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| Words: 438 | Submitted: 07-Apr-2011
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Title: Centripetal force
To measure the centripetal force for whirling a mass round a horizontal circle and compare the result with the theoretical value F= m?²r.
Rubber, glass tube about 15 cm long , weights 0.6 kg, 1.5 m of cotton string, meter rule, stop-watch, triple beam balance
When a mass m attached to a string is whirled round a horizontal circle of radius r, the centripetal force for maintaining the centripetal force for maintain the circular motion is given by F= m?²r, where ? is the angular velocity of the circular motion. This force is provided by the tension of the string. The formula can also be as expressed in terms of the velocity v of the mass, where ?=v / r.
Substituting ?=v / r into the formula for F,
F= m v²/r
However, the string was not horizontal as the rubber bung moved around. In fact, the bung moves in circle of radius r = L sin ? (Fig. A6.3). the tension T thus provides both the centripetal force and a force to support the weight of the bung. But resolving T into its horizontal and vertical components, show that T = m?²L regardless the angle ?.
The tension T in the string is provided by the weight of screw nuts (Mg). Therefore
T = Mg
As there is no vertical motion, the vertical component of tension (T) is balanced by the weight of the rubber bung (mg):
T cos?= mg (1)
The horizontal component of the tension provides the net centripetal forces:
T sin?= m?²r (2)
Substituting r = L sin? into equation (2), we can find the tension(T) in the string:
T sin?= m?²( L sin?)
T = m?²L
1. One end of a string (~1.5 m ) was attached to a rubber bung. A length of say 0.6m of the string from the rubber bung to the glass tube was measured. This length l of the string was marked with the color pen. The free end was passed through a glass tube and then it was attached to some weights (~ 10 times the mass the ...
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