Physics circular motion question?

Question

can you show your solution please? I'm stuck on 2 of my homework questions.

41. In an old-fashioned amusement park ride, passengers stand inside a 5.0 m diameter hollow steel cylinder with their backs against the wall. The cylinder begins to rotate about a vertical axis. Then the floor on which the passengers are standing suddenly drops away! If all goes well, the passengers will "stick" to the wall and not slide. Clothing has a static coefficient of friction against steel in the range 0.6 to 1.0 and a kinetic coefficient in the range 0.4 to 0.7. A sign next to the entrance says "No children under 30kg allowed." What is the minimum angular velocity, in rpm, for which the ride is safe?

2. A 500g ball moves in a vertical circle on a 102cm long string. If the speed at the top is 4.0m/s, then the speed at the bottom will be 7.5m/s
a) What is the ball's weight?
b. What is the tension in the string when the ball at the top?
c. Tension in the string when the ball at bottom?

Answer 1

41.
weight must balance out withe force of friction

W=f
mg=uam or g=ua so a=g/u ( u better be static friction since if they start to slide=> major lawsuit)
Since weight canceled out why was this silly requirement about limiting to over 30 kg.

a= V^2/R
v=srt(aR) = sqrt(gR/u)

Angular velocity
w=V/R or in rpm w=60V/R
w=60sqrt(gR/u)/R=60sqrt(g/(uR ))

2R=D=5.0m
w=60sqrt(2 x 9.81/(5 u))

w1=60 sqrt(3.924/(.6))=153 rpm min for u = .6
w2=60 sqrt(3.924/(1.0))=119rpm min for u = 1.0

2.
(a) W=mg
(b) F(top)=F-W= (V1^2/R)m-mg
b) F(botom)=F+W=(V2^2/R)m-mg
R=1.02 m
V1= 4.0 m/s
V2= 7.5 m/s
m=.500kg
g= 9.81 m/s^2