Lab 7 - Rotational Inertia

Introduction

When you were younger, you might have had toy cars that you rubbed quickly across the floor several times to "rev them up." Then, when you put them down, they would go zipping along for some distance. Or perhaps you have seen an auto accident on television where the wheels of the overturned car continued to turn for a little while. Maybe you have watched a helicopter land, and have noticed that the blades continue to rotate after the pilot turns the engine off. All of these are examples of rotational inertia. The toy car has a little wheel inside, called a flywheel, which is attached to the car's wheels. When you "rev it up" the flywheel begins to spin. Then, when you let the car go, the flywheel continues to spin because of rotational inertia, and so turns the wheels. The wheels on the overturned car and the helicopter blades both continue to turn until friction overcomes their rotational inertia. You have already learned about inertia in linear systems; an object in motion will continue in motion as long as there is no net force acting on it. Rotational inertia is a similar concept applied to objects whose motion is rotational instead of linear. The rotational inertia of an object is that object's resistance to a change in its angular velocity. That is, if an object is spinning, it has a tendency to keep spinning until a net torque acts upon it. This is just another manifestation of Newton's second law.

Discussion of Principles

In Newton's second law,

F = ma 

the mass m of an object is a measure of its inertia. Clearly the smaller the mass, the less force is required to change the object's linear velocity. In rotational motion, it is the rotational inertia, often called the moment of inertia I that determines the torque

τ, 

required to change an object's angular velocity

ω. 

The analogue of Newton's second law for rotational motion is where

τ_net 

is the net torque and

α 

is the angular acceleration. The moment of inertia of an object depends on the shape of the object and the distribution of its mass relative to the object's axis of rotation. A uniform disk of mass m is not as hard to set into rotational motion as a "dumbbell" with the same mass and radius. For a symmetric, continuous body (like a solid disk) that is rotating about an axis of symmetry, e.g., an axle through the center and perpendicular to the disk, the moment of inertia is calculated by carrying out the integral where

dm 

is the tiny bit of mass located a distance r from the axis of rotation. For a collection of n point masses, the moment of inertia is calculated by carrying out the sum where

m_i 

and

r_i 

are the mass and position, respectively of the ith particle. Thus, for the dumbbell

n = 2 

and with mass m at each end, seperated by a distance d, the moment of inertia is For moments of inertia of different symmetrical solids see this list. In this experiment you will use a disk of nearly uniform mass and apply a torque by adding weight to a string attached to a step pulley at the center of the disk. See Fig. 1. Using your results, you will determine the moment of inertia of the disk. Even though the wheel is horizontal and the string passes over a second pulley, in Fig. 1a we show the wheel in a vertical position. As the hanging mass falls, the wheel rotates counterclock-wise. This rotation is due primarily to the torque exerted by the tension force in the string, which arises from the weight of the hanging mass. The wheel has bearings at its axle, and there is some friction in their movement. The torque associated with that friction force f acts to oppose the torque produced by the tension force T in the string. If we consider just the wheel, we can examine the two torques that are acting, that of the friction

τ_f  

and that of the tension

τ_tension.  

The magnitude of the torque due to the tension is where r is the radius of the step pulley. The net torque is the vector sum of these two torques Applying Eq. (1)

τ_net = I α 

 we obtain Here we follow the sign convention that counterclockwise torques are positive and clockwise torques are negative. Now let us consider the hanging mass m. Figure 2 shows the free body diagram for the hanging mass. We see that the tension is opposing the force of gravity. Newton's second law written for the hanging mass is then where a is the acceleration of the hanging mass, and we have chosen the downward direction to be positive. Solving for T, If we assume that the string doesn't slip, then a is also the tangential acceleration of a point on the step pulley. Similarly,

a' 

is the tangential acceleration of a point at the edge of the wheel. This is where we will actually measure the acceleration with the Smart Pulley. The angular acceleration of the wheel is related to these tangential accelerations by With Eqs. Eq. (9)

T = m(g − a) 

 and (10)

α =

a

r

 =

a'

R

 

 we can rewrite Eq. (7)

rT − τ_f = Iα 

 as

Analysis by Linearization

There are two points to consider with regard to Eq. (11)

m(g − a)r − τ_f = I

a

r

 

: it is not linear in m and a and we do not know

τ_f. 

In this analysis, we will make an approximation that will result in Eq. (11)

m(g − a)r − τ_f = I

a

r

 

 becoming linear. We assume that

g >> a, 

and, therefore

T ≈ mg. 

As a consequence, Eq. (11)

m(g − a)r − τ_f = I

a

r

 

 becomes Solving Eq. (12)

mgr − τ_f = I

a

r

 

 for a gives Thus, if we determine the acceleration a for various hanging masses m, and graph the acceleration versus mass, then the slope will be equal to

gr² / I 

and the y-intercept will be

τr_f / I. 

Note: If we determined the accelerations,

a₁ 

and

a₂, 

for only two masses

m₁ 

and

m₂, 

then and therefore

Analysis by the Exact Method

As in the note above, suppose the procedure is repeated with two different masses,

m₁ 

and

m₂, 

resulting in accelerations

a₁ 

and

a₂, 

respectively. Now,

τ_f 

can be mathematically eliminated. Solving the two simultaneous Eqs. (16)

m₁ (g − a₁)r − τ_f = I

a1

r

 

 and (17)

m₂ (g − a₂)r − τ_f = I

a2

r

 

 for I, we have Compare Eqs. (15)

I =

(m2 − m1)gr2

(a2 − a1)

 

 and (18)

I =

(m2 − m1)gr2 + (m1a1 − m2a2)r2

(a2 − a1)

 

.

Objective

In this experiment you will use a wheel with a step pulley to measure the acceleration of the rotating wheel and determine the moment of inertia of the wheel. You will also determine the moment of inertia of an extension that you will add to the wheel.

Equipment

Procedure

The apparatus (see Fig. 1) consists of a wheel with a small step pulley mounted at its center. A string wound around this pulley and attached to a hanging mass provides the necessary torque needed to set the wheel rotating. A small pulley with a photogate is mounted so it makes contact with the rim of the wheel. As the wheel rotates, this Smart Pulley will also rotate. The spokes of the rotating pulley will interrupt the infrared beam from the photogate and the photogate sends these signals to the computer. The computer uses the information to measure the speed at which the pulley turns. The computer measures the speed at which the pulley turns and calculates first the tangential velocity of the edge of the wheel, and then the tangential acceleration

a' 

at the rim of the wheel. From this, you will calculate the tangential acceleration at the step pulley's rim using Eq. (10)

α =

a

r

 =

a'

R

 

. An extension like a hoop or a rectangular plate can be added to the wheel and the moment of inertia of the combination can be measured. Because the extension has the same axis of rotation as the wheel, the moment of inertia of the combination will be From this the moment of inertia of the extension can be determined.

Procedure A: Linearization Method

Procedure B: Exact Method

Procedure C: Determining the Moment of Inertia of the Extension

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