A popular amusement park ride, shown on WB p. 8, operates as follows: riders enter the cylindrical structure when it is stationary with the floor at the point marked "a". They then stand against the wall as the cylinder then begins to rotate. When it is up to speed, the floor is lowered to the position marked "b", leaving the riders "suspended" against the wall high above the floor. Important values for this question:
radius of ride = 1.5 m
μs = 0.53
Include units in all numerical answers.
a. Draw a force diagram for the 82 kg rider after the floor moves down (this will not be collected but should help you answer the rest of the question.
b. What type of force (from your force diagram) is causing the passenger not to slide down the wall?
Correct: Your answer is correct.
c. How big does this force need to be for the person not to fall?
d. What type of force is causing a net force to the center (or centripetal force)?
e. How big (numerical value) does the force in "d" need to be so the force in part "c" is big enough (remember your old equations relating these two types of forces)?
f. What is the minimum tangential speed necessary to make sure that the force in "e" is big enough? In other words, how fast must the ride be moving before the floor drops out?
well, the vertical force of friction has to be greater than person weight.
mu*m v^2/r > mg
mu*v^2/r>g
Your FBD should have weight down, friction up, and a normal force towards the center of the ride. The normal force is equal to the centripetal i.e. v^2/r. Mu (Fn) and as Bob showed this must be greater than mg. And v = (sqrt)(rg/mu). Important idea is that the normal force is causing the circular motion.
a. The force diagram for the 82 kg rider after the floor moves down would include the following forces:
- Weight acting downwards (mg)
- Normal force from the wall acting perpendicular to the wall (N)
- Friction force opposing the downward sliding motion (Ffr)
b. The force from the force diagram that is causing the passenger not to slide down the wall is the friction force (Ffr).
c. In order for the person not to fall, the friction force (Ffr) needs to be equal to or greater than the component of the weight force (mg) parallel to the wall. This can be calculated using the equation: Ffr ≥ μs * N, where μs is the coefficient of static friction and N is the normal force. The normal force is equal to the weight of the rider, which is mg. Therefore, Ffr ≥ μs * mg.
d. The force causing a net force towards the center (centripetal force) is the normal force (N) acting towards the center of the circular path.
e. The force in part "d" (N) needs to be equal to the force in part "c" (μs * mg) for the person not to fall. Therefore, N = μs * mg.
f. The minimum tangential speed necessary to make sure that the force in "e" is big enough can be calculated using the equation for centripetal force, which is Fc = mv²/r, where m is the mass of the rider, v is the tangential speed, and r is the radius of the ride. Rearranging the equation, we get v = √(Fc * r / m). Since Fc = N, we can substitute N = μs * mg into the equation. Therefore, v = √((μs * mg) * r / m).
Remember to substitute the given values and calculate the final numerical answer with appropriate units throughout the steps.
a. To draw a force diagram for the rider after the floor moves down, we need to consider all the forces acting on the rider. The forces acting on the rider are their weight (mg) acting downward and the normal force (N) acting perpendicular to the wall. Since the rider is not sliding down the wall, there must be a frictional force (f) acting upward opposing the gravitational force. The force diagram can be represented as:
- Weight (mg) downwards
|
- - - - - -
| |
| | Normal force (N) perpendicular to the wall
| |
| o < --------------------------------------------- Frictional force (f) upwards
|------------------------------------------------------- Wall
b. The force that is causing the passenger not to slide down the wall is the frictional force (f) acting upward. This frictional force opposes the gravitational force and prevents the rider from sliding down.
c. The frictional force (f) needs to be equal to or greater than the gravitational force (mg) for the person not to fall. This can be expressed as f >= mg. To find the minimum value of f, we need to calculate mg. Using the given mass of the rider (82 kg) and the acceleration due to gravity (9.8 m/s^2), we can calculate mg as:
mg = (82 kg)(9.8 m/s^2) = 803.6 N
Therefore, the frictional force (f) needs to be greater than or equal to 803.6 N for the person not to fall.
d. The type of force causing a net force towards the center (centripetal force) is the normal force (N). The normal force acts inwards towards the center of the circular path, providing the necessary centripetal force to keep the rider moving in a circular motion.
e. The force in part "d" (normal force, N) needs to be equal to the force in part "c" (frictional force, f) for the person not to fall. Since we want the minimum value of tangential speed, we can use the equation:
f = N = mv^2/r
Where m is the mass of the rider, v is the tangential speed, and r is the radius of the ride. Rearranging the equation, we can solve for v:
v^2 = fr/m
v = sqrt(fr/m)
Plugging in the values, we have:
v = sqrt((0.53)(803.6 N)/(82 kg))
Calculating this value, we get:
v ≈ 6.339 m/s
Therefore, the tangential speed needs to be at least 6.339 m/s for the force in part "e" to be big enough.
f. The minimum tangential speed necessary to make sure that the force in part "e" is big enough can be calculated using the same equation as in part "e":
v = sqrt(fr/m)
Since the radius of the ride is given as 1.5 m, we can plug in the values to calculate the minimum tangential speed:
v = sqrt((0.53)(803.6 N)/(82 kg))
Calculating this value, we get:
v ≈ 6.339 m/s
Therefore, the ride must be moving at a minimum speed of approximately 6.339 m/s before the floor drops out to ensure that the force in part "e" is big enough.