@@aryangurusamy6238 There is no mass dependence, because weight and inertia are proportional to each other. Weight governs the minimum static friction needed, and inertia governs the normal force that the rotating ride will generate on the passenger, to keep them rotating with the ride. Both are proportional to mass, and thus cancel out of the equation. It is very common that mass cancels out of an equation.
At the Pike in Long Beach, that ride was called, 'The Rotor.' It was old and had stairs that wound around near to the top. Upon entering the round room we were told to stand as close to the wall as possible. The walls felt like they were made out of hard rubber. The center of the floor had an iron hub-like thing resembling a giant fire hydrant. Looking up, there was no ceiling just a railing that circled around the top. From there, onlookers would watch and laugh at the people spinning around down below. As the room picked up speed, the floor would start decending until it was a good seven or eight feet below us! After a couple of minutes being totally glued to the wall, the floor would start returning back to normal, and the room would slow down enough to make everyone slide down the wall a foot or so and scream. Then the floor would rise up and the room would slow down and stop. I loved that ride......until it made me sick to my stomache when I was 14. I wouldnt ride on it anymore after that. 💫😩💫!!!
BEST PART OF THIS VID: 8:01 AND THE FEW SECONDS PROCEEDING. (When he puts his face perfectly in the Roter Rider, and the stick figure clings on to him (for dear life) in hopes that we all will find the Min speed to save his ass. 🙃👆🤓 #ThingsThatAmuseYouMost@3:45am((after 18hrs of Physics cramming before another exam)) THANK YOU, AS ALWAYS PROFESSOR 👍
Light *4* Love, Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
You are working during your summer break as an amusement park ride operator. The ride you are controlling consists of a large vertical cylinder that spins about its axis fast enough that any person inside is held up against the wall when the floor drops away (Fig. P6.7). The coefficient of static friction between a person of mass m and the wall is ms , and the radius of the cylinder is R. You are rotating the ride with an angular speed v suggested by your supervisor. (a) Suppose a very heavy person enters the ride. Do you need to increase the angular speed so that this person will not slide down the wall? (b) Suppose someone enters the ride wearing a very slip- pery satin workout outfit. In this case, do you need to increase the angular speed so that this person will not slide down the wall?ans me
A) Since "everyone brought mass to the party" in the equation, the mass of the rider is irrelevant to what the minimum angular speed needs to be. Both the traction force and gravitational force are proportional to mass. B) Increase. The lower the static friction coefficient, the more normal force you'll need to generate the same friction force for the same rider. Spinning it faster generates more normal force. I would expect that there is a standard angular speed of operation that covers a broad survey of all possible clothing, so that no matter what the passengers are wearing, the operator doesn't even need to think about this. The coefficient of static friction will be Greek to the ride operator, in more ways than one. Plus, there is a maximum number of g's that the passengers are permitted to experience, so that it is a safe ride, so there is a maximum safe speed as well.
I have 2 clarifications, isn't there also a radial force acting on the person and holding him pinned to the cylinder wall i.e. the centrifugal force. I am aware that centrifugal force is a pseudo force but that's the force we experience when we are are moving in a circle e.g. when sitting in a car that is taking a sharp turn at high speed - and this is a similar situation. In the accelerated frame of reference the person will experience the centrifugal force. Second wrt the wall exerting a normal force on the person, as a concept isn't it abstract. The cylinder wall is fixed and it cannot exert a force on our back. I can understand when we exert a force on a wall, the wall exerts a force on us - Newtons 3rd Law action-reaction pair. Kindly help resolve the confusion Thanks
Sushil, Great questions, thanks. The centrifugal force is a "perceived" force. It feels like you're being pushed outward. The normal force is a real force. It's the wall pushing on you which keeps you moving in a circle. If the wall was not there, you would fly off in a straight line. Hope this helps. You might also like my new website: www.universityphysics.education Cheers, Dr. A
@@yoprofmatt Thank you for your prompt response and clarification. I think the difficulty I am having is in "thinking" or 'visualizing" about the wall pushing on the person. Since wall is an innate, rigid body and its fixed in space & position, the concept of it pushing onto us sounded kinda bit fuzzy. feeling a bit lost.. though the explanation is logical. When we explain Newton's Third Law of Action to high school students, we usually cite the example of a person pushing a wall with his hands (this makes sense, the person is using his muscles to push the wall) - and we further explain that the wall also exerts an equal and opposite force on him. I'll preview thru your new website - thanks for letting me know
How come there is not this 4th force at 4:00? If all forces come in equal and opposite pairs, how can there be the normal force to the left without another force going to the right?
Good question. Newton's 3rd Law says there have to be equal and opposite forces. So where is the equal and opposite force to N? Turns out there IS a force to the right, equal to N, which is acting on the wall. This is Newton's 3rd Force Pair. However, when you draw a free-body diagram, you are only interested in the forces acting on ONE particular body. Our free-body diagram is for the person, and the force that is pushing on the person is the normal force N. If we had drawn a second free-body diagram for the wall, it would have a force N pushing it to the right (along with many other forces like gravity, the ground, other mechanical connections, etc.). Cheers, Dr. A
There was an old ride I remember where the floor would drop, but then the entire ride would tilt on an axis, so it became a 30° loop-to-loop kind of thing also. I don't regret never going on it ha
Hi Matt I am having trouble with this question. "A "gravitron" from amusement park, with minimum safe speed on the surface of Earth 4.21 m/s was brought to the planet X and was set into operation. The safe speed which allowed people not to slide down on the planet X was measured as 7.87 m/s. Find acceleration of free fall on planet X." I mean is it possible to calculate the acceleration of free fall if you don't know the radius of the the gravitron?
Shreya Shah, Use the last equation we found for v and set up a ratio. The unknown variables will cancel out. Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
@@yoprofmatt Thank you so much!!! I will definitely check it out! I set up the ratio, by setting the two coefficients of friction equal to each other. The radii of the gravitron cancelled out and I was able to obtain the gravitational acceleration on planet x. I think my main issue was that I though the coefficient of friction varied with the gravitational acceleration, but I may be mistaken. I still am not a hundred percent why the coefficient of friction can't change due to varying gravitational acceleration on different planets. I guess I will have to do some more digging to understand why this is the case.
if there is dual rotation in opposite directions and centrifugal force is same for both the rotation but the radius and velocity of the rotations are different, so does the force cancel out?
Manav Gupta, I really don't understand your question. But if you're asking if a second cylinder above the first was rotating the opposite way, what would happen? The same thing. The normal force points towards the circle's center and would keep the people moving in a circle. Now a person that was straddling the two rotating cylinders (a difficult trick) would not be rotating, feel no normal force, and fall straight down. Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
Why is the normal force acting in that direction? I drew N in the opposite direction making the assumption that the mass of the person pushes against the wall, causing them to feel pushed up/suspended.
I think I figured it out, for anyone who also doesn't understand, when drawing a free body diagram we are analyzing the forces acting on an object, so the person does exert a normal force on the wall towards the right, and the action-reaction pair is the normal force towards the left acting on the person (relevant N force for the question)
Nia Ambursley, Excellent. Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
Can someone explain to me when Matt was summing the forces in the x direction he left force normal as positive although it is acting in the negative direction? I realize that the person is at some arbitrary point within the ride and on the other side it will become positive, so do we always treat the force normal as positive in this type of problem?
Forget Cartesian coordinates here and work with cylindrical coordinates. Think about the radius of the circle the person is moving in. When we write ΣFr=mv^2/r, the convention is to take a force as positive if it is pointing towards the center of the circle. Since N points towards the center, it's positive. (Note also that the equation wouldn't make sense if N was negative, since everything on the right hand side is a positive quantity.) Cheers, Dr. A
i'm still confused why only three forces acting on the person. if i answered four forces acting on the person, it will be give the same answer. ΣFx=0 ΣFy=0 N-Fc=0 Fs-mg=0 N=Fc Fs=mg N=(mv^2)/R
+indra fitriyanto Indra, good question. Remember that the centripretal force (Fc as you call it) is not a real force. Newton's 2nd law in rotational motion simply says that the sum of real forces in the radial direction give us the centripetal force. The equation should read ΣFr = mv^2 / r. The left side are the real forces, the right side is the centripetal force. So your equation should read N = Fc or N = mv^2 /r. We're of course getting the same answer, but only really forces should appear on the free body diagram (N, Fs, mg). Hope this helps. Cheers, Dr. A
They both will swing to the same angle on the chairoplane (what I call gyro swings). The ratio of the tension in the chain to the rider's weight will be the same, since they both need to have the same centripetal acceleration in proportion to g. This means that the angle between tension and the vertical will be the same for both of them. You can see this if you watch a chairoplane, that the passengers all swing to the same angle. We'd expect a much different range of possible angles, if it were different for the mass of each rider.
Seriously, I think there are 4 forces acting on the person. The 4th force is a friction force on the person's back, moving her to the direction of the room's spinning .
Excellent point. This static friction is certainly there when the person is accelerating (otherwise the wall would just slip by behind them), but once the room is spinning at constant speed, there is no longer a force tangential to the circle (if we have no air resistance). The problem we worked out in this video is during this condition of constant speed, so only three forces are needed. Still, great point. Cheers, Dr. A
