In an amusement park ride called "drop out", riders are spun in a horizontal circles of radius 7.5m, which pins their backs against an outer wall. When they are spinning quickly enough. The floor drops out, and they are suspended by friction. If the coefficient of static friction between the riders and the wall is as small as 0.28, how many revolutions per second must the ride achieve before the floor is allowed to drop out?
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Answered by
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⚫⚫⚫⚫⚫⚫. HEY
◼ NICE QUESTION
.......
HERE IS THE ANSWER
.........
-------. Weight = m * g
-----------. Fc = m * v^2 ÷ r
------------. Ff = μ * Fc
---------------. Ff = μ * m * v^2 ÷ r
_____________________
⚫μ * m * v^2 ÷ r = m * g
⚫v^2 = r * g ÷ μ
⚫ v^2 = 7.5 × 9.8 / 0.28
⚫ v = √262.5
_________________
AS WE KNOW
= C = 2 * π * 7.5 = π * 15
---------------------------
THEREFORE
= t = π * 15 ÷ √262.5
============================
f = √262.5 ÷ π * 15
it is approximately 0.344 rev/s.
_____________________________
# BRAINLY 2222
@ PHYSICS
◼ NICE QUESTION
.......
HERE IS THE ANSWER
.........
-------. Weight = m * g
-----------. Fc = m * v^2 ÷ r
------------. Ff = μ * Fc
---------------. Ff = μ * m * v^2 ÷ r
_____________________
⚫μ * m * v^2 ÷ r = m * g
⚫v^2 = r * g ÷ μ
⚫ v^2 = 7.5 × 9.8 / 0.28
⚫ v = √262.5
_________________
AS WE KNOW
= C = 2 * π * 7.5 = π * 15
---------------------------
THEREFORE
= t = π * 15 ÷ √262.5
============================
f = √262.5 ÷ π * 15
it is approximately 0.344 rev/s.
_____________________________
# BRAINLY 2222
@ PHYSICS
Answered by
1
0.096 rev/s
because 0.28mg = mv^2/r
v= 4.53 m/s
w = 2*pi*f
f = 0.096 rev/s
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