Ic

(5)

2 M r 2 = moment of inertia of the two masses on the bar, Ih Iv Ip Im Ic

= = = = =

moment of inertia of the horizontal bar, moment of inertia of the vertical spindle, effective moment of inertia of the pulley, effective moment of inertia of mass m, I h + I v + I p + I m = constant. Figure 2 Experimental Setup

The dynamics equation of the system is :

τ = Iα

(6)

where α is the angular acceleration of the system and τ is the net torque on the system. The torque is applied using the mass m attached to the string wrapping around the spindle and dropping through a distance y:

-2-

τ = r s mg α = a / rs

and

(7) (8)

where a is the acceleration of the applied mass. Since the mass is released from rest and falls through a distance of y, its acceleration can be calculated using :

a = v 2 / (2y)

(9)

where v is the velocity of the mass at the end of the fall. The angular acceleration of the system can thus be expressed as:

α = v 2 / (2yr s )

(10)

Substituting equations (7) and (10) into (6), the moment of inertia I takes the form:

I = (2r s 2 y m g) / (v 2 )

(11)

Velocity of the mass passing by the photocell t is given by:

v = d/t (12) The experimental value of the moment of inertia can be determined by substituting equations (7) and (10) into (12). It is hence expressed as : I exp = (2r s 2 ymgt 2 ) / d 2

(13)

3.4 TRIFILAR SUSPENSION The trifilar suspension method shown in Fig. 3 can be used to determine the mass and moment of inertia of an object more accurately. The platform of the apparatus was displaced through a small angle θ and set in oscillation. Taking the moments about the z axis :

-r(mp + m)gsinφ - (Ip + I) α = 0

(14)

where φ is the angular displacement of the wires, α is the angular acceleration of the platform, Ip and I are the moments of inertia of the platform and the object respectively. For small angle displacements, say β , sin β ≈ β , ∴ φ = (rθ) / L

-3-

(15)

Substituting equation (15) into (14),

α + (mp + m)gr2 θ = 0 (Ip +I)L Since the motion is simple harmonic, the period of the motion is :

Hence the moment of inertia of the empty platform is:

I p = T p 2 (m p gr 2 ) / 4π2 L where T p is the period of oscillation for the empty platform. The moment of inertia of the experimental objects is then： (18)

-4-

(17)

I = [T 2 (m p + m )gr 2 / 4π2 L] - I p