dynamics of circular motion problem: please show me the solution, i find it really difficult...?

Question

On the ride "Spindletop" at the amusement park Six Flags Over Texas, people stood against the inner wall of a yellow vertical cylinder with radius 2.5 m. The cylinder started to rotate and when it reached a constant rotation rate of 0.60 rev/s, the floor on which people were standing dropped about 0.5 m. The people remained pinned against the wall.

a.) Draw a force diagram for a person on this ride, after the floor has dropped.

b.) What minimum coefficient of static friction is required if the person on the ride is not to slide downward to the new position of the floor?

c.) Does your answer in part (b) depend on the mass of the passenger? (Note: When the ride is over, the cylinder is slowly brought to rest. As it slows down, people slide down the walls to the floor.)

Answer 1

a) there is a normal force from the wall on the person, that provides the centripetal force needed to keep the person in circular motion. Of course there is gravity down, and there is also static friction up.

b) I won't do this for you, but i'll help you get it started. As I said above, normal force is the centripetal force, which is mv^2/r. And you know the equation for static friction and how it relates to normal force. And you know that static friction must equal the weight, since the person is not moving up or down. Play with all those equations to work out the value of the coef.

c) If you did part b without plugging in the numbers first, just doing the algebra, then either your value for the coef. either contains the term for mass or it doesn't.

Answer 2

The answers are written in as much simplicity and detail as possible:
Ans a) The free body diagram can be seen here - http://ccc.domaindlx.com/akinetworks/Freebodyfig.jpg

Ans b) Let Normal reaction = N
Let frictional force = f
Let Weight/gravitational force = W
Let Normal reaction due to the inner wall = N
Let centrifugal force = C
Let the mass of a person = m
Let the coefficient of friction = k
Let the distance between the person and the centre of the centrifuge/spindletop be = r = radius
Let velocity of person = v (note v is perpendicular to centrifugal force in the Z direction.)

Since the people who stood against the inner wall are completely stuck in the same position, we can state that -
Net force in direction of X axis = Fx = 0
Net force in direction of Y axis = Fy = 0

Fx = N - C = 0
Implies, N = C ------------- (1)

Similarly Fy = f - W = 0
Implies, f = W --------------(2)

From (1), N = mv^2/r ------------ (3)

From (2), k x N = m x g
kN = mg ----------- (4)

From (3) and (4), we get,

kmv^2/r = mg
kv^2/r = g

Min value of k = gr/(v^2) ....... ANS(b)

----- Substituting real values in the above eq ---

angular velocity = w = 2 x pi x frequency
= 2 x pi x 0.60
= 1.2 x pi
v = w x r = 1.2 x pi x 2.5
= 3 x pi
Hence, min k = 9.8 x 2.5 / (3 x pi)^2
= 0.276

Minimum value of k = 0.276 ......... Calculated answer.

Ans c) No. The answer in part(B) doesn't depend on mass of the person. When the ride slows down, the Normal reaction would reduce as it depends on the velocity v. As a result the frictional force would reduce and hence the weight/gravitational force would pull the person down so that the potential energy is as minimum as possible. Hence, the person comes back on the floor gradually.

I hope the explanations are satisfactory and serve the purpose. If there is any help required with physics, please don't hesitate to post another qts. There are many out here including me who would love to help you out.

Answer 3

a) Forces are the centrifugal force Fcp, horizontally toward the wall; the force of static friction Ff with the wall, upward; the normal force Fn by the wall, away from the wall; the force of gravity Fg, downward. The horizontal forces are equal in magnitude; the vertical forces are equal in magnitude.

b) The minimum coefficient s is when force of gravity equals max friction force, i.e.
Fg = Ff,max
m g = s Fn

The normal force is equal in magnitude to the centrifugal force:
m g = s Fcp
m g = s m w^2 r
g = s w^2 r
s = g / (r w^2)

With g = 9.81 m/s^2; r = 2.5 m; and w = 2pi * 0.60 = 3.77 rad/s; we find

s = 9.81 / (2.5 * 3.77^2) = 0.276


c) Note that the mass m cancels; the answer does not depend on the mass of the passenger.

Answer 4

There's hardly anything more to say than what has been said by kris in her answer so lucidly. ( a ques for the Yahoo! Answer team - can I say 'congrats' here or will it be considered a chat?)

I may add that the mass cancels out because the frictional force from the wall and the downward pull due to gravity are both proportional to the mass. That's why passengers of various masses , all having the same motion, remain stuck to the wall.

Answer 5

A) force against the wall genterates friction of passenger/wall equal to or greater than gravity.

B) Minimum friction is equivalent to gravity

C) the only variable which comes into play as a variant is the coefficient of friction between the wall and the passenger.

Answer 6

Get the factors you certainly can from the question. Draw a image. Get the in many situations occurring constants. Get the easy values you may from the question. ascertain the values you may desire to renowned. What relationships do you recognize. Constants: Gravity: 9.8m/s/s basic Values: length of rotating arm: 6m length of cable: 4.25m perspective of cable: 40 5 degree Values mandatory: finished radius: rotating arm + x area of cable length & cable perspective answer fee - velocity of rotation. basic relationships: Centripetal tension at 40 5 ranges has the comparable X and Y fee, so the centripetal tension desires to equivalent the tension of gravity. finished radius (available from question) and rotational velocity (unknown fee) = centripetal tension (9.8m/s/s)

Answer 7

For a given rpm there is a minimum frictional coefficient.
f> g/(v^2/r)