- Level: AS and A Level
- Subject: Science
- Word count: 792
Objective To measure the centripetal force for whirling a mass round a horizontal circle and compare the result with the theoretical value given by F= mw2r .
Extracts from this document...
Introduction
School: | Canossa College | Class: | 6B | |
Name: | Hazel Chow Ho Ying | Class no: | 3 | |
Date: | 20-10-2010 | Mark: |
Title
Centripetal force
Objective
To measure the centripetal force for whirling a mass round a horizontal circle and compare the result with the theoretical value given by F= mω2r .
Apparatus
- rubber bung
- glass tube
- screw nuts
- Wire hook
- 1.5m of Nylon string
- Small paper marker
- metre rule
- stop-watch
Theory
When a mass m attached to a string is whirled round a horizontal circle of radius r, the centripetal force for maintaining the circular motion is given by
F = mω2r where ωis the angular velocity of the circular motion.
This force is provided by the tension of the string.
The formula can also be expressed in the terms of the velocity v of the mass, where ω=v/r .
Substituting ω=v/r into the formula for F , F = mv2/r
Middle
Result
Tabulate the results as follows:
Mass of rubber bung m = 0.03491 kg
Mass of screw nuts M = 0.13 kg
⇒ Tension in string T = Mg = 0.13 × 9.8 N = 1.274 N
Length of string L/m | 0.35 | 0.50 | 0.75 | 0.90 |
Time for 50 revolutions 50 t/s | 30.85 | 37.04 | 45.00 | 49.65 |
ω = /rad s–1 | 10.183 | 8.482 | 6.981 | 6.327 |
mω2L/N | 1.267 | 1.256 | 1.276 | 1.258 |
Mean mω2L = 1.264 N
Conclusion
Form the results, we can find that the length of the string L is increasing, the value of angular velocity ω is decreasing. The length of the string is indirectly proportional to the value of angular velocity.
Conclusion
They are almost equal.
Possible sources of errors
- Friction exists between the glass tube and the string.
- The rubber bung is not set into a horizontal circular path.
- The rubber bung does not move with constant speed.
- The length of the string beyond the upper opening is not constant.
θ increases with ω
Vertical components is T cosθ
Horizontal components is T sinθ
The system has no vertical acceleration
∴T cosθ = mg
The horizontal component of tension provides the centripetal acceleration
∴T sinθ = mrω2
Let L be the length of the string
i.e. r = L sinθ
T sinθ = mrω2
T sinθ = m(L sinθ)ω2
T = mLω2
mg/cosθ= mLω2
θ increases with ω
∴When the rubber bung is whirled around with a higher angular velocity ω , the angleθ becomes larger.
Reference
- Level practical physics for TAS p. 28 - 30
- Physics Beyond 2000 p. 40
- http://en.wikipedia.org/wiki/Centripetal_force
- http://www.greenandwhite.net/~chbut/centripetal_force.htm
This student written piece of work is one of many that can be found in our AS and A Level Mechanics & Radioactivity section.
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