Brandon9000 wrote:A phonograph (not a CD player) is used to play a 33 rpm record. A marble with a mass of 10 grams is glued to the record at a distance of 10 centimeters from the center. What is the minimun force which the glue, and any other sources which contribute such as friction, must supply to keep the marble from sliding or rolling off the record? Show your work.
The force required to keep an object in circular motion is known as centripetal force and is directed towards the center of the circle.
Centripetal force = mv^2/r.
First determine the speed of a point 10 centimeters from the center of a 33 rpm record.
Circumference = pi * Diameter = 3.142 * .2 meter = .6284 meter
Such a point makes a complete revolution 33 times per second, thus its speed is:
Circumference * 33 / 60 seconds = .6284 meter * 33 / 60 seconds = .34562 meters/sec
Now plugging back into the forumula for centripetal force:
F = .01 kg (.34562 meters/sec) ^ 2 / .1 meter = .01 kg (.11945 meters^2/sec^2)/.1 meter
= .011945 kg - meter/sec^2 = .011945 Newton
-----------------------------------------------------
Let me do this a different way to make it look more like yours:
F = m * (Circumference/Period) ^ 2/r = m * (2 * pi * r/Period)^2/r
= m * (4 pi^2 r^2/Period^2)/r = m * 4 * pi ^2 * r / Period^2
= .01 kg * 4 * (3.1416)^2 * .1 m / (60 sec /33)^2
= .01 kg * 4 * 9.86965056 * .1m / (1.81818 sec)^2
= .03948 kg-m / (1.81818 sec)^2
= .03948 kg-m / 3.30578 sec^2
= 0.011943 kg-m/s^2 = 0.011943 Newton
-----------------------------------------------------
Now analyzing your solution:
nipok wrote:
netforce = mass * ( ( 4 * pi ^2 * radius) / time ^ 2 )
Formula is correct, although curiously without explanation of where anything comes from.
No, .01 kg, which I regard as a trivial oversight.
nipok wrote:Radius = .1 m
Time = 1.81818 seconds per revolution
Correct.
nipok wrote:.1 * ( 1.25664 / 3.305785 )
.1 * 0.380133
Aside from the factor of 10 in the mass, you have taken 4 * pi, not 4 * pi^2. Also, you should show the units all the way through and not just at the end to show how you get Newtons.
Well, basically the errors are failure to square pi and the factor of 10 in the mass. The original formula you gave is correct, although sort of an odd starting point. Typically arithmetic errors are not regarded as very important. What is extremely curious, though is your failure to give the original formula mv^2/r or discuss the principles involved in setting up the problem in any way, shape, or form. I did log on earlier and know that were you now have multiplication symbols, you originally put plus signs which would have involved adding quantities with unlike units, which is always wrong. You have now corrected that, as evidenced by the notation that you have edited the post. I have my suspicions about this, but I am forced to judge it as an essentially correct answer despite the arithmetic problems.