Amusement park ride gravitation problem

[holmeskaei]

Homework Statement

In an old-fashioned amusement park ride, passengers stand inside a 5.4-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 0.64 to 1.0 and a kinetic coefficient in the range 0.40 to 0.70. A sign next to the entrance says "No children under 30 kg allowed."

Homework Equations

f(fric)=(us)(n)
Fr=n=mv^2/r
Fz=f=mg
Ft=0
w=v/r

The Attempt at a Solution

f=(0.64)(9.8*30)=188.16N
188.16N=(30)(v)^2/2.7, v=4.1151m/s
4.115m/s/2.7=1.524rad
1.524rad/1s=1rev/2piradians=60sec/1min=143.65rpm
This answer is not correct, however. Any guidance is appreciated. I don't want the answer, just what I am doing incorrectly.

 

[PhanthomJay]

Homework Statement

In an old-fashioned amusement park ride, passengers stand inside a 5.4-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 0.64 to 1.0 and a kinetic coefficient in the range 0.40 to 0.70. A sign next to the entrance says "No children under 30 kg allowed."

Homework Equations

f(fric)=(us)(n)
Fr=n=mv^2/r
Fz=f=mg
Ft=0
w=v/r

The Attempt at a Solution

f=(0.64)(9.8*30)=188.16N use your 3rd relevant equation, f(fric)= mg. You are incorrectly assuming that the Normal force is the person's weight.
188.16N=(30)(v)^2/2.7, v=4.1151m/s
4.115m/s/2.7=1.524rad
1.524rad/1s=1rev/2piradians=60sec/1min=143.65rpm
This answer is not correct, however. Any guidance is appreciated. I don't want the answer, just what I am doing incorrectly.

Please see above in red

 

[nlightner]

Your attempt at solving this problem is on the right track, but there are a few issues with your equations and calculations. Let's break it down step by step.

First, we need to calculate the normal force (n) acting on each passenger. This is equal to their weight (mg), which in this case is 30kg * 9.8m/s^2 = 294N.

Next, we need to calculate the maximum frictional force (f) that can act on the passengers. This is given by the formula f = u_s * n, where u_s is the static coefficient of friction. Since the range of u_s is given as 0.64 to 1.0, we can use the average value of 0.82. Therefore, f = 0.82 * 294N = 241.08N.

Now, we can use the fact that the passengers are not sliding to set up an equation for the centripetal force (F_c) acting on them. This force is provided by the frictional force, so we have F_c = f = 241.08N. We can also use the formula F_c = m * v^2 / r, where m is the mass of each passenger and r is the radius of the cylinder (2.7m). Since we have the mass and radius, we can rearrange this equation to solve for the required speed (v) for the passengers to not slide:

v = sqrt(F_c * r / m) = sqrt(241.08N * 2.7m / 30kg) = 4.115m/s

This is the same result you got in your attempt, so this part is correct.

Next, we need to calculate the angular velocity (w) of the cylinder. We can use the formula w = v / r, where v is the tangential velocity we just calculated and r is the radius of the cylinder. Therefore, w = 4.115m/s / 2.7m = 1.524rad/s.

Finally, we can use the fact that the ride completes one revolution every 2 seconds (given by the information that the ride rotates at 143.65rpm) to set up an equation for the angular displacement (theta) in 2 seconds. This is given by the formula theta = w * t, where t is the time in seconds. Therefore, theta