Centripetal force (from Latin centrum "center" and petere "to seek"[1]) is a force that makes a body follow a curved path: its direction is always orthogonal to the velocity of the body, toward the fixed point of the instantaneous center of curvature of the path. Centripetal force is generally the cause of circular motion. In simple terms, centripetal force is defined as a force which keeps a body moving with a uniform speed along a circular path and is directed along the radius towards the centre.[2][3] The mathematical description was derived in 1659 by Dutch physicist Christiaan Huygens. Isaac Newton's description was: "A centripetal force is that by which bodies are drawn or impelled, or in any way tend, towards a point as to a centre."[4] Contents [hide] * 1 Formula * 2 Sources of centripetal force * 3 Analysis of several cases * 3.1 Uniform circular motion * 3.1.1 Calculus derivation * 3.1.2 Derivation using vectors * 3.1.3 Example: The banked turn * 3.2 Nonuniform circular motion * 3.3 General planar motion * 3.3.1 Polar coordinates * 3.3.2 Local coordinates * 3.3.2.1 Alternative approach * 3.3.2.2 Example: circular motion * 4 See also * 5 Notes and references * 6 Further reading * 7 External links| -------------------------------------------------

[edit]Formula
The magnitude of the centripetal force on an object of mass m moving at tangential speed v along a path with radius of curvature r is:[5]

where is the centripetal acceleration. The direction of the force is toward the center of the circle in which the object is moving, or the osculating circle, the circle that best fits the local path of the object, if the path is not circular.[6] The speed in the formula is squared, so twice the speed needs four times the force. The inverse relationship with the radius of curvature shows that half the radial distance requires twice the force. This force is also...

...To be able to understand and verify the relationship centripetalforce, mass, velocity, and the radius of orbit for a body that is undergoing centripetal acceleration.
Background Information
An object moving in the same direction is not necessarily undergoing acceleration. If the object changes speed while moving in the same direction there is acceleration (or deceleration). On the other hand, if the object moves at a constant speed in the same direction, there is no acceleration. This does not mean that constant speed always indicates no acceleration, however. An object that moves at a constant speed and changes direction is also experiencing acceleration even though its speed never changes. Both the acceleration produced by changing speed and the acceleration produced by changing direction require a net force. This force that is produced in called the centripetalforce and the acceleration that causes a change in direction is called centripetal acceleration.
Centripetalforce means “center seeking.” It is the force responsible for keeping an object in circular motion. If there were no centripetalforce the object would fly off at a tangent because of Newton’s First Law. This is demonstrated by spinning an object on a string. If the string were to break or be cut, the...

...Title: CentripetalForce
Tools and Equipments: nylon cord, different weighing hanging masses, stopwatch, meter stick.
Purpose: To be able to determine the relationship between centripetalforce, mass, velocity, and the radius of orbit for a body that is undergoing centripetal acceleration. To investigate the dynamics of uniform circular motion. Specifically the relationships among thecentripetalforce, the accelerated mass and the radius of rotation.
Procedure:
THEORY:
DATA AND CALCULATION:
PART A (Constant radius) r = .20 m
Frequency = (30 Revolutions)/〖t 〗_(av (30 revolution)) → f = 30/(26.27 s) = 1.142 Hz
V= 2πrf = 2π (.20)(1.142) = 1.44 m/s
F = 〖mv〗^2/r = ((.456)(1.44)²)/(.20) = 4.73 N
Radius
(m) Mass
(kg) F_exp
(N) 〖Time〗_1
(s) 〖Time〗_2
(s) 〖Time〗_av
(s) Frequency
(Hz) Velocity
V=2πrf
(m/s) F_Th
F=〖mv〗^2/r (N)
.20 m .456 kg 4.46 N 26.21 s 26.33 s 26.27 s 1.142 Hz 1.44 m/s 4.73 N
.20 m .506 kg 4.46 N 27.65 s 27.71 s 27.68 s 1.084 Hz 1.36 m/s 4.68 N
.20 m .556 kg 4.46 N 28.68 s 28.65 s 28.67 s 1.046 Hz 1.31 m/s 4.77 N
.20 m .606 kg 4.46 N 29.59 s 29.98 s 29.78 s 1.007 Hz 1.27 m/s 4.88 N
% Error F_Th vs F_exp
% Error = (F_(Th-) F_exp)/F_TH × 100 → = (4.73-4.46)/4.73 ×100 → = 5.71 %
F_TH F_exp % Error
4.73 N 4.46 N...

...Physics Lab Report
Experiment M3 CentripetalForce
School: La Salle College
Class: 6G
Group members (Group 7): Carson Ho, Tang Yui Hong, John Yu, Justin Kwong
Date: 1 / 10 / 2014
Report is written by: Tang Yui Hong 6G (27)
Title
CentripetalForce
Objective
To verify the equation for centripetalforce
Apparatus
Instrument
Descriptions
1 rubber bung
circular, cylinder
screw nuts and wire hook
/
1 small paper marker
/
1 rule
1 metre
safety goggles
/
adhesive tape
/
1 glass tube
~ 15 cm
1 nylon string
1.5 m, inextensible
1 stop-watch
/
1 triple beam balance
/
scissors
/
Sketching of the set-up
Theory
Centripetalforce F is the net force causing the centripetal acceleration of an object performing uniform circular motion. Its magnitude is given by the equation: F = mrω2
When an object is whirled in horizontal circular motion in mid-air with a piece of string (as shown in figure 1 above), the centripetalforce on the object is provided by the horizontal component of tension in the string:
2
=>2 (since r = L)
Procedure
1 A triple beam balance is used to find the mass of the rubber bung, the screw nuts and the wire hook respectively. The total weight of the screw nuts and the wire hook provided the tension T in the nylon thread.
2 The...

...Title : Centripetalforce
Objective
To measure the centripetalforce for whirling a mass round a horizontal circle and compare the result with the theoretical value given by F = m(2r .
Apparatus
12 slotted weights with hanger (0.02kg each)
1 rubber bung with nylon string about 1.5m
1 glass tube about 20cm long
1 triple beam balance
1 meter rule
1 stop watch
Several small paper markers
Theory
When a mass m attached to a string is whirled round a horizontal circle of radius r, the centripetalforce for maintaining the circular motion is given by F = m(2r , where ω is the angular velocity of the circular motion. The centripetalforce is provided by the tension of the string.
The string will always make an angle( to the horizontal instead of lying on the plane of the horizontal circle described by the mass m. Thus, the centripetalforce is given by the horizontal component of the tension. It is shown that the tension T = m(2L regardless of the angle(.
Procedure
1. Measure the mass m of the rubber bung by the triple beam balance.
2. Attach the rubber bung with nylon string thread through the glass tube and a number of weights. First start with M = 0.12kg.
3. Measure the length of L from the rubber bung to the glass tube (i.e. 0.8m). Adjust the position of a small paper marker 1cm below the glass tube. The set...

...Centripetal Acceleration
Experiment # 7
Abstract
In this experiment we studied the centripetal acceleration of an object while it was moving in a constant circular motion.
Introduction
In this experiment we studied the centripetal acceleration of an object at different constant motions. We found that when an object is traveling at a constant circular motion, the object does not stop unless an outside force acts upon it to change its direction.
To start we measured the diameter of the vertical shaft and calculated the radius of the circle path which is the radius of the shaft and the distance between the shaft and indicator rod, which was initially set to between 14 and 15 cm. We then measured the force for (r), which was the weight of the mass, (M).
Using a stopwatch we then timed 50 revolutions of the mass, producing three trials and averaging the results for 1 revolution (T). We then used our data to compute centripetal acceleration using this equation:
(1)
We then calculated the percent difference such that :
|Fm-Fe|/ ½(Fm + Fe) (2)
We repeated the previous procedures for different values of (r) by moving the indicator stick back farther at different intervals.
Results
The initial measurements are :
Table 1
Diameter of shaft (m)
0.0139
Radius of shaft (m)
0.00695
Mass (kg)
0.4606
The results from our trials are as...

...Centripetalforce
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Not to be confused with Centrifugal force.
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Roller coaster cars are forced through a loop by the track applying acentripetalforce on them. A reactive centrifugal force is applied to the track by the cars.
Centripetalforce (from Latin centrum "center" and petere "to seek"[1]) is a force that makes a body follow a curved path: its direction is always orthogonal to the velocity of the body, toward the fixed point of the instantaneous center of curvature of the path. Centripetalforce is generally the cause of circular motion.
In simple terms, centripetalforce is defined as a force which keeps a body moving with a uniform speed along a circular path and is directed along the radius towards the centre.[2][3] The mathematical description was derived in 1659 by Dutch physicist Christiaan Huygens. Isaac Newton's description was: "A centripetalforce is that by which bodies are drawn or impelled, or in any way tend, towards a...

...AL Physics CentripetalForce(1ST Lab Report)
Objective :
To measure the centripetalforce by whirling it around a horizontal circle, then compare the result with theoretical value FC = m(2r.
Apparatus :
1Rubber bung
1Glass tube (About 15 cm long)
1Slotted weights, with hanger 12 × 0.02 kg
1Nylon thread 1.5 m
1Paper marker
1Adhesive tape
1Metre rule
1Stop watch
1Safety goggles
Set-up:
Procedure:
1. Attach one end of a 1.5 m length of nylon thread to a rubber bung and thread the other end through a glass tube, a paper marker and a number of weights as shown.
2. First adjust the position of the marker so that it is about 1 cm near one end of the glass tube, and the length of the thread L from the other end of the glass tube to the rubber bung is, say, 0.8 m. Fix the position of the marker using adhesive tape if necessary. First start with M = 0.16 kg (i.e. 160 g).
3. Holding the glass tube vertically, whirl the bung around above your head in a horizontal circle. Increase the speed of the bung gradually and allow it to move out (i.e. let L increases) until the marker is about 1 cm below the lower end of the glass tube.
4. Try to keep the angular speed constant so that the marker is always about 1 cm below the tube throughout. Ask your partner to time 20 revolutions of the bung using a stop watch. Remember to start the stop watch at 0 and stop it at 20. Take one more confirmatory...