Saturday, December 1, 2012

Can a lighter object be accelerated sufficiently to lift a heavier object?

By Ali Pesch

The focus of my second physics news is centered on the question, “can a lighter object be accelerated sufficiently to lift a heavier object?” To attempt to answer this question, I viewed a YouTube video (link below) that depicted the question I am asking. In the video, a string is pulled through a drinking straw and then attached to a small AA battery to one end and 3 batteries to the other end. Holding the straw firmly, the single battery is hurled into rotation, orbiting around the axis of the straw. With increasing speed, the orbital radius increases. At some point, the centrifugal forces exceed the gravitational pull of the 3 AA batteries, and they lift off.


The experiment described above uses the formula a = υ 2 r -1 with “a” signifying acceleration, “υ” representing constant speed, and “r”  being the radius of the circle. In addition, we need to use the equation F = ma, as it follows that the force is the acceleration multiplied by the mass of the battery. The forces in play include the centrifugal pull exerted by the orbiting single battery, that is opposed by the force of tension of the string connected to the weight of the 3 AA batteries attached to the other end. As Newton’s Third Law of Motion suggests, the increasing rotational speed augments the centrifugal vector. At some point, it exceeds the centripetal vector exercised by the 3 AA batteries and these lift off. The force needed to lift off the 3 AA batteries follows Newton's Second Law of Motion F = ma, where “F” stands for force, “m” symbolizes mass and “a” signifies acceleration. In other words, acceleration is produced when a force acts on a mass; heavier objects require more force to move the same distance than lighter objects. The motion in this experiment is not linear but is circular. We are only interested in the force that is necessary to lift the 3 AA batteries up higher into the air. The orbiting battery is being accelerated along a circular path with steady speed while constantly changing its direction tangentially to the orbital circle. Here, the centripetal acceleration is a radial acceleration and can be calculated as ac = υ2 r -1 where “υ” is the constant speed of the object along the circular path and “r” equals the radius of the path. In the experiment, the average weight of a single AA battery is 23 grams (0.023 kg). Below are my calculations:

m1 = 0.023 kg

m2 = 0.069 kg

m2 = 3m1

r = 0.20 m.

Solution

(0.023 kg)(ac) = (0.069 kg)(9.8m/s2)

ac = 29.4 m/s2

ac = υ2 r -1

(.2)(29.4) = υ2

υ = 2.42 m/s


If we want to take it a step further we can find θ.

Solution

Fc = mac

Fg2 = 3Fg1 = Fc

m1ac = m2g

tan(θ) = 3x/1x = 3

tan-1(3) = θ

θ = 71.6o

Alternative

a/g = m2/m1

υ2/ (g)(r) = m2/m1

In this case, the 3 batteries have 3 times the mass of 1 battery, so ac must equal 3g. Overall, a smaller object can accelerate sufficiently to lift a heavier object if the magnitude of the centrifugal acceleration exceeds ag by a factor equal to the factor that the magnitude of the larger mass exceeds the smaller mass by.

This principle is used in commercial applications, such as the centrifugal clutch in vehicle transmissions.

Fg1 = Fg2

θ = tan-1(m2/m1)

http://www.youtube.com/watch?feature=player_embedded&v=DbUU-ueiPUc

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