More on Uniform Circular Motion

Circular motion is a new topic for me and I am still struggling through it. I read through the whole chapter on circular motion in my textbook, but still left me with tons of questions. I hope someone can explain and help me out. I know you are all keen physics experts.

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I really appreciate for your help!

1a) A coin is place 15 cm from the axis of a rotating turntable whose rotational frequency can be varied. As the turntable increases its frequency the coin remains on its surface until the turntable reaches 66 revolutions/minute. At the rate, the coin slides off. Find the coefficient of static friction between the coin and the turntable.
[Can I say (coefficient of static friction)(normal force)=4(pi^2)mr(f^2) and then solve for the unknown (coefficient of static friction)? I am having a problem with this method because 4(pi^2)mr(f^2) is the centripetal force NEEDED to keep the coin in circular motion when f=66 revolutions/minute. But in the context of the question, the coin ALREADY slid out.......how can I solve it, then?]


1b) How would the path followed by the coin as seen by an observer at rest above the turntable look like?
And how would the path followed by the coin as seen by an observer sitting on the turntable exactly at its centre look like?


2) What minimum speed must a roller coaster be travelling at when upside down at the top of a vertical loop so that the passengers do not fall out? The radius of the loop is 8m.
[The problem is that I don't know how the free body diagram would look like when the passenger is at the top of the loop. Force of gravity is definitely there, but is there any other foce? (normal force?)]


3) In a "ROTOR-RIDE" at the amusement park, riders are pressed against the inside wall of a vertical cylinder of radius 2.8m. When the ride reaches a rotational speed of 3.2 rad/s, the floor that the passengers are standing drops away. What is the minimum coefficient of friction requried in order that the passengers remain stationary against the wall, and do not slide down?
[Why would the passengers drop away and slide down?
Should I just let the frictional force equal to the centripetal force NEEDED for the passenger to stay in circular motion. I got the answer as 2.9 by doing it this way. This answer seems unreasonable to me, because 2.9 is such a high value for coefficient of friction, so maybe this is not the right approach...]

4) A ferris wheel of radius 16m rotates in a vertical circle uniformly once every 20s. What is the "apparent weight" of a 45kg passenger at the highest point of the ride?
[First I would need to know what apparent weight is. My textbook defines "apparent weight" as the net force exerted on an accelerating obejct in a noninertial frame of reference. This wording is too complicated for me to understand.
In simple terms, is "apparent weight" simply the force of gravity on the passenger?]

 

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1a)

The force of static friction provides the centripetal force.

You are right except you need to divide by r.

1b) use relative velocity equations.
Vct = Vcg - Vtg

3) again force of static friction offsets gravity. normal force provides the net (centripetal) force. there ya go.

4) both gravity and normal force contribute to the net (centripetal) force. you can figure out the velocity from the information given: v= 2(pi)(f)r

the 'apparent weight' is the same as the normal force.

 

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1) Why divide by r??

2) I think the question requires sketching out the path instead of calculations...

3) How can static friction offset gravity? Shouldn't the static friction be the centripetal force? (The force of gravity and normal force balances out)

4) But normal force isn't a weight, right? Weight=force of gravity? My textbook says apparent weight is the net force on the body, I don't understand...

 

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Don't divide by r. You used the equation a = r * (angular velocity)^2, SouthStar was probably thinking of a = v^2 / r.

At the minimum speed, there's no normal force. Gravity alone is causing the acceleration.

The friction is preventing the person from sliding down the wall. The normal force is causing the centripetal acceleration.

Very loosely, apparent weight is how heavy you feel. You feel heavy on the ground because there's a normal force that, combined with gravity, is compressing you. When you're in free-fall, you feel weightless. Your apparent weight is related to how much your acceleration differs from g, and is given by the net external force (not counting gravity) acting on you.

 

Hi przyk,

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1) a=v^2/r is the equation for centripetal acceleration, right? I used this one for my calculations...I have never seen this equation: a = r * (angular velocity)^2.....

2) Under what condition will the passenger fly out, then?

3) I sketched a free body diagram of a passenger, with gravity pointing downward, normal force pointing upward, and friction force (between the floor and the shoes of the passenger) pointing towards the center of the circle and calculate using centripetal force = friction force. Is this wrong? If so, how can the normal force provide the centripetal acceleration and how can I calculate how large it is? and how can I find the minimum coefficient of friction needed to avoid sliding?

4) So is apparent weight the net force, excluding force of gravity, on the object?
In the context of the question, we simply need to find normal force (normal force = apparent weight), right?

 

Its easy to get one equation from the other: v = ωr (with ω measured in rad/s), so a = v²/r = (ωr)²/r = ω²r.

The substitution ω = 2πf gets you the RHS of your original formula.

The passenger will fly out when the rollercoaster is accelerating downward at less than g. If there's nothing holding the passenger in the rollercoaster, the passenger only stays in their seat when the coaster accelerates downward as much as the passenger.

I think you've misunderstood the question. The passenger is against the wall of the spinning cylinder, but the floor has dropped away. The passenger is no longer in contact with the floor, and is being held off the floor by friction between the person and the wall.

Yes.

 

1) Why should we use radians?

I will show you how I solve it:

f=66 rev./min.=1.1Hz
r=0.15m

Fc=4(pi^2)mr(f^2)
Ff=4(pi^2)m(0.15)(1.1)^2 [Fc=Ff]
(mu)*mg=0.726(pi)^2 * m
mu= 0.7312
Thus, the coefficient of static friction is 0.73.

Am I on the right track?
Besides, I am having a problem with this method because 4(pi^2)mr(f^2) is the centripetal force NEEDED to keep the coin in circular motion when f=66 revolutions/minute. But in the context of the question, the coin ALREADY slid out.......how can I solve it, then?

2) I did it this way.

At min. speed, Centripetal force=Force of gravity
So Fc=Fg
mv^2/r=mg <----------I get an answer but I am not sure what I am calculating. (sad...

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) What is "m" actually representing here? mass of passenger or mass of roller coaster? Let's consider the case where the force of gravity exceeds the centripetal force needed for it to maintain circular motion, by saying "the passenger falling out", will the passenger fall out and separate from the roller coaster with the roller coaster staying on the track? or will both the passenger and the roller coaster fall out at the same time while having both obejcts in contact?
v^2/r=g
v^2/8=9.8
So v=8.8544m/s=9m/s

3) I think I did misread the question. So the floor disappeared when it reaches a rotational speed of 3.2 rad/s. The free body diagram should have force of gravity downward, force of friction upward, and a normal force pointing to the center. And the gravity and friction balances out. Will there be a force of friction existing in the hozitonal direction preventing the passenger from sliding sideways? If so, how can I account for that?

4) Do we actually "feel" our weight because of normal force? (not force of gravity?) and that's called apparent weight?

At the highest point of the ferris wheel, should the free body diagram of the passenger have normal force and force of gravity, with the second force being greater??

Thanks for all your help!

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1) a = v^2/r or whatever you want.

2) at the instant of release the coin shoots off in a tangent to the circle. someone from above will see this (the coin moving in a straight line in whatever tangential direction). Since there is no work done on the coin it will continue to have its original velocity.

someone sitting at the center will see the coin recede backwards in a straight line with respect to his own vision. accounting for gravity, he will also see it fall straight down as it goes backwards.

3) you will not account for any force of friction in the 'sideways' direction. if there was friction in that direction, there would have to be a net force in that direction and then the guy would have to be sliding in the direction of the force of friction. See?

4) Normal force is the apparent weight. This will become apparent if/when you are asked to explain why astronauts feel weightless.

 

You don't need to, if you have the frequency. a = v²/r = &omega;²r = 4&pi;²r. Its all the same equation, depending on whether you relate a to the velocity, angular velocity, or frequency. Don't worry about it - I was just guessing at why SouthStar was suggesting you divide by r.

Your method is completely correct. And you do use f=66 revs/min. What you'd want to use was the highest frequency at which the coin stayed in place, which is less than 66 revs/min by an infinitesimal amount. In practice, it actually makes very little sense to ask whether the coin actually starts slipping or not at 66 revs/min.

First of all, method's correct, except that since all you have to do is deal with accelerations there's no need to bring in m in the first place (you can use the mass of the car or the mass of the car+passenger - it doesn't matter since the car and passenger aren't exerting any forces on one another.

What's happening is that the person at the top of the loop is actually in free-fall. They are accelerating down toward the Earth at 9.8 m/s².

There's no acceleration acting tangentially in uniform circular motion (the acceleration is directed toward the centre of the circle), so no horizontal friction is needed. I think its safe to assume the cylinder is spinning at a constant rate.

We feel heavy because of internal pressure: the floor exerts a normal force on your feet which compresses them. Your feet apply a force on your legs, which in turn push on your body above you, etc.

With gravity, every particle in your body is trying to accelerate at the same rate, so none of your atoms are getting pushed together or pulled apart if gravity is the only force acting on you. This is partly because the gravitational field does not vary much with distance most of the time (its a different story near black holes - where strong tidal forces can pull you apart), and because the same m that appears in F = ma also appears in F = GmM/r² (this eventually led to one of the postulates of Einstein's general theory of relativity).

If you work out what the gravitational force is on an astronaut in orbit around the Earth, you'll find its still pretty significant. They feel and appear weightless because they are in free-fall.

It should have both the normal force and gravity. The two sum up to give the force responsible for the centripetal acceleration. It looks like the magnitude of the normal force is almost the same as the gravitational force.

A similar exercise would be to calculate the relationship between the "real" and "apparent" g at the equator because of the Earth's daily spin around its own axis.

 

hi,

1b) For an observer at rest from above, shouldn't he see the coin spiral out with increasing radius instead of flying out tangentially?

For an observer sitting at the center, shouldn't he see the coin moving away from him in a straight line? (I am not sure if the path seen by this observer would be straight or not because the coin is not maintaining a constant radius...)

These are some of my initial thoughts...

 

hello kingwinner,

the coin will not spiral out. Remember: when it leaves the surface, there is NO FORCE (apart) from gravity downwards acting on the coin. Hence there cannot possibly be centripetal motion.

Given that and assuming that Fgravity points in the -y direction, you can expect that the x-velocity component stays constant (Newtons 1st law).

Hence an observer looking from above will see exactly this. The coin will be moving in its tangential direction without changing (horizontal) direction. Of course, it will also be falling down.

You should figure out for yourself that the observer at the center will be spinning at the same rate as the coin is moving in the x-velocity through the principles of rotational motion or just through 'intuition'. Once you get this I can explain the rest.

Just in case anyway, (I'm assuming you accept that the x-velocity is constant), think of the observer as looking straightforward at all times. at time t1 let there be a position vector r1 drawn from his eye to the coin and the same for time t2 with a position vector r2 of equal magnitude to r1. do you think the coin will be to the left, to the right, or on the observer's line of vision (the vector r2)?

 

3) Fnet.vertical=0
Ff - mg = 0
(mu)N - mg =0
N = mg/(mu)

Fc = 4(pi^2)mr(f^2)
N = 4(pi^2)mr(f^2)
mg/(mu) = 4(pi^2)mr(f^2)
g/(mu) = 4(pi^2)r(f^2)
Substituting the given information,
I get mu = 0.34

I am not sure if this is true because of a number of potential problems...
Will Ff still equal to (mu)N in this case? (The normal force is horizontal here, not vertical, will that affect things?)
And from the above, we can see that N = mg/(mu) which means that the normal force depends on the mass of the passenger. But for a passenger with a given mass, he can also PRESS HARDER (can he?) on the wall to increase the normal force, which in turn increases the friction force, right? This would not give a unique value of (mu) because the normal force of a passenger with a given mass can have a "range" of values...How can I account for that?

 

Hi §outh§tar,

I thought that the question is looking for the path of the coin when it's still ON the turntable...If the turntable is big enough, what will the observers see before the coin flys out?

 

4) For the ferris wheel problem at highest point of the ride, we know that Fc=Fg+N! Should the normal force be pointing upward or downward?

 

No. Why should it matter whether the friction is horizontal or vertical?

The total normal force wouldn't change, just its distribution around the persons body (more pressure on the hands, and less on the rest of the body).

The &mu; you calculated is an average &mu; for the whole person. Some parts of the person would slide more easily that others. Changing the normal force distribution might affect this average, but you haven't been given the information to work out by how much. This doesn't change the problem, since the minimum average &mu; required only depends on the normal force, which doesn't change.

 

The wording of the problem seems to suggest that the coin leaves the table the instant it starts to slip.

 

Well you can work out F_c and F_g. Both of these point downward, so they have the same sign. All you have to do is see if F_N = F_c - F_g is positive or negative, and compare this sign with F_c and F_g.

 

Then the normal force must be upward because the person is not sitting upside down. The free body diagram of the person should have force of gravity pointing downward, and normal force pointing upward.
And Fc=Fn+Fg.
-71.06=Fn+(45)(-9.8)
Fn=369.94N so the normal force is clearly pointing upward.

How would the free body diagram for the box of a ferris wheel (the box where a passenger sits) look like? Would both the force of gravity and the normal force be pointing downward in this case? (because I recall that when a ball revolving in a vertical circle, at the top the force of gravity and tension force both act downward to sum up to the centripetal force.)

 

Yes.

The normal force would still point up. The car has to be pulled upward to prevent it from falling down at 9.8 m/s². (The revolving ball had to be pulled down to give it an acceleration greater than g).

 

Oh, I see!

How can I do question 1 if the turntable is tilted at 30 degrees above the horizontal? Which way will the friction be pointing? And what provides the centripetal force?