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Answered: - hey can anyone help me with question 3 and 4 please for PHY thank

hey can anyone help me with question 3 and 4 please for PHY thank you?

Rotational Motion

Introduction

In this lab a constant torque will create a constant angular acceleration for a rigid body rotating

about its center of mass. You will see that the moment of inertia depends on the rotation axis for

a given object and finally you will look at rotational motion from an energy viewpoint.

Applications of the moment of inertia concept include the design of crankshafts, rotary rides at a

fair, or the baton twirling of a marching band leader. Text Reference: Young & Freedman ? 9.15.

Theory

1

g?t 2 . If we convert this equation to the

2

1

angular domain by considering a disk turning about some axis it says ?? = ??t 2 .

2

A stone dropped into a well falls a distance ?y =

In more detail, a rigid object free to rotate about a principle axis with moment of inertia I will

accelerate according to

(1)

? = I?

where ? is the net torque (given by the sum of r ? F for each force applied to the object), and ?

is the angular acceleration in rad/sec2. Recall ? radian = 180?. Take care that ? , ?, and I are in

consistent units. The moment of inertia I is readily calculated for symmetric objects, and has the

form I = ?MR2, where ? is a dimensionless fraction between 0 and 1 that depends on the shape of

the object and the rotation axis. A 3-dimensional object has three principal values of I,

corresponding to three independent possible axes of rotation. These values would all be the

same for a sphere through its center, but all three values would be different for a low symmetry

object. Thus, for example, a rectangular block of dimensions a?b?c has moment of inertia

(

)

1

M a2 + b2

12

for rotation about the axis normal to the a?b face and passing through its center.

I=

(2)

In this lab, we use the arrangement shown in figure 1. A rectangular block of mass M and

dimensions a?b?c is mounted with a rotation axis through the center of one face. A mass m is

tied to a string wound around a massless pulley of diameter 2r mounted on the same axis. The

string exerts a torque on M given by

? = rT

(3)

where T is tension in the string which can be found using Newton?s second law and is

T = m(g ? a) = m(g ? ?r ).

Note we have used a = ?r, and both g and ? are positive. Combining these equations we have

(4)

? g ? r? ?

(5)

I d = mr ?

?

? ? ?

where we have written Id (dynamic moment of inertia) to distinguish it from the static moment of

inertia Is determined from the geometry and mass, as in equation 2. These are entirely equivalent

in principle.

M

2r

rotation axis

m

Figure 1. Apparatus for the dynamic measurement of rotational moment of inertia.

It is also useful to look at this problem from an energy viewpoint. Thus we have

?K + ?U = 0

(6)

and for our specific problem this can be written as

(

)(

)

1

I s + mr 2 ? 2 ? ?i2 + mgr (? ? f + ? i ) = 0

f

2

(7)

where ? and ? are the angular velocity and position of the pulley. The terms in m and ? are due

to the kinetic energy of the small mass as it falls. To get the signs right, note that the change in

height ?h is negative, and the change in potential energy is negative. Note that the kinetic

energy includes both the spinning block (M) and the falling mass (m), while the potential energy

involves only the falling block, and all terms can be written in terms of pulley angle, as above.

Think about the signs of the two terms in equation 7 needed to keep the sum zero.

Procedure

Measure the mass and dimensions of the steel block estimating the errors. The pulley radii are Rs

= 0.48 cm and Rm = 1.43 cm (considered exact, but see below regarding errors).

Open the setup file for the Data Studio (DS) data acquisition program (see whiteboard for file

location). The file will be preset to record ?(t) and ?(t).

Part I: Mount the block through its shortest axis onto the rotary encoder (do not over tighten the

screw!). Put the string on the medium pulley and wind it up completely in a direction so it will

unwind CCW (counterclockwise) when facing the pulley. Do not tie the string onto the pulley as

this jars the sensor. Start the data acquisition (?record?) then release the pulley carefully. Check

that your data looks okay, otherwise repeat the collection process. You may over-write Run #1

or save as Run #2, etc., then use the best one for analysis. You may wish to delete unnecessary

data before exporting. Do not attach this raw data file (table) in your report. Use the recorded

data to determine Id using equation 5.

For part I only find ?K and ?U and ?E for the complete unwinding motion. Use these results to

find the average torque due to friction by energy methods. See notes provided on the white

board.

Part II: Mount the block through its longest hole and repeat.

Part III. Repeat the experiment when the block mounted to its shortest and longest axes and

measurements repeated for the medium and smallest pulleys.

In the Lab report calculate the static value of moment of inertia Is for the block for both axis

(shortest and longest) based on its dimensions and mass. Error analysis is messy, so discard

small terms. Show explicitly how you reject terms. Take the error in the mass measurement

based on the scale of the provided balance. The error in the measured dimensions of the block is

to be 0.05 mm.

Calculate and report value of dynamic moment of inertia (Id )using linear fits to ?(t) for parts I,

II and III. Report errors for Id again explicitly showing how you reject small terms, if any. How

well your values of Id and Is agree with each other? How did you arrive at your conclusion?

Friction undoubtedly plays a role in this experiment. Does friction cause your value of Id to be

too high or too low? Explain.

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