Angle of FlightDate: 11/27/97 at 20:58:59 From: Joel Subject: Physics A fly is sitting at 12 o'clock on a frictionless clock. He takes a step to one side and begins to slide down the clock. At what angle, relative to 12 o'clock, does the fly "fly" off the clock? (assume he doesn't use his wings). My math teacher gave us this one for fun, even though it's really a physics problem. I think the angle I need to find is where the normal force is equal to the centripetal force, but I can't get to an equation where enough of the variables cancel each other out. Any tips on how to go about this question would be great. I really want to find out the answer and my math teacher says he doesn't remember enough physics to solve it. Thank you :) Date: 12/01/97 at 09:22:19 From: Doctor Mark Subject: Re: Physics Hi Joel, You're right--this is really a physics problem. In answering this, I will have to assume that you know some physics. Suppose the fly has traversed an angle A from the vertical. That is, A is the angle (taken from the center of the clock) between the position of the fly on the circumference of the clock, and the vertical. We'll also let the radius of the circle (i.e., the clock) be R (though it will turn out not to matter what R is). If you set up a force diagram for the fly, you find that there are two forces on it: the normal force N (which points radially outward from the center of the clock), and the gravitational force, mg, (where m is the mass of the fly) which points straight down. Now as long as the fly is still traveling along the circumference of the clock, the fly is undergoing circular motion, so the net force on the fly must be (mv^2)/R, directed toward the center of the clock. Now if you set up a coordinate system at the fly's position so that the x axis points along a tangent, and the y axis points radially outward, then we see that (taking components) mg cosA - N = (mv^2)/R, where N is the normal force. Now the condition for the fly to "fly" off is that it not be in contact with the surface. This means that N must be zero! Thus, we find the condition to be mg cosA = (mv^2)/R. Now we need to find v. To do this, we need to use conservation of energy. This will give total energy at the top = total energy at the position of the fly where total energy = gravitational potential energy + kinetic energy of the fly. If we take the gravitational potential energy (PE) to be 0 at the 12 o'clock position, then the gravitational potential energy at the position of the fly will be - mg*(vertical distance the fly has moved). From the geometry of the situation (you really need to draw a picture here!), we see that the vertical distance the fly has moved is R - R cosA = R(1 - cosA) so the gravitational PE of the fly will be: - mgR(1 - cosA) The total energy of the fly at the top is [PE at the top (= 0)] + [KE at the top (= 0, since the fly starts from rest)] = 0+0 = 0. The total energy at the present position of the fly is [PE at the present position of the fly] + [KE at the present position of the fly] = [- mgR(1-cosA)] + [(mv^2)/2] Equating the two total energies, we get: 0 = (mv^2)/2 - mgR(1 - cosA) =====> (mv^2)/2 = mgR(1 - cosA). Now comes the mathematics! We want to solve the following pair of equations for cosA: mg cosA = (mv^2)/R (mv^2)/2 = mgR(1 - cosA) To do so, it is most elegant to solve the second equation for mv^2, then substitute into the first equation: mv^2 = 2mgR(1 - cosA), whence we find: mg cosA = [2mgR(1 - cosA)]/R = 2mg(1 - cosA). Divide both sides by mg to find that cosA = 2(1 - cosA), whereupon one finds that cosA = 2/3. Then use a calculator to find the angle A. Now that you know the angle A, you have to convert that to the proper time on the clock. That's pretty easy, isn't it? I find 22 seconds after 1:36 P.M. This problem is often given as involving a person sliding down an igloo. The most interesting thing about the answer is that it doesn't depend on the mass of the object doing the sliding, and it also doesn't depend on what g is (as long as g is not 0--why?). That means that the fly will "fly" off the clock at the same angle whether the clock is on the earth, the moon, or on Uranus (which would give an interesting twist to the question: What time is it? "Let me look.") -Doctor Mark, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 12/01/97 at 18:18:37 From: Wiskey123 Subject: Re: Physics Thank you for your help :) |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/