In an amusement ride, fun seeking people enter a 7.4 meter diameter cylinder and stand up against the wall. The cylinder spins fast enough so that when the bottom drops down 2 meters, the people remain pinned to the wall. The cylinder continues to rotate for awhile but then slows down and the people slowly slide down the wall. Assuming that a person has a mass of 75 kg and the coefficient of static friction is 0.56 with the wall, at what rpm will that person begins to slide down?
?confused...
rpm= 34.68
Thanks but how do you do it?
To find the RPM (revolutions per minute) at which the person begins to slide down, we need to calculate the critical angular velocity at which the static friction between the person and the wall is overcome.
Let's break down the solution step by step:
1. Start with the formula for centripetal force:
Fc = (m * v^2) / r
where Fc is the centripetal force, m is the mass of the person, v is the velocity, and r is the radius of the circular path (in this case, half of the cylinder's diameter).
2. Rearrange the formula to solve for velocity:
v = sqrt((Fc * r) / m)
3. In this case, the vertical force acting on the person is the weight (mg) minus the normal force (N), which is the force pushing outward due to the rotation of the cylinder. This is given by:
N = m * g - m * a
where g is the acceleration due to gravity (approximately 9.8 m/s^2) and a is the centripetal acceleration (v^2 / r)
4. The maximum static friction force (Fs max) that can act between the person and the wall is given by:
Fs max = μ * N
where μ is the coefficient of static friction.
5. At the point when the person begins to slide, the maximum static friction force (Fs max) is equal to the force due to gravity minus the centripetal force:
Fs max = m * g - (m * v^2) / r
6. Set the expression for Fs max equal to the formula for Fs max and solve for v:
μ * (m * g - m * a) = m * g - (m * v^2) / r
7. Simplify the equation:
μ * (g - a) = g - (v^2 / r)
8. Rearrange the equation to solve for v:
v^2 = r * (g - a) * (1 - μ)
v = sqrt(r * (g - a) * (1 - μ))
9. Since we know that a = g - 2 (the acceleration when the bottom drops down 2 meters), we can substitute it into the expression:
v = sqrt(r * (g - (g - 2)) * (1 - μ))
10. Finally, calculate the angular velocity (ω) in radians per second and convert it to RPM:
ω = v / r
RPM = ω * (60 / 2π)
By plugging in the given values (diameter = 7.4 meters, mass = 75 kg, coefficient of friction = 0.56), you can calculate the RPM at which the person will begin to slide down.