The question is that there is a amusement park and in one of the rides the riders stand with their backs against the wall of a spinning vertical cylinder. The floor falls away and the riders are held up by friction. The acceleration of gravity is 9.81m/s^2. The radius of the cylinder is 8m and I have to find the minimum number of revolutions per minute necessary when the coefficient of static friciton between a rider and the wall is 0.5. I have to answer is rev/min. I asked this question before but with different values but I don't get how the person got the answer (1/2π)*√(9.8/0.6*10) ≈ 0.064rev/s when I do it I get 0.2.
2007-11-12 12:15:15 · 1 answers · asked by glance 3 in Science & Mathematics ➔ Physics
The angular speed required is
m*w^2*r*.5-m*g=0
w=sqrt(2*9.81/8)
w=1.566 rad/sec
RPM=rad/sec * 60/(2*pi)
=1.566*60/(2*3.14)
I get 15 RPM
j
2007-11-12 12:41:53 · answer #1 · answered by odu83 7 · 0⤊ 0⤋