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Circumference is pye times twice the radius or by the diameter

3.1417 X 6 ( diameter or 2 times the radius )= 18.8496

90 degrees is 1/4 of the circumference

18.8496 divided by 4 = 4.7124 cms

arc of 90 degrees =4.74124 cms

Posted on May 03, 2017

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Posted on Jan 02, 2017

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https://www.scribd.com/doc/109721558/Bifilar-Suspension

https://www.scribd.com/doc/109721558/Bifilar-Suspension

THEORY OF MACHINES AND MECHANISNMS I LAB MANUALEXPERIMENT NO. 2 TITLE : RADIUS OF GYRATION BY BIFILAR SUSPENSION METHOD

AIM:

To determine the radius of gyration of given bar using bifilar suspensionmethod.

APPARATUSUSED:

Uniform steel bar, cylindrical weights, stop watch.

DESCRIPTION

Fig 1 shows the general arrangement for carrying out experiment. A

SET UP

uniform rectangular section bar is suspended from support frame by twoparallel cords.

Two small chucks fitted in the frame and other endssecured in bifilar bar. Ends of these cords pass through a chuck so as tofacilitate for change in length of the cord it is possible to adjust length of chord by loosening the chuck.The suspension may also be used to determine radius of gyrationof any body,In this case the body under investigation is bolted to thecenter. Radius of gyration of combined bar and body is then determined.

PROCEDURE

1) Suspend the bar from chuck and adjust length of chord ' L' convenientlynote the suspension length of each cord must be same.2) Allow the bar to oscillate about the vertical axis passing through center of gravity and measure periodic time 'T' by knowing time for 20oscillations.3) Repeat the experiment by mounting the weights one equal distancefrom the center.

THEORY

Consider bar AB whose moment of inertia is to be determine issuspended by two parallel strings as shown in figure 2.Let, L = length of each string.m = mass of bar.K = radius of gyration about the vertical axis through G.I = mass moment of inertia about the vertical axis through G.2r = distance between two stringsT = tension in each string

?

= angle of twist of bar.

?

= angle of twist of string

?

= angular accelaration of the bar.Initial position of bar is AGB. Now, if bar is set gently in vibration inhorizontal plane, it will start oscillating about vertical axis through its masscenter G. when bar is twisted about its C.G, through an angle

?

, bar is inposition A'GB' as shown. When bar is getting

?

as angular displacement,then string is getting

?

as angular displacement about vertical, distancemoved by point A is arc AA'

arc AA' = r

?

=L

?

?

= r

?

L

( 1)PIMPRI CHINCHWAD COLLEGE OF ENGG. MECHANICAL ENGG DEPT

https://www.scribd.com/doc/109721558/Bifilar-Suspension

THEORY OF MACHINES AND MECHANISNMS I LAB MANUALEXPERIMENT NO. 2 TITLE : RADIUS OF GYRATION BY BIFILAR SUSPENSION METHOD

AIM:

To determine the radius of gyration of given bar using bifilar suspensionmethod.

APPARATUSUSED:

Uniform steel bar, cylindrical weights, stop watch.

DESCRIPTION

Fig 1 shows the general arrangement for carrying out experiment. A

SET UP

uniform rectangular section bar is suspended from support frame by twoparallel cords.

Two small chucks fitted in the frame and other endssecured in bifilar bar. Ends of these cords pass through a chuck so as tofacilitate for change in length of the cord it is possible to adjust length of chord by loosening the chuck.The suspension may also be used to determine radius of gyrationof any body,In this case the body under investigation is bolted to thecenter. Radius of gyration of combined bar and body is then determined.

PROCEDURE

1) Suspend the bar from chuck and adjust length of chord ' L' convenientlynote the suspension length of each cord must be same.2) Allow the bar to oscillate about the vertical axis passing through center of gravity and measure periodic time 'T' by knowing time for 20oscillations.3) Repeat the experiment by mounting the weights one equal distancefrom the center.

THEORY

Consider bar AB whose moment of inertia is to be determine issuspended by two parallel strings as shown in figure 2.Let, L = length of each string.m = mass of bar.K = radius of gyration about the vertical axis through G.I = mass moment of inertia about the vertical axis through G.2r = distance between two stringsT = tension in each string

?

= angle of twist of bar.

?

= angle of twist of string

?

= angular accelaration of the bar.Initial position of bar is AGB. Now, if bar is set gently in vibration inhorizontal plane, it will start oscillating about vertical axis through its masscenter G. when bar is twisted about its C.G, through an angle

?

, bar is inposition A'GB' as shown. When bar is getting

?

as angular displacement,then string is getting

?

as angular displacement about vertical, distancemoved by point A is arc AA'

arc AA' = r

?

=L

?

?

= r

?

L

( 1)PIMPRI CHINCHWAD COLLEGE OF ENGG. MECHANICAL ENGG DEPT

Feb 25, 2018 | Homework

Assume the central angle is 35 degrees and the radius is 1.

The circumference of the whole circle is 2 x pi x radius. Since the radius is 1, the circumference will be 2 x pi.

Now a full circle is 360 degrees.

Now we can set up a ratio of 35 degrees is to 360 degree as x is to 2pi.

35 x

---- = -------

360 2xpi

Cross-multiply and isolate your variable.

Good luck.

Paul

The circumference of the whole circle is 2 x pi x radius. Since the radius is 1, the circumference will be 2 x pi.

Now a full circle is 360 degrees.

Now we can set up a ratio of 35 degrees is to 360 degree as x is to 2pi.

35 x

---- = -------

360 2xpi

Cross-multiply and isolate your variable.

Good luck.

Paul

Mar 25, 2015 | Office Equipment & Supplies

About 146 and a half degrees.

If this is homework, make sure to show your work.

If this is homework, make sure to show your work.

Oct 22, 2013 | Office Equipment & Supplies

About 146 and a half degrees.

If this is homework, be sure to show your work.

If this is homework, be sure to show your work.

Oct 22, 2013 | Office Equipment & Supplies

You can not do it unless you know the measure of the central angle sustending (supporting) the arc. If the angle is known, you use the proportionality relation that follows:

If angle is in degrees

(length of arc) / circumference=(measure of central angle sustending arc)/360.

Here the circumference is 2*PI*radius.

If angle is in radians , the relation is somewhat simpler,

**arc length= (radius length)* (angle measure in radians)**

It is clear that in the last relation, the unit for the arc length is the same as the unit for the radius.

If angle is in degrees

(length of arc) / circumference=(measure of central angle sustending arc)/360.

Here the circumference is 2*PI*radius.

If angle is in radians , the relation is somewhat simpler,

It is clear that in the last relation, the unit for the arc length is the same as the unit for the radius.

Jul 15, 2011 | Casio FX-300MS Calculator

Hello and Welcome to FixYa!

Area of pizza (Combined Bread + Crust) = Area=pie * (radius)^2. If divided in 8 equal parts, Area of 1 piece=Area/8.

Area= 3.1415* (0.1524)^2 Area of 1 slice= Area/8*Concerned.*

Area of pizza (Combined Bread + Crust) = Area=pie * (radius)^2. If divided in 8 equal parts, Area of 1 piece=Area/8.

Area= 3.1415* (0.1524)^2 Area of 1 slice= Area/8

Jun 28, 2011 | Computers & Internet

A regular hexagon is made up of six equilateral triangles. This means that the length of any one side of the hexagon is equal to the radius.

Jun 02, 2011 | Texas Instruments TI-83 Plus Calculator

If an arc is 4╥, it is 12.56.

You can use 12.56 as a starting point and figure out what circumference that is 40° of.

Circumference = 2 • π • radius = π • diameter

Reverse this and you will have your answer.

You can use 12.56 as a starting point and figure out what circumference that is 40° of.

Circumference = 2 • π • radius = π • diameter

Reverse this and you will have your answer.

Oct 12, 2009 | Computers & Internet

Hi,

The arc length can be obtained by:

(pi x r x Angle) / 180

or

(3.1415 x 5 x 30) / 180 = 2.618 in.

Hope this helps...

The arc length can be obtained by:

(pi x r x Angle) / 180

or

(3.1415 x 5 x 30) / 180 = 2.618 in.

Hope this helps...

Sep 15, 2009 | Vivendi Math Blaster Geometry for PC, Mac

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