Title: Uniform Circular Motion
Objective: To investigate the relationship between FnetT² and radius Proposed Hypothesis: FnetT² is directly proportional to the radius Manipulated variable: Radius of the circular motion
Responding variable: The time taken for 20 rotations
Controlled variables: The mass of the rubber stopper, the mass of the weight hanger, the total weight of the slotted weight, the length of the PVC tube Apparatus and Materials: rubber stopper, stopwatch, weight hanger, slotted weights,

crocodile clip, metre rule, thread, PVC tube
Diagram:

Procedure: 1. Weigh and record the masses of weight hanger and rubber stopper.
2. Tie the thread to the rubber stopper.
3. Pass the thread through the PVC tube.
4. Tie a node at the end of the thread and hang the weight hanger which is with
0.08g slotted weights on it.
5. Measure the 0.1m radius from the bottom of PVC tube and mark it.
6. Zheng Yie starts to rotate the thread with an acceleration until the bottom of
PVC tube is reached the mark.
7. Keep the speed of rotation constant so that the bottom of PVC tube is
always touched the mark.
8. After the speed is kept at constant speed, Adeline starts to studies the time
taken for 20 rotations by using a stopwatch.
9. Adeline is also responsible to record the time taken for 20 rotations.
10. Step 5 until step 9 are repeated by using different radii such as 0.2m, 0.3m,
0.4m, 0.5m, 0.6m, 0.7m, 0.8m, and 0.9m.
11. The experiment is repeated once again and take the average of the two of
the results as the final result.
Data:
Radius, r (m)| Time for 20 rotations, (s)| Time for 1 rotation, T (s)| T² (s²)| FnetT² (Ns²)| | Reading 1| Reading 2| Average| | | |
0.1| 05:37| 05:69| 05:53| 0:28| 0.078| 0.076|
0.2| 06:37| 06:94| 06:66| 0:33| 0.109| 0.106|
0.3| 07:44| 08:19| 07:82| 0:39| 0.152| 0.150|
0.4| 08:37...

...INVESTIGATING CIRCULARMOTION 11/3/04
AIM
To examine some of the factors affecting the motion of an object undergoing uniformcircularmotion, and then to determine the quantitative relationship between the variables of force, velocity and radius.
APPARATUS
Rubber bung Metre rule 50 gram slot masses
Glass tube 50-gram mass carrier 50-gram slot masses Metre rule
Stopwatch Sticky tape Metre rule String
THEORY
As in Jacaranda HSC Science Physics 2 p.54
In this experiment when the rubber bung is moving in a circularmotion and the string it is tied to moves neither up or down a constant radius is being maintained. For this to be true the centripetal force must equal the gravitational force hence
Mv"/r = mg from this
v"/r =mg/M and v" ∞ r therefore as v increases so does r and vice versa.
Where
m = Mass of mass carrier + masses (kg)
g = acceleration due to gravity 9.8 m/sec"
M = mass of object in motion (kg)
v = instantaneous velocity of mass (m/sec)
r = radius of circularmotion (m)
METHOD
As in Jacaranda HSC Science Physics 2 p.54
However instead of measuring the time for 10 revolutions, the time for 20 revolutions was measured, this allowed more accurate results to be obtained. Furthermore the lengths given in the book were used as merely guidelines and not followed precisely also 50 and 100-gram masses...

...Exploration Guide: UniformCircularMotion
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 uniformcircularmotion.
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...

...UniformCircularMotion – a constant motion along a circle; the unfirom motion of a body along a circle
Frequency (f) – the number of cycles or revolutions completed by the same object in a given time; may be expressed as per second, per minute, per hour, per year, etc.; standard unit is revolutions per second (rev/s)
Period (T) – the time it takes for an object to make one complete revolution; may be expressed in seconds, minutes, hours, years, etc.; standard unit is seconds per revolution (s/rev)
Note: Period and frequency are reciprocals: T = 1/f; f = 1/T.
Sample Problems:
1. Suppose the rear wheel makes 5 revolutions in 1 minute. Find the wheel’s period and frequency.
2. As a bucket of water is tied to a string and spun in a circle, it made 85 revolutions in a minute. Find its period and frequency.
3. * An object orbits in a circularmotion 12.51 times in 10.41 seconds. What is the frequency of this motion?
Tangential Speed (v or vs) – average speed; rotational speed; speed of any particle in uniformcircularmotion; standard unit is meters per second (m/s); v = Cf = C/T = 2πrf = 2πr/T = rω
Sample Problems:
3. What is the rotational speed of a person standing at the earth’s equator given that its radius is 6.38*106 m and that it takes 365 days for the earth to complete a revolution?
4. A ball...

...changing. Because the direction is changing, 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...

...CircularMotion and Gravitation
Circularmotion is everywhere, from atoms to galaxies, from flagella to Ferris wheels. Two terms are frequently used to describe such motion. In general, we say that an object rotates when the axis of rotation lies within the body, and that it revolves when the axis is outside it. Thus, the Earth rotates on its axis and revolves about the Sun.
When a body rotates on its axis, all the particles of the body revolve – that is, they move in circular paths about the body’s axis of rotation. For example, the particles that make up a compact disc all travel in circles about the hub of the CD player. In fact, as a “particle” on Earth, you are continually in circularmotion about the Earth’s rotational axis.
Gravity plays a large role in determining the motions of the planets, since it supplies the force necessary to maintain their nearly circular orbits. Newton’s Law of Gravity describes this fundamental force and will analyze the planetary motion in terms of this and other related basic laws. The same considerations will help you understand the motions of Earth satellites, of which there is one natural one and many artificial ones.
Angular Measure
Motion is described as a time rate of change of position. Angular velocity involves a time rate of...

...BCA- I Physics Assignment 1 Unit I Laws of Motion
1. A man of mass 70kg stands on a weighing machine in a lift which is moving
(a) upwards with a uniform speed of 10m/s.
(a) downwards with a uniform acceleration of 5m/s2 .
(a) upwards with a uniform acceleration of 5m/s2.
What would be the readings in each case?
(d)what would be the reading if lift mechanism failed and it hurtled down freely under gravity?
2. A shell of mass 0.2kg is fired by a gun of mass 100kg. If the muzzle speed of the shell is
80m/s, what is recoil speed of the gun?
3. A mass of 6kg is suspended by a rope of length 2m from the ceiling. A force of 50N in the
horizontal direction is applied at the midpoint P of the rope. What is the angle the rope
makes with the vertical in equilibrium.( take g=10m/s2, neglect mass of rope)
Q3. Q4.
4. What is the acceleration of the block and trolley system,if the coefficient of kinetic friction
between the trolley and the surface is 0.04? What is the tension in the string? (neglect
mass of the rope, take g=10m/s2.)
5. Explain the following with appropriate reasons:
(i) What is the source of centripetal force when earth revolves around the sun?
(ii) For uniformcircularmotion, does the direction of centripetal...

...of the Pantheon in Paris with a wire that was over 200 feet long. The ball was used as a pendulum, and it could swing more than 12 feet back and forth. Beneath the ball he placed a circular ring with sand on top of it. Attached to the bottom of the ball was a pin, which scraped away the sand in its path each time the ball went by. To get the ball started on a perfect plane, the ball was held to the side by a cord until it was motionless. At that point, the cord was burned, which started the ball swinging. As the ball continued to swing as a pendulum, the path the pin carved into the sand changed, as the floor itself, as well as the rest of the Earth, was moving beneath it.
Essentially, the Foucault pendulum demonstrates the rotation of the Earth. The Foucault pendulum is not forced to stay in a fixed plane like Newton’s pendulum, also known as Newton’s cradle, which means it can move freely in response to the Coriolis force. The Coriolis force, also known as the Coriolis effect, occurs when masses above the Earth’s surface, such as a bullet or rocket, appear to be deflected from their trajectory, meaning they don’t reach their intended location straight ahead of them. In fact it is our frame of reference, the Earth, which is changing. Our frame of reference changes due to our uniformcircularmotion around the Earth. As the Earth is not a perfect circle (elliptical), the closer to the equator you are, the further...

...ASP 0501
EXERCISES – circularmotion
1 A car travels at a constant speed around a circular track whose radius is 2.6 km. The car goes once around the track in 360 s. What is the magnitude of the centripetal acceleration of the car?
2 An astronaut in a chamber moves on a circular path, much like a model airplane flying in a circle on a guideline. The chamber is located 15 m from the center of the circle. At what speed must the chamber move so that the astronaut is subjected to 7.5 times the acceleration due to gravity?
3. A child is twirling a 0.0120-kg ball on a string in a horizontal circle whose radius is 0.100 m. The ball travels once around the circle in 0.500 s. (a) Determine the centripetal force acting on the ball. (b) If the speed is doubled, does the centripetal force double? If not, by what factor does the centripetal force increase?
4. Car A uses tires for which the coefficient of static friction is 1.1 on a particular unbanked curve. The maximum speed at which the car can negotiate this curve is 25 m/s. Car B uses tires for which the coefficient of static friction is 0.85 on the same curve. What is the maximum speed at which car B can negotiate the curve?
5. A curve of radius 120 m is banked at an angle of 18°. At what speed can it be negotiated under icy conditions where friction is negligible?
6. . A racetrack has the shape of an inverted cone, as the drawing shows. On this surface the cars race...