PURPOSE:
In this experiment we will verify that linear momentum is conserved for both elastic collisions and inelastic collisions.
INTRODUCTION :
One of the most widely used principles in physics is the conservation of linear momentum.
The utility of this principle lies in the fact that regardless of the complexity of the actual collision, the total momentum before a collision is equal to the total momentum after a collision: m
1 v
1i
+ m
2 v
2i (Before Collision) = m 1 v
1f
+ m
2 v
2f (After Collision)
Two types of two-body collisions will be investigated using the air track. The first is one in which almost no energy is dissipated; this is called an elastic collision. The second is one in which a fraction of the kinetic energy is dissipated; this is called an inelastic collision.
Remember that velocities and momenta are vector quantities and consequently, in one dimension, add algebraically. That is to say, momenta in the right-hand direction are positive and momenta in the left-hand direction are negative.
EQUIPMENT & SUPPLIES:
Air Track & accessories
Electronic balance
2 gliders
2 photogate timers
2 cardboard flags
Masking tape
PROCEDURE :
A. ELASTIC COLLISIONS
1. Level the air track by placing the glider at the middle of the track and adjusting the leveling screws until the glider is motionless. Place two independent photogate timers near 70 cm and 130 cm.
2. Place a rubber-band bumper at the front of glider 1 and place a counter mass at the back of glider 1. Tape a cardboard flag to the top of each glider, with the longer side horizontal.
Determine and record the effective length L to the nearest millimeter of each cardboard rectangle by measuring the glider’s positions when the LED on top of the photogate turns on, then off. Determine the mass of each glider with attachments and record the mass of each.
3. Set the photogate timers to gate mode and turn on the memory feature. Launch the gliders with moderate speed from opposite ends of the air track. The cardboard flags will pass through the photogates and a time will be displayed on each photogate. From these times the initial velocity of each glider can be calculated. After impact, the gliders will pass back through the photogates, adding to the times in memory. Subtract the displayed time from the memory time to arrive at the actual time for the final velocities. Be sure to stop the
- 33 -

gliders after they have passed through the photogates the second time. Use negative signs for the velocities where appropriate.
4. Repeat the above procedure two more times so that you will have three trials altogether.
Change the mass on the gliders for these two trials by adding masses symmetrically to the sides of one of the gliders.
5. Calculate the momentum for each trial before and after the collision. Is linear momentum conserved? Calculate the total kinetic energy both before and after collision. Is kinetic energy conserved?
Glider 1
Photogate 1
Photogate 2
Glider 2 m
1
Fig. 1: Experimental Setup for Elastic Collision m
2
B. INELASTIC COLLISION
1. Repeat the above procedure for a completely inelastic collision. Use the wax plug and needle on the gliders so that they stick together, and place a counter mass at the back of glider 2. After the collision, stop the gliders after one rectangle of cardboard has passed through a photogate.
Only one timing is necessary, as v
1f
= v
2f
.
2. Calculate the momenta both before and after collision. Is momentum conserved for the inelastic case? Calculate the total kinetic energy both before and after collision. Is kinetic energy conserved?
Glider 1
Photogate 1
Photogate 2
Glider 2
Needle
Wax Plug m
1
Fig. 2: Experimental Setup for Inelastic Collision
- 34 - m
2

1
2
3
DATA SHEET: ELASTIC & INELASTIC COLLISIONS
Table A. Elastic Collision
Trial m
1
(kg)
L
1
= ____________________ m ,
Glider 1
 t
1i
(s)
 v
1i
(m/s) p
1i
(kg
 m/s)
Before Collision m
2
(kg)
Glider 2
 t
2i
(s)
 v
2i
(m/s)
L
2
= ____________________ m p
2i
(kg
 m/s)
Total
Momentum
(kg
 m/s)
1
2
3
Trial
3
Trial m
1
2
3
Table B. Inelastic Collision
Trial m
1
(kg)
Glider 1
 t
1i
(s)
 v
1i
(m/s)
1
2 m
1
(kg)
1
(kg)
Glider 1
 t
1f
(s)
 v
1f
(m/s)
Glider
 t
1f
(s)

1 v
1f
(m/s) p
1f
(kg
 m/s) p
1i
(kg
 m/s) p
1f
(kg
 m/s) m
After Collision
2
(kg)
Glider 2
 t
2f
(s)
 v
2f
(m/s)
Before Collision m
2
(kg)
Glider 2
 t
2i
(s)
 v
2i
(m/s) m
After Collision
2
(kg)
Glider 2
 t
2f
(s)
 v
2f
(m/s) p
2f
(kg
 m/s) p
2i
(kg
 m/s) p
2f
(kg
 m/s)
Total
Momentum
(kg
 m/s)
Total
Momentum
(kg
 m/s)
Total
Momentum
(kg
 m/s)
Total
Energy
(J)
Total
Energy
(J)
Total
Energy
(J)
Total
Energy
(J)
- 35 -

PURPOSE:
To determine the center of mass (CM) of objects of uniform density and thickness.
INTRODUCTION :
One way to locate the center of mass of an object of uniform thickness is to balance the object on a pivot. Another way is to drop a plumb line from one corner and trace out the plumb line, then repeat this for another corner. The point of intersection of the two lines is the center of mass.
A third way to locate the center of mass is to divide an object into rectangles, and imagine the mass of each rectangle to be concentrated at the rectangle’s center. Each rectangle can be weighted by area instead of mass, as the object has uniform surface density and thickness. Then the center of mass of the object is the average of the positions of these centers, weighted according to area.
For example, suppose that the upper bar of Object A in Fig. 2 has an area of A
1
= 10 cm 2 extending from y = 10 cm to y = 14 cm, so y
1
= 12 cm, and suppose the vertical stem has an area of A
2
= 15 cm 2 and extends from y = 0 cm to y = 10 cm, so y
2
= 5 cm. Then,
Y
CM
=

10

12
10
 

15
15

5

=
195
25
= 7.8 cm.
Generally,
X cm
=

A i x

A i i Y cm
=

A

A i y i i
EQUIPMENT & SUPPLIES:
Cardboard sheet
Scissors
Ruler
Plumb line
String
Nail
PROCEDURE :
1. Cut out the following geometric shapes from a sheet of cardboard of uniform thickness.
Object A Object B
Fig. 1
Object C
- 36 -

2. Puncture one corner of Object A with a nail, let it hang freely, and trace the path of the plumb line across the object, starting at the nail. Repeat this from another corner, and find the point where these two lines intersect. Mark this point, and label it as the “experimental” center of mass.
3. Verify that this is the center of mass by balancing the object with your index finger.
4. Calculate the position of the center of mass for object A, by imagining object A as two rectangles of unequal mass, each replaced by point masses at their geometric centers.
5. Mark and label the "calculated" position of the center of mass on the object by placing it against a sheet of graph paper. See Fig. 2. y y y x x
O x
O
O
Object A Object B Object C
Fig. 2.
6. Compare the "experimental" and the "calculated" center of mass positions.
7. Repeat procedure for object B.
8. With object C, place it against the coordinates and locate the calculated X
CM
position. See
Fig. 3. y x
O
X
CM
Fig. 3.
9. Draw a line parallel to the y-axis through the calculated X
CM
. Tape a string along this line.
10. Experimentally locate Y
CM
by sliding your index finger along the mounted string to find the point of balance. Mark this point "exp" Y
CM
. Compare with "calculated" Y
CM
.
11. Verify Y
CM
by using the plumb lines from two different corners.
- 37 -

DATA AND CALCULATIONS: CENTER OF MASS
Calculated Object
X
CM
_______ cm Y
CM
_______ cm
X
CM
_______ cm Y
CM
_______ cm
X
CM
_______ cm Y
CM
_______ cm
Experimental
X
CM
_______ cm Y
CM
_______ cm
X
CM
_______ cm Y
CM
_______ cm
X
CM
_______ cm Y
CM
_______ cm
- 38 -

PURPOSE :
The object of this experiment is to use the method of balancing torques to determine the center of mass of a non-homogeneous meter stick, and to determine the unknown mass of an object.
INTRODUCTION :
If a rigid object is in rotational equilibrium, the net torque acting on it, about an axis, is zero.
This equilibrium condition can be stated as:
 
= 0 where

= Fd, F is the applied force, and d is the distance from the horizontal axis of rotation to the point where the downward force is applied. The plus sign {+} corresponds to a counterclockwise torque and the negative sign { } corresponds to a clockwise torque.
The center of mass, denoted here by CM, is the point at which the mass of the object can be considered to be concentrated. The position x of the CM of a non-homogeneous meter stick can be determined by balancing the torque of the stick on one side of the fulcrum with the torque of a known mass on the other side of the fulcrum.
Fulcrum at midpoint of stick x d
1 d
2
CM
 counter-clockwise
 is a positive torque
F
1
= m
1 g m
2
F
2
= m
2 g  clockwise is a negative torque

= F
1 d
1
+ (

F
2 d
2
) = 0
Fig. 1
Having established the position of the CM and knowing the mass of the stick, the same procedure can be used to determine the unknown mass of another object.
SUPPLIES & EQUIPMENT:
Weighted meter stick
1 knife-edge clamp
Electronic balance Scissors & light string
Knife-edge stand Hooked masses
Metal cube
- 39 -


PROCEDURE :
A. CENTER OF MASS OF A NON-UNIFORM METER STICK
1. Record the mass of the non-uniform meter stick m
1
indicated on the electronic balance.
2. Set up the apparatus as shown in Fig. 1, making sure that the fulcrum is at the midpoint of the stick, and using one of the hooked standard masses suspended from a loop of string as m
2
. Slide m
2
in along the stick until the stick is in equilibrium. Record m
2
and d
2
. Be sure to include the mass of the string in the mass of m
2
.
3. Use Eq. 1 to obtain your first estimate of x, the distance of the CM from the weighted end of the stick. This equation was obtained from the equilibrium condition:
 counter-clockwise
+
 clockwise
=
F
1 d
1
- F
2 d
2
= 0
F
1 d
1
=
F
2 d
2 d
1
=
F
2
F
1 d
2
= m
2 g m
1 g d
2
d
1
= m m
2
1 d
2
= distance of CM from fulcrum
x = position of CM from weighted end
= fulcrum position minus d
1
Eq. 1
4. Move the fulcrum 5.0 cm away from the midpoint, toward the weighted end of the stick as shown in Fig. 2. Slide m
2
to establish equilibrium. Record m
1
and m
2
and the new value of d
2
. Use Eq. 1 to calculate the new value of d
1
and subtract this from the position of the fulcrum to obtain your second estimate of x, the position of the CM.
Fulcrum Midpoint of stick x d
1 d
2
CM m
2
F
2
= m
2 g
F
1
= m
1 g
Fig. 2
5. To obtain your third estimate of x, remove m
2
and balance the stick on the knife-edge clamp. Record x as the position of the fulcrum.
- 40 -

B. DETERMINATION OF AN UNKNOWN MASS
1. Having calculated the position of the center of mass on the previous page, set up the apparatus as shown in Fig. 1, only this time m
2
, a metal cube, will be an unknown mass.
Using a loop of string, hang this unknown mass on the stick and slide it along the stick to balance.
2. Record the new value of d
2
. Use Eq. 2 to obtain your estimate of the unknown mass, m
2
.
 
= 0 , so F
1 d
1
+ (

F
2 d
2
) = 0 and F
2
=
F
1 d
2 d
1
, giving m
2 g = m
1 g d
2 d
1
. m
2
= d
1 d
2
m
1
Eq. 2
3. Weigh the metal cube on the electronic balance and find the percent difference between the two measurements of m
2
.
C. MULTIPLE-TORQUE SYSTEM: DETERMINATION OF THE MASS OF A METAL CUBE
1. The equilibrium condition can be used even when there are several torques involved. Set up the apparatus as shown below, using the same m
2
as in Part B, and using hooked standard masses suspended from loops of string as m
3
and m
4
:
Fulcrum at midpoint of stick d
2 d
4 x d
1 d
3 m
2 m
3 m
4

= 0 , so F
F
2
F
1
1 d
1
+ F
2 d
2

F
3 d
3
- F
4 d
F
3
4
= 0 giving F
2
=
F
4
Fig. 3
2. Use Eq. 3 to obtain another estimate of the unknown mass m
2
.
F
3 d
3

F
4 d
4 d
2

F
1 d
1
m
2
= m
3 d
3
 m
4 d
4 d
2
 m
1 d
1
Eq. 3
3. Find the percent difference between this measurement and the value obtained directly from the electronic balance.
- 41 -

DATA SHEET: TORQUE AND CENTER OF GRAVITY
Data Table A: Determination of the Center of Gravity
Fulcrum Position m
1
(stick)
(kg) m
2
(kg) d
1
*
(m) d
2
(m) x
(m)
At midpoint
(Steps 1 – 3)
Fig. 1
At 5.0 cm from midpoint
(Step 4)
Fig. 2
At CM
(Step 5)
Data Table B: Unknown Mass m
2 m
1
(stick)
(kg)

= 0
(Steps 1-2)
Fig. 1
Unknown mass from weighing
(Step 3) d
1
*
(m) d
2
(m) m
2
(kg)
Percent Difference _________________
Data Table C: Multiple Torque System m
1
(kg) d
1
*
(m) m
3
(kg) d
3
(m) m
4
(kg) d
4
(m) d
2
(m)

= 0
(Step 2)
Fig. 3
Unknown mass from weighing
*Notice that d
1
is the distance of the center of mass (CM of the meter stick) to the fulcrum.
Percent Difference ___________________ m
2
(kg)
- 42 -

INTRODUCTION :
When an object is caused to rotate with uniform speed, the inward force that keeps it in a circular path is called the centripetal force. The centripetal force produces a radial acceleration towards the center of the circle. Applying Newton's Second Law F = ma to rotational motion:
Eq. (1) F = mv r
2 v =
2
 r
T because a = v r
2
. where T = period, or time for one revolution
SUPPLIES & EQUIPMENT:
Centripetal force apparatus
Hooked masses, 50g, 100g & 200g Slotted masses
Electronic balance
# 5 Rubber stopper
PROCEDURE, PART A:
Thistle tube
Stop clock
String & scissors
Red felt marker
Meter stick
The 0930 Centripetal Force Apparatus is intended for the study of uniform circular motion. It consists of a revolving object of mass 0.458 kg supported from a cross arm attached to a vertical shaft. The shaft is supported in a metal housing on a wooden base. Radial and thrust bearings permit the shaft to rotate freely without wobbling, if the cross arm is correctly counterbalanced. A position-adjustable vertical rod mounted on the base serves as a radius vertical rod. A ball bearing pulley mounted on a rod near one end of the base is used in measuring the force exerted by a spring with which the mass is coupled to the shaft. A coiled spring, a 50-gram mass hanger and a hook are supplied.
1. Use the bubble level to level the apparatus. Measure r, the distance from the tip of the vertical rod to the center of the vertical shaft.
2. A sheet of white paper located to provide a light background is helpful in seeing that the revolving object passes exactly over the vertical rod. Rotate the system by applying torque with the fingers on the knurled portion of the shaft. Make sure that the revolving object is hanging vertically when it is directly over the vertical rod. Warning! Students with long hair should be very careful not to get their hair caught in the revolving apparatus. With a little practice, the rate of revolution can be adjusted to keep the mass passing directly over the vertical rod.
3. Time 25 revolutions with the stop clock. Repeat twice and average the three results.
4. Check the actual centripetal force exerted by the spring by hanging the weight holder over the pulley and adding masses until the pointed tip of the object is just above the vertical rod.
- 43 -

5. Enter the data in Table 1. Calculate the percent difference between the theoretical value
F = mv 2 /r and the measured value F = Mg.
6. Shift the vertical rod to another position. Adjust the cross arm to allow the revolving object to hang vertically when it is directly over the vertical rod, and adjust the position of the counterweight if necessary to prevent wobbling. Repeat steps 1
– 5.
PROCEDURE, PART B:
In this part of the experiment we will again study the motion of an object traveling in a circular path. A small object of known mass will revolve in a circular path. The centripetal force will be determined directly and then calculated from measurements of the radius and the velocity.
The following relation will be verified:
F = mv
2 r
PROCEDURE:
1. Determine the mass of a stopper. Tie a 1.5 m length of string to the stopper, then thread it through the thistle tube. Tie a 0.150 kg mass to the other end of the string. The weight of this mass creates the tension in the string that provides the centripetal force on the stopper.
2. To help you maintain the radial distance, use a dot of red ink as a marker at the top edge of contact with the thistle tube.
3. Using the stop clock, measure the total time for 25 revolutions for two different values of radial distance.
Try values close to 0.500 m and
0.750 m. The time for one revolution is the total time divided by
25.
4. Maintain a steady horizontal swing. r
Revolving mass,
Thistle Tube
Mark
Hanging mass, M m
(Actual Centripetal Force = Mg)
5. The velocity is given by the equation: v = time circumfere nce for one revolution
=
2
 r
T
where r is the radius of the circular path and T is the time for one revolution.
6. Repeat the above procedure for a 0.200-kg mass attached to the string.
7. Calculate the percent difference between the theoretical value F = mv 2 /r and the measured value F = Mg.
- 44 -

DATA SHEET: CENTRIPETAL FORCE
Data and Calculations Table for Part A:
Mass of Revolving Object, m
Radius, r
Time for 25 Revolutions
(kg)
(m)
(s)
Time for 1 Revolution, T
Velocity, v = 2
 r/T
A. Experimental F = mv
2
/r
Hanging Mass, M
B. Centripetal Force from F = Mg
Percent difference between centripetal forces A and B, above
(s)
(m/s)
(N)
(kg)
(N)
0.458
____________
- 45 -
0.458
____________

Data and Calculations Table 1 for Part B:
Mass of Stopper, m
* Radius, r
Time for 25 Revolutions
(kg)
(m)
(s)
Time for 1 Revolution, T
Velocity, v = 2
 r/T
A. Experimental F = mv
2
/r
Hanging Mass, M
B. Centripetal Force from F = Mg
Percent difference between centripetal forces A and B, above
*Approximately 0.500 m
(s)
(m/s)
(N)
(kg)
(N)
____________
0.150
- 46 -
0.200
____________

Data and Calculations Table 2 for Part B:
Mass of Stopper, m
* Radius, r
Time for 25 Revolutions
(kg)
(m)
(s)
Time for 1 Revolution, T
Velocity, v = 2
 r/T
A. Experimental F = mv
2
/r
Hanging Mass, M
B. Centripetal Force from F = Mg
Percent difference between centripetal forces A and B, above
*Approximately 0.750 m
(s)
(m/s)
(N)
(kg)
(N)
____________
0.150
- 47 -
0.200
____________

INTRODUCTION :
In the case of linear motion, an unbalanced force F acting on an object gives it an acceleration a and by Newton’s second law, F = ma, where m is the mass of the object. For rotational motion, an unbalanced torque

acting on a body causes it to rotate, i.e., gives it an angular acceleration
 . The rotational analog of Newton’s second law is 
= I

where I is the moment of inertia of the body and is the rotational analog of mass.
The moment of inertia of a body depends on its mass distribution and shape. For a symmetrical, homogeneous object, the moment of inertia can be derived theoretically using calculus methods and is expressed in terms of the object’s total mass and dimensions. By determining these quantities, the moment of inertia can be calculated.
However, the moment of inertia of an object can also be determined experimentally from the dynamical equation

= I

by measuring

and

or from energy considerations where the rotational kinetic energy is ½ I 
2 .
In this experiment the theoretical and experimental values of some common, regularlyshaped objects will be determined and compared. The experimental procedures will provide good insight into the concepts of rotational motion and the moment of inertia.
SUPPLIES & EQUIPMENT:
Moment of inertia apparatus
1- & 2-meter sticks
2 Right angle clamps
Transfer caliper
PROCEDURE:
Stop clock
Support rod
Mass holder
2 Rod pulleys
Metric ruler
Masking tape Foam pad String & scissors
Slotted masses
Table clamp
Vernier caliper
Bubble level
1. Set up the equipment as shown in the diagram. Make sure the string runs vertically and horizontally, and the pulleys are aligned with the string.
The falling weight should travel about
1.5 meters before striking the floor.
Make sure the string is long enough to exert torque on the wheel for the full length of the weight’s travel.
2. Use a total mass m = 0.250 kilograms for the pulley. Measure the distance that the weight has to fall, to the nearest millimeter. Always start the weight from the same position.
- 48 -

3. Make three timings of how quickly the 0.25-kilogram mass falls through your distance d ~ 1.5 meters when the platform is empty, and find the average time t of these three timings. The average velocity of the mass is v ave
= d/t. Since the acceleration is uniform, the final velocity of the mass just before it hits the floor is v = 2v ave
. Since the initial velocity is zero, the acceleration of this object is a = v/t. The final velocity of the mass must also be the final velocity of the string wrapped around the axle of the platform, so the angular velocity of the axle (and the platform) is

= v/r. Similarly, the angular acceleration of the platform is

= a/r.
The tension in the string is T = m(g - a), with g = 9.80 m/s/s, and this tension exerts a torque on the axle of

= T
 r
 sin(90 o ). The moment of inertia of the empty platform is I =

/

.
4. Repeat procedure 3, but with the ring lying on the platform. After you calculate the moment of inertia of the platform and ring, subtract the moment of inertia of the platform to get the moment of inertia of the ring. Calculate the theoretical moment of inertia of the ring, and calculate the percent difference between the theoretical and the measured values.
5. Repeat procedure 4, using the solid disk instead of the ring.
6. Repeat procedure 4, using one or two of the cylinders instead of the ring. You may use one cylinder standing on edge, one cylinder lying on its side, or two cylinders standing on edge at the rim of the platform. Masking tape may be needed to hold the cylinders in position. Specify which arrangement you used.
- 49 -

DATA SHEET: MOMENT OF INERTIA
Radius r of axle _______________
Mass of falling weight ___________________
Distance d through which weight falls __________________
Fill out the table below, entering the appropriate units and being careful to use the appropriate number of significant figures.
Empty Platform Ring Disk Cylinders
Time of fall of mass:
1
2
3
Average
Average velocity of falling mass
Final velocity of falling mass
Acceleration of falling mass
Angular acceleration
Tension in the string
Torque acting on the rotating assembly
Moment of inertia of object and platform
Moment of inertia of object
Theoretical moment of inertia
% difference
- 50 -

PURPOSE :
To determine the spin angular velocity (
 s
) and precessional angular velocity (
 p
) of a gyroscope.
INTRODUCTION :
A spinning object such as the rotor of a gyroscope posesses angular momentum (L). When a torque is applied, changing the angular momentum, precession occurs.
SPIN TORQUE y

Spin
Axis
L d d x
 s
= spin angular velocity
L = angular momentum z
F = m g
 
= d X F
PRECESSION y
 p
L



L z

=

L/L

L = L
 x

 
=

L
 t
 
=
L
 
 t
= L
 p
 p
=

L
  p
 d
I
X F
 s
 p
= mgd
I
 s
Eq. (1)
 s
= spin angular velocity
This angular velocity can also be directly measured by timing the period of each revolution, T p
.
 p
=
2

T p
Eq . (2)
- 51 -

EQUIPMENT:
Gyroscope
Stop clock
Large caliper
PROCEDURE:
Small torque mass
Large torque mass
Vernier caliper
Photogate timer
Masking tape Ruler
Balance
String
1. Balance the rotor statically by screwing the pointer of the representative L vector in or out until the pointer is stationary, when pointed horizontally.
2. Weigh and record the torque masses.
3. Physically measure the rotor angular velocity by the following method: a) Tape a short piece of string to the edge of the rotor as shown in the diagram to the right. b) Set the photogate to the "pulse" mode. Plug in the gyroscope and hold the arrow in a vertical position. Place the photogate beam in the path of the rotating string, and press
Rotor
Direction of rotation
Photogate the reset button on the photogate timer.
This will give you the time for one revolution directly. Warning! Students with long hair should be very careful not to get it tangled up in the gyroscope! c) Take six consistent time readings by pressing the reset button and calculate the average time for one revolution, T s
.
 s
is then 2

/T s
.
String
4. Use the stop clock to measure the time for one precessional revolution, using the small torque mass clipped into the notch just behind the pointer, pointed horizontally. Then repeat the measurement for the large torque mass. Do this six times and calculate an average time for each mass. This will give you a value for the precessional frequency. Measure the mass of these torque masses, and the length of the moment arm (the distance from the center of the rotor to the center of the torque mass).
5. Calculate the moment of inertia of the rotor itself by the method of superposition, i.e., the moment of inertia of whole is equal to the moment of inertia of the constituent parts. Assume that the rotor is an ideal disk of density 7.0 X 10
3
kg/m
3
.
6. Calculate the percent difference between your two rotor precessional angular velocities.
- 52 -

DATA SHEET: THE GYROSCOPE
Data and Calculations Table 1: Rotor Angular Velocity
Rotor period, T s
Rotor angular velocity,
 s
= 2

/T s
Trial (s) (rad/s)
3
4
1
2
5
6
___________________________
Average
Mass of small torque mass __________________ kg
Mass of large torque mass __________________ kg
Length of moment arm,d ______________________ m
Data and CalculationsTable 2: Precessional Angular Velocity
Trial
Small Torque Mass
T p
(s)
 p
= 2

/T p
(rad/s)
1
2
3
4
5
6
Average
__________________
Large Torque Mass
T p
(s)
 p
= 2

/T p
(rad/s)
__________________
- 53 -

Moment of Inertia Calculations:
Moment of Inertia of Entire Disk, (I
1
):
Rotor Outer Radius, r
1
= ________________ m
Rotor Density,

= 7.0 X 10 3 kg/m 3
Rotor r
1
Rotor Outer Thickness, h
1
= ________________ m h
1
I
1
= ½ m
1
r
1
2 = ½ 
V
1
r
1
2 = ½  h
1
 r
1
2 r
1
2 = ½  h
1
 r
1
4 = __________ kg ·m 2
Moment of Inertia of "Missing" Part of Disk, (I
2
):
Rotor Inner Radius, r
2
= ________________ m
Rotor Density,

= 7.0 X 10 3 kg/m 3
Rotor Inner Thickness, h
2
= ________________ m
Rotor r
2 h
2
I
1
= ½ m
2
r
2
2 = ½ 
V
2
r
2
2 = ½  h
2
 r
2
2 r
2
2 = ½  h
2
 r
2
4 = __________ kg ·m 2
Moment of Inertia of Composite Disk:
I = I
1
- I
2
____________________ kg ·m 2
Small torque:

1
= F
1 d = m
1 gd ______________________
Large torque:

2
= F
2 d = m
2 gd ______________________
Data Table 3: Precessional Angular Velocity
Precessional angular velocity Small Torque Mass
From Eq. (1)
 p
= mgd/I
 s
*
From Eq. (2)
 p
= 2

/T p
**
Percent difference
*
 s
from Table 1
** T p
from Table 2
Large Torque Mass
- 54 -

PURPOSE:
Archimedes' Principle will be used to determine: a) the density of a symmetrically-shaped object; b) the density of an irregularly-shaped object; and c) the specific gravity of a liquid.
INTRODUCTION:
Archimedes' Principle states that an object that is submerged in a fluid is buoyed up by a force that is equal to the weight of the fluid displaced by the object. This force is called the buoyant force, or the buoyancy. The buoyant force can be determined experimentally with the following setup:
Beam Balance Beam Balance
Paper Clip
T
1
T
2
B m m a mg mg
Lab Jack
Fig. 1 Fig. 2
T
1
= W o
(Weight of object in air) T
2
= W aw
(Apparent weight of object in water)
The force diagrams in Figures 1 and 2 both have the same downward force, which is the weight of the object, W o
= m o g. The difference between T
1
and T
2
measured by the balance is the upward push by the fluid, the buoyant force, F
B
. Since F
B
= T
1
- T
2
, the buoyant force can be determined experimentally by measuring the difference in weight of the object in air and when it is submerged in a fluid such as water.
F
B
= W o

W aw
(Eq. 1)
Key to symbols:
m o
= mass of object in air
m w
= mass of water displaced
  w
= density of water = 1000 kg/m 3
W o
= weight of object in air m aw
= apparent mass of object in water (fluid) W aw
= apparent weight of object in water
W w = weight of water displaced
V w
= volume of water displaced
V o
= volume of object
- 55 -

THE DENSITY OF AN OBJECT
When an object is totally submerged in water (a fluid), the volume of water displaced is equal to the volume of the object.
(Volume of submerged object) V o
= V
Since the volume of an object is: V o
= m then, m o
/
 o
= m
And,
 o
= m w o
/
 w
/
 o
(Volume of water displaced) (Eq. 2)
, and the volume of fluid displaced is V w o
 w
/m w
= m o
g
 w
/m w g w
= m w
/
 w
As m w g is the weight of the water (fluid) displaced, W w and Archimedes’ principle states that the buoyant force is equal in magnitude to the weight of the fluid displaced,
F
B
= W w
= m w g. (Eq. 3)
The density of the object can be determined by measuring the mass of the object in air and when submerged in a fluid, and using the following expression
  o
=
W o  w
(Eq. 4)
F
B

SUPPLIES & EQUIPMENT:
Double pan balance
600-ml beaker
Unknown fluid
Short support rod
String & scissors
150-ml beaker
500-ml graduated cylinder
Vernier caliper
Table clamp
Small paper clips
Lab jack
Rock sample
Hydrometer
Overflow can
Metal cube
PROCEDURE:
A. DENSITY OF A METAL CUBE
1. Measure the length of one side of the metal cube. Calculate the volume of the cube V o
.
2. Hang a paper clip and a string from the bottom of the left side of the beam balance mounted on a support rod as in Figure 1. Make sure all sliding weights are set at their zero positions and to zero the balance, rotate the small calibration weight until the needle lines up on the center vertical line. Suspend the cube from the beam balance with the attached string, as shown in Figure 1, and determine its mass, m o
.
3. Immerse the suspended cube in a beaker of water as in Figure 2. Determine the apparent mass of the cube in water, m aw
.
4. Determine the buoyant force (F
B
= W o

W aw
) in newtons. (Eq. 1)
5. Determine the density of the cube from
 o
=
W o  w
. (Eq. 4)
F
B
B. DENSITY OF AN IRREGULARLY-SHAPED ROCK
1. Suspend a rock from the beam balance and determine its mass, m o
in grams ____________, and convert to kilograms ____________.
2. Immerse the suspended rock in a beaker of water as in Figure 2.
- 56 -

3. Determine the apparent mass of the rock immersed in the fluid, m aw
in grams ____________, and convert to kilograms ____________.
4. Determine the buoyant force, F
B
= W o

W aw
, in Newtons. (Eq. 1)
5. Determine the density of the rock from
 o
=
 o
= m o g
 w
. (Eq. 4)
F
B
6 Determine the mass of a 150-ml beaker m b
= _______________ kg.
7. Place the displacement can on a level surface near the edge of a sink. Fill it with water, and let the excess drain off into the sink.
8. Slowly lower the rock into the water, allowing the displaced water to flow into the small beaker. Weigh the beaker with displaced water. m b+w
= _______________ kg.
9. Determine the mass of water displaced, m w
= m b+w
- m b
____________ kg.
10. Determine the weight of water displaced, W magnitude to the buoyant force F
B w
= m w
g ____________ N. This is equal in
, according to Archimedes’ principle.
11. Determine the density of the rock by applying (Eq. 4), ______________ kg/m 3 .
C. SPECIFIC GRAVITY OF AN UNKNOWN LIQUID.
1 With the same cube used in Part A, determine the buoyant force, F
B (fluid )
on the metal cube by immersing it in an unknown fluid.
F
B (fluid)
= W o

W af
. W o
= weight of object in air.
F
B (fluid)
= m o g

m af
g W af
= apparent weight of object in fluid.
2. Calculate the density of the fluid using equation (5).
In Water:
 o
= m o
F g
B
 w
=
W o

F
B ( w water )
In Fluid:
 o
= m o g
 f
F
B
=
W o
 f
F
B ( fluid )
Therefore,
W o
 f
F
B ( fluid )
=
W o
 w
F
B
, and
 f
=
F
B ( fluid )

F
B ( water ) w
( Eq. 5) where
 o
= density of metal cube in air,
 w
= density of water and
 f
= density of Fluid
3. Calculate the specific gravity, S.G. =
 f
/
 w
.
4. Fill a tall measuring cylinder with the “unknown” fluid. Use a hydrometer to measure the specific gravity of the fluid.
- 57 -

DATA SHEET: ARCHIMEDES' PRINCIPLE
A. Metal Cube
Mass of cube
(i) Apparent mass of cube in water
Buoyancy F
B
= m o g

m aw g
Density of cube
 o
=
W o
F
B
(ii) Length of side
 w
Volume of cube
Density of cube
 o
= m o
V o
(iii) Density (known)
B. Rock
Mass of rock
Weight of rock
(i) Apparent mass of rock in water
Buoyancy F
B
= m o g

m aw g
Density of rock
 o
=
W o
F
B
 w
(ii) Mass of beaker
Mass of beaker with water
Mass of water displaced
Weight of water displaced
Density of rock
 o
= (W o
/F
B
)

W w w
= m m = __________ kg
V m m aw
F


 m
W m
F
 m m o aw w
= __________ kg
= __________ kg w g = F
B
= __________ N

B o

= __________ kg
= __________ N
= __________ kg/m
L = __________ m o o o
B o b o b+w

= __________ m
= __________ kg/m
= __________ kg/m
= __________ kg
= __________ kg
= __________ N
= __________ kg/m
= __________ kg
= __________ kg
3
= __________ kg /m
3
3
3
3
3
- 58 -

C. Specific Gravity
(i) Mass of cube (from part A)
Apparent mass of cube in fluid m m o af
= __________ kg
= __________ kg
Buoyancy F
B(fluid)
= m o g

m af g F
B
= __________ N
Density of fluid
 f
=
F
B ( fluid )
F
B ( water )
 w

= __________ kg/m
Specific gravity
 f
/
 w
S. G. = __________
(ii) Specific gravity from hydrometer reading S. G. = __________
3
Percent error __________
- 59 -

PURPOSE :
The purpose of this experiment is to measure the coefficient of linear expansion for various metals and to compare the results with the known values.
INTRODUCTION :
In most cases, when materials are heated or cooled, they undergo expansion or contraction respectively. From the standpoint of materials science, this process must be taken into account when designing structures that are subjected to temperature variations. Otherwise, tensile or compressive stresses might develop which could destroy the structure.
The linear (one-dimensional) coefficient of expansion is defined as the fractional increase in length divided by the temperature change. This coefficient is designated by the Greek letter alpha (

), and is found to be almost constant over a wide range in temperature. In equation form, the definition of

is:

=
L

L o

T where

L is the change in length, L o
is the original length, and

T is the temperature change in degrees Celsius.
In this experiment, the value of the linear coefficient of expansion of several rods of common metals will be determined. The length of the rod is measured at room temperature, then steam is passed over the rod with the resulting temperature increase causing it to expand. The amount of expansion is measured with a dial indicator. The coefficient is then determined using the data gathered.
SUPPLIES & EQUIPMENT:
Linear expansion apparatus
Aluminum, copper and steel rods
0 - 100 o C Thermometer
Dial indicator
Electric steam generator
Meter stick
Glycerine
PROCEDURE :
1. Measure and record the initial length of the rod L o
, to the nearest millimeter. Determine and record the ambient temperature (room temperature).
- 60 -

2. Set up the apparatus as shown in Figure 1. The steam jacket for the rod has an opening for steam, thermometer, and rod ends, and an outlet for the condensed steam. Fill the steam generator about 2/3 full of water and turn on the generator, but do not connect the generator to the expansion apparatus as yet. Insert the rod in the apparatus until it just makes contact with the dial indicator probe and is in firm contact with the screw at the other end.
Steam inlet tube
Thermometer
Rod
Steam
Generator
Dial indicator
Steam outlet tube
To sink
Fig. 1. Linear Expansion Apparatus with Associated Equipment
3. Make sure that the dial indicator is firmly screwed onto its holder and that the graduated ring is tightened down. See Fig. 2. Record the initial reading of the dial indicator, to the nearest
0.01 mm (= 0.00001m).
Secure movable ring firmly
20 30
READING THE DIAL INDICATOR:
0.01 mm per div .
0
10
90
80
0
1 2
9
0
1
8 7
6
2 3
4
5
60
50
70
40
1 mm/div
1 cm/div
Example:
The gauge to the left indicates
0.07 mm
The gauge below indicates 0.14 mm
10
20
0
Fig. 2 Dial Indicator (Micrometer Gauge) 0
1
0
1 2
3
4
- 61 -

4. When the generator is generating steam briskly, connect the steam tube to the inlet on the apparatus. Warning! Be careful not to scald yourself.
5. Allow the steam to warm up the rod to a constant maximum temperature, T max
. When the rod stops expanding, record the final reading of the dial indicator.
Calculate

T = (T max

T ambient
).
6. Calculate the change in length,

L = Final reading - Initial reading.
7. Calculate the coefficient of expansion and record it on the data sheet. Compare your values with the known values of the coefficient of linear expansion by calculating the percent difference.
8. Repeat the above procedure for two other rods. Be careful not to burn yourself on the hot metal. When finished, dry the equipment thoroughly.
- 62 -

DATA SHEET: COEFFICIENT OF LINEAR EXPANSION
Ambient Temperature _________________ o C
Data and Calculations Table:
Aluminum Type of Rod
L o
(m)
Initial Reading of Dial Indicator (m)
Final Reading of Dial Indicator (m)
T max
( o
C)

T ( o C)

L (m)

( o C -1 )
Known

( o
C
-1
)
Percent error
Copper
- 63 -
Steel

PURPOSE :
The value of the latent heat of fusion for water and the latent heat of vaporization for water will be determined by the method of calorimetry.
INTRODUCTION :
When a substance such as water undergoes a change of state from the solid phase to the liquid phase, not all of the heat energy that is added to the system is reflected in a change of temperature of the substance. Some energy is needed to break the permanent bonds between the molecules of the substance and this energy is called the latent heat of fusion of the substance.
In today's experiment the latent heat of fusion will be determined by the method of mixtures and by applying the principle that the heat lost is equal to the heat gained (conservation of energy).
For the heat of fusion, ice cubes are placed into a measured amount of warmed water and is left to melt, cooling the water in the process. By noting the temperatures before and after melting, the heat of fusion can then be calculated.
For the heat of vaporization, steam is introduced into a measured amount of cooled water warming the water in the process. By noting the temperatures before and after the introduction of the steam, the heat of vaporization can be calculated.
SUPPLIES & EQUIPMENT:
Double-walled calorimeter
Thermometer
Baster
Ice cubes
Electronic balance
Pinch clamp
Steam Generator
Water trap
Ring Stand
PART A: HEAT OF FUSION
HEAT LOST: by water = (mass of water) X (1.00 cal / g
.
o C ) X (temperature change)
Q w
= m w c w
(T o
- T f
) c w = Specific heat of water. by calorimeter = (mass of calorimeter) X (0.22 cal / g
.
o C ) X (temperature change)
Q c
= m c c c
(T o
- T f
)
HEAT GAINED: c c = Specific heat of calorimeter. by ice cube = (heat needed to melt the ice) + (heat for warming the melted ice)
Q i
= m i
L f
+ m i c w
(T f
- 0 o C)
- 64 -

CONSERVATION OF ENERGY:
Heat Gained = Heat Lost
Q i
= Q w
+ Q c
Eq. (1)
m i
L f
+ m i c w
(T f
- 0 o C) = m w c w
(T o
- T f
) + m c c c
(T o
- T f
)
m i
L f
= m w c w
(T o
- T f
) + m c c c
(T o
- T f
) - m i c w
(T f
- 0 o C) Eq. (2)
L f
= m w c w
( T o

T f
)
 m c c c
( T o

T f
)
 m i c w
( T f

0 o
C )
Eq. (3) m i
PROCEDURE: Part A: Heat of Fusion
1. Determine the mass of the inner cup and stirrer of the calorimeter to the nearest tenth of a gram. Do not include the plastic collar in any measurements of mass.
2. Fill the inner cup of the calorimeter to about ⅔ full with warm water at about 40 o Celsius. Use the baster to transfer hot water from the steam generator to reach this temperature. You need to start the experiment with water hotter than room temperature because the calorimeter is not a perfect insulator. Some heat leaks out at the beginning of the experiment, but leaks back in near the end of the experiment as the temperature drops below room temperature. By starting with heated water, these heat transfers approximately cancel each other out.
3. Determine the mass of the inner cup, stirrer, and water. Calculate the mass of water in the cup.
4. Place the cup, stirrer, collar, and water into the outer calorimeter jacket and record the initial temperature just before the ice cube is placed in the water, to the nearest tenth of a degree
(yes, I know it’s difficult).
5. Wipe all the water from the surface of one large ice cube or three small ice cubes and place the ice carefully into the calorimeter cup.
6. Stir the contents occasionally while constantly observing the ice cubes. As soon as the ice cubes are completely melted, record the final temperature.
7. Determine the mass of the cup, stirrer and contents. The mass of the ice cubes can now be calculated.
8. Use Eq. (3) to compute the heat of fusion of water, and compare this to the accepted value by calculating the percent difference.
- 65 -

DATA SHEET: HEAT OF FUSION OF WATER
Data and Calculations Table A:
Mass of inner cup and stirrer of calorimeter, mc (g)
Mass of inner cup, stirrer and water, mc + mw (g)
Mass of water, mw (g)
Initial temperature of water, To (oC)
Final temperature of contents,Tf (oC)
Mass of cup, stirrer and contents, mc + mw + mi (g)
Mass of ice cubes, mi (g)
Heat of fusion of water, Lf Eq. (3) (cal/gram)
Known value of the heat of fusion of water
Percent error
- 66 -
79.7 cal/gram

PART B: HEAT OF VAPORIZATION
HEAT LOST: by steam = (heat released in the condensation of steam) + (heat released in cooling of condensed steam)
Q s
= m s
L v
+ m s c w
(100 o C - T f
) c w = Specific heat of water .
HEAT GAINED: by water = (mass of water) X (1.00 cal / g
.
o
C ) X (temperature change)
Q w
= m w c w
(T f
- T o
) by calorimeter = (mass of calorimeter) X (0.22 cal / g
.
o
C ) X (temperature change)
Q c
= m c c c
(T f
- T o
) c c = Specific heat of calorimeter.
CONSERVATION OF ENERGY:
Heat Lost = Heat Gained
Q s
= Q w
+ Q c
Eq. (4) m s
L v
+ m s c w
(100 o C - T f
) = m w c w
(T f
- T o
) + m c c c
(T f
- T o
)
m s
L v
= m w c w
(T f
- T o
) + m c c c
(T f
- T o
) - m s c w
(100 o C - T f
) Eq. (5)
L v
= m w c w
( T f

T o
)
 m c c c
( T f

T o
)
 m s m s
PROCEDURE: Part B: Heat of Vaporization
1. Set up the apparatus as shown in c w
( 100 o
C

T f
)
Eq. (6) the diagram to the right.
2. Determine the mass of the inner cup and stirrer of the calorimeter.
3. Fill the inner cup to about ⅔ full with cold water at about 15 o below room temperature.
4. Determine the mass of the inner cup, stirrer and water. Calculate the mass of water in the cup.
5. Place the cup, stirrer, collar and water into the outer calorimeter jacket and record the initial temperature just before steam is introduced into the water.
6. Stir the contents occasionally while constantly observing the rise in temperature. Drain the water from the water trap by occasionally opening the clamp on the hose that drains into the sink. Only steam, and no water, should enter the hose leading to the calorimeter. When the temperature has increased to about 15 o above room temperature, remove the steam inlet and record the final temperature.
- 67 -

7. Determine the mass of the cup, stirrer and contents. The mass of the steam which was introduced and condensed into water can now be calculated.
8. Use Eq. (6) to compute the heat of vaporization of water L v
, and compare this to the accepted value by calculating the percent difference.
DATA SHEET: HEAT OF VAPORIZATION OF WATER
Data and Calculations Table B:
Mass of inner cup and stirrer of calorimeter, mc (g)
Mass of inner cup, stirrer and water, mc + mw (g)
Mass of water, mw (g)
Initial temperature of water, To (oC)
Final temperature of contents,Tf (oC)
Mass of cup, stirrer and contents, mc + mw + ms (g)
Mass of steam, ms (g)
Heat of vaporization of water, Lv Eq. (6) (cal/gram)
Known value of the heat of vaporization of water
Percent error
540 cal/gram
- 68 -