Kater_s Pendulum Topic

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KATER’S PENDULUM

Introduction

Kater’s pendulum (Figure1) consists of a long metallic rod R of circular cross section

weighted at one end so that the center of mass is much nearer to one end. To do this, aheavy metallic cylinder W1 and an identical wooden cylinder W2 are placed at the two

ends of the rod. It has two movable knife edges K 1 and K 2. A much smaller metallic

cylinder W is kept at the middle of the rod. All the five can be moved along the rod andfixed at any position using screws. The labeled diagram of Kater’s pendulum is shown in

Figure 1. This configuration ensures that the Center of gravity lies near one of the knife

edges.

Figure 1: Kater’s Pendulum

l 2

W2

K 2

R

W

G

l 1

W1

K 1



• G - Center of gravity

• R - Metallic rod

• K 1, K 2 - Movable knife edges

• W1 - Metallic cylinder

• W2 - Wooden cylinder

• W1, W2 are to be placed beyond K 1and K 2 and can be moved along R • W - metallic weight to be kept near the center of rod and can be slided for adjustments

Apparatus

• Kater’s Pendulum

• Telescope

• Stop watch

• Meter scale

• Sharp wedge

• Rigid support

Theory

Kater’s pendulum is a Compound pendulum based on the principle that the centre of

suspension and centre of oscillation are interchangeable. The movable cylinders, knife

edges and the metallic weight are so adjusted such that the time periods of the pendulum

about the two knife edges situated asymmetrically with respect to the center of gravity are

exactly equal. Then, the distance between the knife edges is equal to the length of equivalent simple pendulum whose time period is given by (refer to Equation (5), Bar

pendulum)

2

24

T

L g

π =and

Hence, g may be calculated.

g

LT π 2=

We resort to Bessel’s approximation where we require making the two time periods to be

nearly equal because it is quite difficult and time-consuming to set the Kater’s pendulumso that the time period becomes exactly equal about two points.

If T1 and T2 represent two nearly equal time periods (in sec) for positions of K 1 and K 2 distant l 1 and l 2 (in cm) from C.G., then we can write(refer to Equation (3), Bar pendulum)

and .1

22

11 2

gl

k l T

+= π

2

22

22 2

gl

k l T

+= π



Hence, 22

12

2

11

4k l

T gl +=

π and 22

22

2

22

4k l

T gl +=

π

Subtracting and rearranging we obtain

( ) ( )21

2

2

2

1

21

2

2

2

1

28

l l

T T

l l

T T

g −

−+

+

+=

π

Since T1~T2 and positions of K 1 and K 2 are asymmetrical about C.G, l 1-l 2 is fairly large.

Hence, the second term in the denominator is negligibly small and thus, an approximate

value of l 1-l 2 is sufficient.Therefore,

( ) ( )21

2

2

2

1

21

2

2

2

1

28

l l

T T

l l

T T g

−

−

+

+

+

=π

(1)

where

g = Acceleration due to gravity in cm/s2

T 1 = Time period about K 1 in seconds

T 2 = Time period about K 2 in seconds

l 1 = Distance of K 1 from C.G. in cml 2 = Distance of K 2 from C.G. in cm

Learning Outcomes

This experiment will enable you:

1. To determine the acceleration due to gravity (g) using Kater’s pendulum.

2. To verify that there are two pivot points on either side of the centre of gravity(C.G.) about which the time period is the same.

3. To determine the length of the equivalent simple pendulum.

Pre-lab Assessment

Choose the correct answer:

(1) Example/s of compound pendulum is/are

a) Bar pendulum

b) Kater’s pendulumc) both a and b

d) none of the above(2) The oscillatory motion is

a) always periodic

b) periodic as well as bounded

c) only boundedd) none of the above



(3) About how many points, the time period of a compound pendulum is same?

a) 2 b) 3

c) 4

d) 5

(4) The factors that cause damping is/area) friction at the point of suspension

b) viscosity of air

c) none of the aboved) both (a) and (b)

(5) The value of g

a) changes randomly

b) changes definitely according to some relation

c) is constant everywhered) can not be predicted

Procedure

Method 1 (Equalization of time periods)

1. Determine the middle point of the rod and fix the smaller metal weight W there. Fix

the brass weight W1 near one end of the Kater’s pendulum (5 cm from end 1) and the

knife edge K 1 just below it (at a distance of about 2 cm).2. Similarly, adjust the wooden weight W2 and the knife edge K 2 at the other end (end 2)

of the pendulum with the same symmetry. The metallic and wooden cylinders are

placed at different ends to eliminate viscous drag of air and to make the C.G.asymmetrical about the knife edges .Screw all the five tightly. Knife edges must be

sharp, horizontal and parallel to each other so that the oscillations are confined to a

vertical plane3. Suspend the pendulum vertically about K 1 and focus the telescope at the tip of its

lower end. Set it oscillating with amplitude of about 4-5 degrees for the motion to

remain simple harmonic. Note the time for 30 oscillations using a stop watch.

4. Now suspend the pendulum vertically about K 2 and repeat step 3.This time will bequite different from that about K 1.

5. Keep moving K 1 and K 2 towards W by small distance (approx. 1 cm) and repeat steps

3 and 4 till the difference in time about K 1 and K 2 is less than one second. If at anystage the time difference increases, then K

1and K

2should be moved towards W.

6. Now, move the weight W and repeat step 5 to reduce the time difference to about 0.5

second.7. The apparatus is ready to record the measurements. Suspend the pendulum about K 1

and K 2 vertically and record the time taken for 100 oscillations. Repeat this 5 times

each.

8. Remove the pendulum from support and place it horizontally on a wedge. Balance itand find the C.G. of the system.



9. Measure the distances l1 and l2 from C.G. to the knife edges K 1 and K 2.

Method 2 (Graphical method)

10. Here, after the initial adjustments (steps1-2), measure the time for 30 oscillationsabout K 1 and K 2. Balance the pendulum on the wedge to mark the C.G. Then measure

l1 and l2.

11. After this, move K 1 and K 2 by 2 cm each towards the center of the rod.12. Again measure the time for 30 oscillations about K 1 and K 2, Mark the C.G. and

measure l 1 and l 2.

13. Repeat the process 8-9 times by moving K 1 and K 2 towards each other in steps.Repeat step 12.

Observations

Least count of meter scale = ------ mm

Least count of stop watch = ------ sec

Method 1 (Equalization of time periods)

Table 1

S.No No. of oscillations(n)

Time (t1)about K 1 (s)

Time (t2)about K 2 (s)

T1=t1/n(s) T2=t2/n(s) (T1-T2)(s)

1. 30 (t1-t2~0.3 s)

2. 50 (t1-t2~0.5 s)

3. 100 (t1-t2~1s)

4. 100

5. 100

6. 100

7. 100




Table 2

S.No. No. of oscillations

(n)

Time(t1)

aboutK 1 (s)

Time(t2)

aboutK 2 (s)

T1=t1/n(s) T2=t2/n(s) Distance of K 1from CG

(l 1)

Distance of K 2from CG

(l 2)

1. 30

2. 30

3. 30

4. -

5. -

6. -

7. -8. -

9. -

10. 30

Calculations

Method 1 (Equalization of time periods) (when T1 ~ T2)

T1 = ------ sec

T2 = ------ secl 1 = ------ cm

l 2 = ------ cmSubstitute in the Equation (1) and obtain the value of g .




Draw a graph (Figure2) of l1 v/s T1 and l2 v/s T2 on the same graph sheet.

Graph of l1 v/sT1 and l2 v/sT2

1.7

1.75

1.8

1.85

25 30 35 40 45 50 55 60

Distance from CG (l1, l2 in cm)

T i m e p e r i o d ( a b o u t k 1 ,

k 2 i n s e c . )

T1x

T1y

l1x l2xl1y l2y

Line 1x

Line 1y

Figure 2: Graph between time period and distance from c.g.

Draw two horizontal lines (line 1x and line 1y, as shown in the graph in Figure 2) on the

graph intersecting the two experimental lines.

To find g 1, use T1 = T2 = T1x, l1 = l1x and l2 = l2y in Equation (1) and obtain g1.

To find g 2, use T1 = T2 = T1y, l1 = l1y and l2 = l2y in Equation (1) and obtain g2.

Find mean g= ( g 1 + g 2) / 2

Estimation of error

Maximum log error



( ) ( )21

2

2

2

1

21

2

2

2

1

l l

T T

l l

T T g

−

−+

+

+

=

28π

Differentiating logarithmically, we have

( )Y X

Y X

g

g

+

+=

δ δ δ (2)

where

( )21

2

2

2

1

l l

T T X

+

+=

and

( )21

2

2

2

1

l l

T T Y

−

−=

( ) ( )

Since T1 ~T2. Here δT corresponds to the smallest division of the stop watch and δl tothe smallest division of the meter scale. Thus, δ X can be evaluated.

Also,

Using above equation Y δ can also be calculated. The values of X, Y, δ X and Y δ may

then be used to find g g δ using Equation (2) and hence, the log error.

Percentage error

The percentage error can be calculated as

Standard value – calculated value

Percentage error = ----------------------------------------------ⅹ100

standard value

where

Standard value = 981 cm/s2

Calculated value = g

21

22

l l

l

T

T

+

+≈

21

21

2

2

2

1

2

2

2

1

l l

l l

T T

T T

X

X

+

++

+

+=∴

δ δ δ δ δ

21

2

2

2

1

1 24

l l

l

T T

T T

Y

Y

−

+

−

=δ δ δ



Result

The value of acceleration due to gravity g as calculated in the lab is (---------± max. log

error) cm/s2

Glossary

Acceleration: The rate of change of velocity with time.

Acceleration due to gravity: The acceleration imparted to freely moving bodies towardsthe center of earth by the attractive gravitational force of the earth; its value varies with

latitude and elevation.

Center of gravity: The point in a material body through which the resultant force of gravitational attraction acts.

Center of mass: The point in a material body or system of bodies which moves as thoughthe system’s total mass was concentrated at that point and all external forces were applied

at the point. It is also known as the center of inertia.Center of oscillation: The point in a physical pendulum, on the line through the point of suspension and the center of mass, which moves as if all the mass of the pendulum were

concentrated there.

Center of suspension: The intersection of the axis of rotation of a pendulum with a plane

perpendicular to the axis that passes through the center of mass.Compound pendulum: A rigid body of any shape capable of oscillating about a

horizontal axis passing through it.

Gravitational force: The attractive force between any two objects that have mass.

Least count: The smallest measurement which can be accurately made using an

instrument.

Moment of inertia: The sum of the products formed by multiplying the mass of eachelement of an object by the square of the distance of the element from a specified axis.

Oscillation: Any effect that varies periodically back and forth between two values.

Radius of gyration: The square root of the ratio of the moment of inertia of a body about

a given axis to its mass. Rigid body: A solid body in which the particles are arranged so that the inter-particle

distances remain fixed and are not disturbed by any external forces applied

Simple pendulum: A device consisting of a small massive body suspended by aninextensible object of negligible mass from a fixed horizontal axis about which the body

and suspension are free to rotate.

Simple harmonic motion: A periodic motion about an equilibrium position for which

the displacement is a sinusoidal function of time. The acceleration of the object is alwaysdirected towards the equilibrium position and is proportional to the displacement from

that point.

Telescope: An assemblage of lenses or mirrors or both that enhances the ability of theeye to see distant objects with greater resolution.

Time period: Time taken to complete one oscillation of a periodic motion.



Post-lab Assessment

Choose the correct answer

(1) When the distance of the point of suspension from the centre of gravity increases, the

time period of a compound penduluma) increases

b) decreases

c) remains constant

d) first decreases then increases(2) As we go down below the surface of earth, the value of g

a) increases b) decreases

c) first increases then decreasesd) does not change

(3) The difference in the value of g at equator and poles is equal to

a) radius of earth (R)

b) angular velocity of earth (ω)c) R ω

d) R ω2

(4) If we interchange the center of suspension and center of oscillation, the time period of compound pendulum

a) increases

b) decreases

c) remains samed) cannot be predicted

(5) The length of an equivalent simple pendulum is equal to

a) the distance of the point of suspension from C.G. b) radius of gyration (k)

c) (k 2+l

2)/l

d) infinite(6) Which instrument is expected to give more accurate value of g?

a) simple pendulum

b) Kater’s pendulumc) bar pendulum

d)

mass spring system(7) The time period of a compound pendulum about its C.G. is

a) zero b) infinite

c) finite

d) unpredictable(8) The time period of compound pendulum will be minimum when

a) l = k



b) l = k 2

c) l > k d) none of the above

(9) With altitude the acceleration due to gravity ‘g’

a) increases

b)

decreasesc) remains constant

d) cannot be predicted

Answers to Pre-lab Assessment

1. c2. b

3. c

4. d

5. b

Answers to Post-lab Assessment

1. d

2. b3. d

4. c

5. c

6. b

7. b8. a

9. b

Kater_s Pendulum Topic

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