Asked by Jennifer
In a "Rotor-ride" at a carnival, people are rotated in a cylindrically walled "room." (See Fig. 5-35.) The room radius is 4.6 m, and the rotation frequency is 0.7 revolutions per second when the floor drops out.
What is the minimum coefficient of static friction so that the people will not slip down?
What is the minimum coefficient of static friction so that the people will not slip down?
Answers
Answered by
bobpursley
The force holding them up is mu*fn= mu*mv^2/r= mu*m*w^2 r
set that equal to the force of weight...
m g= mu*m w^2 r
solve for mu.
change .7 rev/sec to rad/sec by multiplying by 2PI.
set that equal to the force of weight...
m g= mu*m w^2 r
solve for mu.
change .7 rev/sec to rad/sec by multiplying by 2PI.
Answered by
j
Help me anyone please
Answered by
Haley
Change frequency to period T=1/f. Find the velocity v=2(pi)r.
Find radial acceleration a=v^2/r.
Then divide gravity by the radial acceleration that you found. g/a
Find radial acceleration a=v^2/r.
Then divide gravity by the radial acceleration that you found. g/a
Answered by
eileen
rotor ride a cylinder of radius 3m is rotation at a linear speed of 15 m/s. What minimum coefficient of static friction is required to keep the participant from slipping?
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