2.6.1 Draw a vector diagram to show that the acceleration of a particle moving with uniform speed in a circle is directed toward the centre of the circle. Review of basic kinematics: If the acceleration and velocity of an object are parallel (or anti-parallel) then the object's speed will increase (decrease). If the acceleration and velocity of an object are perpendicular then only the direction of the velocity will change and the speed (i.e. the magnitude of the velocity) will remain constant. If a ball is attached to the end of string and swung at a constant speed (i.e. only the direction of the velocity is changing not the magnitude) then there must still be an acceleration. The acceleration is directed towards the center of the motion. This acceleration is call centripetal acceleration!

2.6.2 State the expression for centripetal acceleration.
The acceleration of any object moving in a circle at a constant speed is given by the equation: (1)
a⃗ =v2r
It is important to note that centripetal acceleration is very special. It is the acceleration required for an object to move in a circle at a constant speed. The reverse is also true if an object's acceleration is equal to v2r (and perpendicular to the velocity) then the object must be going in a circle. If an object is moving in a circle at with a changing velocity, then the overall acceleration is not equation to the centripetal acceleration. However the acceleration perpendicular to the velocity (that is the part changing the direction) is still equal to v2r 2.6.3 Identify the force producing circular motion in various situations Sometimes people will make reference to the "centripetal force." This is not a real force, its a pseudo-force. In general the centripetal force is made up of many other forces and is the sum of those forces. This is not unlike the idea of a net force which is also generally the sum of multiple forces. If you have a ball on the end of a string and you swing it in a vertical circle...

...
E105: UNIFORM CIRCULARMOTION
NADONG, Renzo Norien D.
OBJECTIVE
The purpose of this experiment is to quantify the centripetal force on the body when one of the parameters is held constant and to verify the effects of the varying factors involved in circularmotion. Mainly, horizontal circular type of motion is considered in this activity.
Circularmotion is defined as the movement of an object along the circumference of the circle or the manner of rotating along a circular path. With uniform circularmotion it is assured that the object traversing a given path maintains a constant speed at all times. Centripetal force is a force that tends to deflect an object moving in a straight path and compels it to move in a circular path.
MATERIALS AND METHODS
This experiment was divided into three parts in order to further study and observe the factors that affect the centripetal force of a body. The concept of this experiment is the same on all parts, which is getting the centripetal force given with three different conditions. Every part of the experiment was executed just the same. Mass hanger plus a desired mass of weights were hanged over the clamp on pulley to determine a constant centripetal force which will act as the actual value. But on the third part of this experiment, aside from the centripetal...

...II Uniform CircularMotion
A. Nomenclature
1. Speed – magnitude of an objects rate of motion (no direction, scalar quantity)
2. Velocity – speed and direction of an objects motion (vector, mag & direction)
3. If a car’s speed is constant but direction is changing, velocity is changing.
4. 2 ways to change velocity (change speed or change direction).
5. acceleration – change in speed over time (vector quantity) TWO types;
a. Linear acceleration – speed up or slow down
b. Centripetal acceleration – change direction
B. Centripetal acceleration (ac) – acceleration changes due to change in direction.
1. Centripetal means center seeking
2. ac is always directed toward the center of the curved path (circle)
3. If an object is moving in a circle it will always have a centripetal acceleration
4. ac = v2/r v=velocity tangent to the circle (m/s) r = radius of the circle (m)
C. Centripetal Force – the force that causes and maintains circularmotion
1. Centripetal Force – Fc – psuedo-force (various forces act as center seeking force)
2. Fc – direction always toward the center.
3. Fc=mac (sub ac = v2/r)
4. Identify Fc
a. Rope over your head
b. Car rounds a corner
c. Earth – Moon
d. Gravitron machine (Fn)
e. Loop de loop (Fn Fg)...

...there is a ∆v and ∆v = vf
- vi
, and
since velocity is changing, circularmotion must also be accelerated motion.
vi
∆v vf
-vi
vf2
If the ∆t in-between initial velocity and final velocity is small, the direction of ∆v
is nearly radial (i.e. directed along the radius). As ∆t approaches 0, ∆v becomes
exactly radial, or centripetal.
∆v = vf
- vi
vi
vf
vf
∆v
-vi
Note that as ∆v becomes more centripetal,
it also becomes more perpendicular with vf
.
Also note that the acceleration of an object depends on its change in velocity ∆v;
i.e., if ∆v is centripetal, so is ‘a’.
From this, we can conclude the following for any object travelling in a circle at
constant speed:
The velocity of the object is tangent to its circular path.
The acceleration of the object is centripetal to its circular path. This type of
acceleration is called centripetal acceleration, or ac
.
The centripetal acceleration of the object is always perpendicular to its
velocity at any point along its circular path.
v
ac
ac
v 3
To calculate the magnitude of the tangential velocity (i.e., the speed) of an
object travelling in a circle:
• Start with d = vavt where ‘vav’ is a constant speed ‘v’
• In a circle, distance = circumference, so d = 2πr
• The time ‘t’ taken to travel once around the circular path is the...

...Uniform CircularMotion
PES 115 Report
Objective
The purpose of this experiment is to determine the relationships between radiuses, mass, velocity and centripetal force of a spinning body. We used logger pro to accurately measure the orbital period of the spinning mass and used these measurements to determine the interrelated interactions of the specified properties and viewed the results graphically.
Data and Calculations
The black markings on the string are about 10 cm apart in length, measured from the center of the spinning mass.
Part A: Factors that influence CircularMotion
Velocity versuse Centripetal Force
Fill out the table holding the Spinning mass (M) and the radius (R) constant.
Figure 1: Experimental setup for the lab
Which Spinning Mass did you select _hook with foam wrapping_ (Tennis ball, etc..)
What is the mass of the Spinning mass _0.0283_ kg.
What Radius did you select _0.30_ m (around 20 cm is a good choice).
Fill out the tables for five different hanging mass values.
Hanging Mass (m) [kg]
0.1001 kg
0.1992 kg
0.2992 kg
0.4000 kg
0.4997 kg
Revolution Number and Time per Revolution (T) [sec]
1
0.61337 s
0.413210 s
0.367288 s
0.316510 s
0.271455 s
2
0.613087 s
0.403737 s
0.370600 s
0.310189 s
0.274200 s
3
0.613727 s
0.393689 s
0.374100 s
0.316308 s
0.273700 s
4
0.611319 s
0.39364 s
0.368047 s
0.309619 s
0.279400 s
5
0.618954 s
0.388600 s
0.365853 s
0.300742 s
0.282000 s
6
0.589000 s...

...CircularMotion
Uniform circularmotion is the movement of an object or particle trajectory at a constant speed around a circle with a fixed radius. The fixed radius, r, is the position of an object in uniform or circularmotion relative to to the center of the circle. The length of the position vector of the circle does not change but its direction does as the object follows its circular path. In order to find the object’s velocity, one needs to find its displacement vector over the specific time interval. The change in position, or the object’s displacement, is represented by the change in r. Also, remember that a position vector is a displacement vector with its tail at the origin. It is already known that the average velocity of a moving object is ᐃd/ ᐃt, so for an object in circularmotion, the equation is ᐃr/ ᐃt. IN other words the velocity vector has the same direction as the displacement, but at a different length. As the velocity vector moves around the circle, its direction changes but its length remains the same. The difference in between two vectors, ᐃv, is found by subtracting the vectors. The average acceleration, a = ᐃv/ ᐃt, is in the same direction as ᐃv, that is, toward the center of the circle. As the object moves around the circle, the direction of the acceleration vector changes, but its length remains the same.One should...

...I noticed that I have not described the rule of F=ma in either the last email or this one. Where would you suggest it be described?
Somehow the details of adding forces and balanced forces were missed in the last email and also it did not make perfect sense for me to note. As far as I am concerned the khan academy does not lecture it so I am not too sure in what to do about this.
I am assuming finding velocity is the sole purpose of applying the law of conservation of momentum. Is this true?
I also would like to note that a graph could not be drawn in some situations again due to me lacking the technology to send photos of handwritten notes. Hence there is sadly no examples of a problem for translational equilibrium and for the force-time graph in which impulse can be identified.
I also have referred to explosions as divisions. Is this appropriate?
Newton's First Law of Motion:
A body will remain at rest or moving with constant velocity unless acted on by an unbalanced force.
Example:
• Q: while traveling in train if one throws a ball up it lands on his palm though the train is moving. my doubt is that though the ball is detached from motion how does it manage to land on his palm though he is moving along with the train?
• A: he ball lands on your hand because the ball is, in reality, traveling at the same velocity as the train, you, and everything else on,...

...Motion
NCERT Chapter Questions and Answers and other Q & A
Q1: An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Answer: Yes an object can have zero displacement even though it has moved through a distance. It happens when the object moves back to its original position i.e. final position coincides with the starting position.
Example: Suppose an object travels from O to C and then comes back to original position O.
Total distance traveled = actual path covered = OC + CO = 25 + 25 = 50m
Total displacement = shortest distance between final position and initial position = 0m
Q2: What do you mean by a body in rest?
Answer: A body is said to be at rest, if it does not change its position with respect to a fixed point in its surroundings.
Q3: Are motion and rest absolute or relative? Explain with an example.
Answer: No these terms rest and motion are relative. For example, a person inside a car, carrying a ball in his hand will see the ball is at rest. While for another person, outside the car will see the ball is also moving.
Q4: What is meant by scalars and vectors?
Answer:
* Scalar Quantities: Quantities that require magnitudes only to specify them are called scalar quantities or scalars. Mass, length, time, temperature, angle, area, speed, distance, volume and density are examples of scalar quantities.
* Vector Quantities: Quantities...

...Exploration Guide: Uniform CircularMotion
Go to www.explorelearning.com and login. Please type or write your answers on a separate sheet of paper, not squished in the spaces on these pages. When relevant, data collected should be presented in a table.
Objective: To explore the acceleration and force of an object that travels a circular path at constant speed. Motion of this kind is called uniform circularmotion.
Part 1: Centripetal Acceleration
1. The Gizmotm shows both a top view and a side view of a puck constrained by a string, traveling a circular path on an air table. Be sure the Gizmo has these settings: radius 8 m, mass 5 kg, and velocity 8 m/s. Then click Play and observe the motion of the puck.
a. The puck in the Gizmo is traveling at a constant speed, but it is NOT traveling at a constant velocity. Explain why.
b. Because the velocity of the puck is changing (because its direction is changing), the puck must be experiencing an acceleration. Click BAR CHART and choose Acceleration from the dropdown menu. Check Show numerical values. The leftmost bar shows the magnitude of the acceleration, or |a|. (The other two bars show the x- and y-components of the acceleration, ax and ay.) What is the value of |a|? Jot this value down, along with radius = 8 m, so that you can refer to it later.
c. Keeping velocity set to 8 m/s, set radius to 4 m....