Ben B.
asked • 02/18/13forces physics problem - Please Help ASAP
You are on an amusement park ride with your back against the wall of a spinning vertical cylinder. The floor falls away and you are held up by static friction. Assume your mass is 72 kg.
(a) Draw a free-body diagram of yourself. (Do this on paper. Your instructor may ask you to turn in this work.)
(b) Use this diagram with Newton's laws to determine the force of friction on you.
(c) If the radius of the cylinder is 4.8 m and the coefficient of static friction between you and the wall is 0.59. What is the minimum number of revolutions per minute necessary to prevent you from sliding down the wall?
rpm
Does this answer hold only for you?
yes or no
Will other, more massive, patrons fall downward?
yes or no
Explain.
More
1 Expert Answer
Michael B. answered • 02/19/13
Tutor
New to Wyzant
Harvard Grad eager to help! A.P., Pre-med, MCAT and Philosophy Tutor
A) First of all, for the free-body-diagram: Gravity will be acting, so draw a downward force due to this. If you aren't falling, this means that there is a net force of zero in the vertical direction, so the force of static friction is equal in magnitude and opposite in direction to the force of gravity (draw an arrow up). Now, we know that the Force of friction = Normal Force x coefficient of kinetic friction (μ). So if there is a static friction force, there is a normal force. A normal force is just a force that is perpendicular to a surface, so this normal force will point inwards towards the center of the spinning cylinder. These are the only forces acting on you--remember, there is no "centrifugal" outward force!
B) Now for the second part of the question: Rather than answer this for you, let me just pose you a couple of questions. Newton's first law states that an object at rest will remain at rest unless acted on by an outside force. If you aren't falling, then at least in the vertical direction you are "at rest," not accelerating. But this only happens when there is no net force pulling you up or down. So this means that the upward force of friction must be exactly equal to the downward force of gravity. So what is the downward force of gravity for a man of 72 kg? I recommend using Newton's 2nd law equation: Force=mass x acceleration.
C) Now this is getting more tricky. In order to solve this problem, we need to determine the minimum rate of rotation that would still create a normal force that would in turn create enough frictional force to keep a person in place. The key here is to find the right equations and then set them equal to each other. Since we're dealing with centripetal motion, one of those equations is going to be Force = mass x velocity squared all over radius, or F=(m*v^2)/r. Now this equation will give us the centripetal force. As we can see from our free body diagram, the centripetal force is the same as the normal force. In part B we solved for the force of friction that is necessary to hold you up. We know that this Friction Force = Normal Force x Coefficient of Static Friction, so if we rearrange the equation we can see that Normal Force = Friction Force over Coefficient of static friction. Now we have two equations for Normal Force and we can set them equal to each other, so that: Friction Force/Coefficient of friction = (m*v^2)/r. Rearrange this equation to solve for velocity using the numbers that are given in the problem. Once you have velocity, divide this number by the circumference and make sure that you convert it to the correct units (rpm, not rotations per second, since your linear velocity that you solved for will be in m/s!). That should give you your answer!
Rather than answer the last two questions, I would just challenge you to think about this fact: A more massive person will have a stronger gravitational force acting them, which is why we might be tempted to say that it will be different. But the Normal Force that they feel will also be bigger. So when we set the Force of Gravity equal to the Force of Friction (Which is equal to the Normal Force x μ), will the increase in the force of gravity increase by the same amount as the force of friction? If you can't see the answer just by looking at the equations, try to put in hypothetical numbers and test it yourself. Does a fat person's gravitational force increase by the same factor as that frictional force? If so, then the same conditions will apply to everyone, and the answer to both questions is No.
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Michael B.
A) First of all, for the free-body-diagram: Gravity will be acting, so draw a downward force due to this. If you aren't falling, this means that there is a net force of zero in the vertical direction, so the force of static friction is equal in magnitude and opposite in direction to the force of gravity (draw an arrow up). Now, we know that the Force of friction = Normal Force x coefficient of kinetic friction (µ). So if there is a static friction force, there is a normal force. A normal force is just a force that is perpendicular to a surface, so this normal force will point inwards towards the center of the spinning cylinder. These are the only forces acting on you--remember, there is no "centrifugal" outward force!
B) Now for the second part of the question: Rather than answer this for you, let me just pose you a couple of questions. Newton's first law states that an object at rest will remain at rest unless acted on by an outside force. If you aren't falling, then at least in the vertical direction you are "at rest," not accelerating. But this only happens when there is no net force pulling you up or down. So this means that the upward force of friction must be exactly equal to the downward force of gravity. So what is the downward force of gravity for a man of 72 kg? I recommend using Newton's 2nd law equation: Force=mass x acceleration.
C) Now this is getting more tricky. In order to solve this problem, we need to determine the minimum rate of rotation that would still create a normal force that would in turn create enough frictional force to keep a person in place. The key here is to find the right equations and then set them equal to each other. Since we're dealing with centripetal motion, one of those equations is going to be Force = mass x velocity squared all over radius, or F=(m*v^2)/r. Now this equation will give us the centripetal force. As we can see from our free body diagram, the centripetal force is the same as the normal force. In part B we solved for the force of friction that is necessary to hold you up. We know that this Friction Force = Normal Force x Coefficient of Static Friction, so if we rearrange the equation we can see that Normal Force = Friction Force over Coefficient of static friction. Now we have two equations for Normal Force and we can set them equal to each other, so that: Friction Force/Coefficient of friction = (m*v^2)/r. Rearrange this equation to solve for velocity using the numbers that are given in the problem. Once you have velocity, divide this number by the circumference and make sure that you convert it to the correct units (rpm, not rotations per second, since your linear velocity that you solved for will be in m/s!). That should give you your answer!
Rather than answer the last two questions, I would just challenge you to think about this fact: A more massive person will have a stronger gravitational force acting them, which is why we might be tempted to say that it will be different. But the Normal Force that they feel will also be bigger. So when we set the Force of Gravity equal to the Force of Friction (Which is equal to the Normal Force x µ), will the increase in the force of gravity increase by the same amount as the force of friction? If you can't see the answer just by looking at the equations, try to put in hypothetical numbers and test it yourself. Does a fat person's gravitational force increase by the same factor as that frictional force? If so, then the same conditions will apply to everyone, and the answer to both questions is No.
02/19/13