Free Report On Experiment 1: Uniform Circular Motion

Lab Report: Circular motion

Introduction
The uniform circular motion experiment investigates the continuous motion of a body rotating about an axis. A body in circular motion has a dynamic velocity vector. Even though the rotation speed of the body is constant, the change in vector means that the body is in constant acceleration. The force that is responsible for the acceleration is referred to as centripetal force. The centripetal force keeps the body in its circular path and if retracted, the body will leave the path in a tangential direction relative to the circular path. Centripetal acceleration is the main difference between linear and circular motion.

Objectives

The objectives of the uniform circular motion experiment were to study:
Velocity in angular motion
The relationship between angular velocity and regular velocity in circular motion
Acceleration in uniform circular motion
Newton’s second law of motion in relation to uniform circular motion.
Theory
The position of a body in circular motion can be described by its angle θ in reference to an arbitrary line running across its path of rotation. If the body is rotating in uniform circular motion, its position at time t is given by the angle θ or by the length of cure S. after a duration of time dt, the body will have moved a distance ds through an angle d θ.
Figure 1: Diagram showing displacement of an object in circular motion through angle d θ and distance s.

Centripetal acceleration

Δvv=Δr/r Therefore, a=ΔvΔt=vΔrrΔt=v2/r
Figure 2: Image showing change in velocity vector due to centripetal force.
According to newton’s second law o motion, F=ma which gives F=mv2r where F is the centripetal force.

Experiment equipment

Sargent- Welch centripetal force apparatus
Rubber bands
Assorted springs
Stopwatch
Plastic pail and washers
String and paper clips
Ruler
Procedure
Figure 3: Diagram showing the Sargent-Welch centripetal force apparatus set up
The equipment is set up as shown above. A horizontal arm slides through the post and supports a counterweight on one side and a bob on the other. An elastic band or spring attaches the bob to the post and represent the centripetal force by keeping the bob in a circular path during rotation. A pointer indicates the position of the bob when it is vertically aligned. The procedure is as follows:
The centripetal force was determined by measuring the fore of the spring necessary to keep the bob aligned to the pointer during rotation. Washers were added to pail until its mass M pulled the bob vertically above the pointer.

The mass of the bob, m, was measured then reattached back to the horizontal arm

The pointer was attached midway along the length of the base and the arm adjusted so that the bob was vertically aligned to it when free hanging.
When the arm was freely balanced by the counterweight and the bob, it was fastened tight on the post and the distance r running from the center of the pointer to the center of the bob was recorded.
With the plastic pail with the washers detached from the bob, the spring pulled the bob towards the post. The base was then spun until the bob was realigned with the pointer.

The average time t taken to complete two rotations was recoded and used to calculate the periodic time, T.

The spring, rotation speed, and r were then varied to give several readings.
Results and discussion questions
Figure 4: Graph showing variation of velocity with increasing centripetal force
The velocity is given by the equation: V=2πr/T where T is the periodic time
The values of velocity are likely to be influenced by the errors in in r as it will lead to a false circumference hence distance travelled by bob.

Theoretically, the graph of F against V is supposed to be a straight line according to the equation:

F=mv2r.
The graphs above do not depict a straight line because of experimental errors, either in recording of the periodic time or when taking measurements.

The graph of F against V2 shows the relationship between acceleration and the centripetal force.

Discussion questions
The units of the gradient of the F against V2 curve are:
Nv2=Nms2=Ns2/m2
The plot of F against v2 should pass through the origin because when F=0, v2=0.
According to the diagram below, when the marble leaves the semicircular curve, it will follow path B. This is because the centripetal force applied in form of reaction force by the semi-circular object is no longer present so it shoots off in a tangential path.

In order to keep a constant centripetal acceleration when the velocity is doubled, the radius should be squared.

EXPERIMENT 2: CONSERVATION OF MOMENTUM IN COLLISIONS
Introduction
Bodies which are in contact with each other exert equal and opposite forces on each other. Newton’s law of conservation of momentum stipulates that the combined momentum of two bodies when they collide is equal to the sum of their individual momentums before the collision. Collisions are either elastic or inelastic. In elastic collisions, both the kinetic energy of the bodies and momentum are conserved while energy is lost but momentum in conserved in inelastic collision.
Objectives

The objectives of the experiment was

Trace the motion of the center of mass of a two body system.
Theory
If two objects of equal masses m1 and m2 collide on a frictionless surface, they will exert opposite forces on ach oter whoichcan be given by:
m1a1=-m2a2

This can also be written as:

m1Δv1Δt=-m2Δ Δv2/Δt
Thus m1Δv1=m2Δv2

Where Δt is an infitesmal change in time and Δv1 is the change invelocity of m1.

The aove equation can alsobe expressed as
m1v1+(Δt-v1(t))=-m2v2+(Δt-v2(t))

Which further gives

m1v'1+m2v2'=m1v1+m2v2
Where the primed velocities are the initial velocities before collision an the unprimed Vs denote velocity after collision. Momentum of a moving object is given as the product of mass and velocity, p=mv.

Thus, the equation above can be expressed as

p1'+p2'=p1+p2
This denots that the sum of momentum of two bodies at time t is equal th some of the same objects at time (t+Δt).

Equipment

Air table apparatus
Pair of velco rings
Two plain paper sheets
Two indentical clear plastic bags
Clear plastic protractor
Clear plastic ruler
Procedure 1: Elastic collision
Figure 5: Air table apparatus showing positioning o the two pucks
The air table apparatus has spikes on the central electrodes under the puke. The sparks from the spark generator momentarily close the circuit through the lack carbon paper and the white paper. The generated sparks creates black carbon markings under the white paper along the path of the pucks.

The air was turned and the masses of the two pucks were recorded as m1 and m2.

The two pucks were then set on adjacent corners on one side of the air table apparatus such that they met at the middle of the apparatus, collided and moved to the opposite corners.
The white paper was then removed and the pucks’ paths were studied. Each path was marked 1,2, 1’, and 2’. The primed number denotes the pucks’ paths after collision.
The sparks marks along the pucks’ paths were marked, after every four sparks i.e. 0, 4, 8 etc.
The velocity vector in the four tracks was the outlined by the use of a ruler and the distanced between three consecutive sparks was determined. The velocity vector magnitude was then determined by dividing the distance by the time between the three spar points.

The velocities were then added up graphically through the use of triangles.

Discussion
Figure 6: Diagram showing velocities o the two pucks before and after the collisions, the second diagram shows the vector addition of the velocities

The mass of the pucks are the same so,

v1+v2=v1'+v2'
The sums of the two vectors will not be equal in a practical situation due to friction caused by the rubber hoses. Also, the collision is also not perfectly elastic which will lead to loss in kinetic energy.

In perfectly elastic collision, both momentum and kinetic energy are conserved.

Inelactic collisions
In inelastic collision test, the same procedure as the the leastic test is repeated. The collosion is made inelastic by wrapping the pucks with velcro strips. The two pucks stick together after collision leading to loss in kinetic energy.

Motion of the centre of mass

The motion of the centre of mass between two objects fall perfectly mid distance between them if they are of the same weght. For masses with dissimilar weght, the center of mass is given by
x=(m1x1+m2x2)/M

Where M is the combined mass of the two objects.

The pucks were placed on th air table apparatus and the timer set to 50ms.
The pucks were then made to make one revolution arund each other
The sparck marks for each puck were numered and corresponding ponts was joined by a line.
The midpoint between two corresponding points, which gave the centre of mass for the two pucks, was determined.
Results analysis
Elastic collision
V’1=8.3cm/5*30ms=0.055cm/ms, V1=30.8cm/20*30=0.051cm/ms
V’2 =10cm/6*30ms=0.055cm/ms, V2=21cm/21*30ms=0.033 cm/ms
Percentage error is 0.027/0.084*100 = 32.14%

Thi shows the collision was not perfectly elastic as velocity before collision is not the same as velocity after collision.

Inelastic collision
V1=4.8cm/5*30ms=0.32cm/ms V2=8.5cm/5*30ms=0.057cm/ms
V’1=3.9cm/5*30ms=0.026cm/ms V’2=4cm/5*30ms=0.027cm/m
(V’1+V’2)-( V1+V2) = (0.032+0.057)- (0.026+0.027) = 0.036
Percentage error: 0.036/0.053*100= 67.92%
This shows the collision was not completely inelastic since the velocities of both plucks was not the same after he collisions.

Center of mass

Figure 7: Graph of angles of centers of mass between consecutive puck markings
Discussion questions
Energy loss is directly proportional to changes in velocity after collision. Theoretically, inelastic collision is supposed to exhibit a higher energy loss than elastic collision. The above results are in line with this principle as elactic collision led to 32.14% loss in energy while inelastic collision led to 67.92% losses.
The pointsof center of mass do not lie on a straight line. This is because the movement of the pucks is not linear and tend to assume a rotary motion.

The plot of angle plots against time is not linear because the pucks travel different distances per each spark.

In the center of mass experiment, the momenta of each puck is conserved. It is depicted as the momentum of the centre of mass of the two weights.