2006-12-20 12:17:55 · 5 answers · asked by beast 1 in Science & Mathematics ➔ Physics
Can someone hep me with this problem:
If the room radius is 4.6 m, and the rotation frequency is 0.4 revolutions per second when the floor drops out, what is the minimum coefficient of static friction so that the people will not slip down?
I got 1.18 as the coefficient.
2006-12-20 12:23:24 · update #1
no. But the most or all of the question you will deal with will be in that range, unless you are dealing with rocket technology or fluids.
So unless you are working with NASA, the answer is yes.
2006-12-20 12:20:41 · answer #1 · answered by el k 1 · 0⤊ 0⤋
The result you obtained (1.18) is correct for the information given.
The question is if this is possible in a real world.
I remember this was an unanswered question in Resnik-Halliday's physics book. I asked this same question many years ago to my physics teacher, and he came with this setup:
If you consider an object on a flat surface and you tilt the surface (change the angle relative to the horizontal) until the object starts sliding, it is easy to prove that the coefficient of friction equals the tangent of the angle of the surface with the horizontal.
Since tg (45) = 1, if you can find an object that you can place on some surface and you have to tilt it more than 45 degrees to make it slide, you have found a coefficient of friction higher than 1.
Try with a rubber eraser or a rubber band on a piece of cardboard.
.
2006-12-20 13:31:33 · answer #2 · answered by Eng_helper 2 · 1⤊ 0⤋
There are indeed situations where the coefficient of friction is greater than one. One example is high-performance tires, which can have coefficients of friction around 1.5 (perhaps a little higher - I forget just exactly).
2006-12-20 16:02:07 · answer #3 · answered by Tim F 2 · 0⤊ 0⤋
yes
2006-12-20 12:20:33 · answer #4 · answered by WxEtte 5 · 0⤊ 0⤋
yes
2006-12-20 12:19:25 · answer #5 · answered by sportsfreak 2 · 0⤊ 0⤋