Physics question...more like logic affirmation.
 10022005, 05:14 PM
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Physics question...more like logic affirmation.
Okay, I'm officially stuck...
Re: NonUniform Circular Motion & Force
A pail of water is rotated in a vertical circle of radius 1.00 m. What is the minimum speed of the pail, upside down at the top of the circle, if no wter is to spill out.
I got the correct answer, I just need to know that my logic is correct.
Here is what I did.

r = 1.00m
note: a = centripital accelaration = v^2/r
sumFx= mgcos(theta) + Tension = ma = mv^2/r
In order for the water to stay in, there should be tension on the string. So, at the point where there is zero tension, the water pail will fall.
So,
mgcos(theta) = mv^2/r
rgcos(theta)=v^2
v=sqrt[rgcos(theta)]
*when the bucket is straight u and down, it is at 90degree, therefore there is no angle theta (it is zero) and the cos(0) is 1.
v=sqrt(rg)=sqrt(1m*9.8m/s^2)=3.13 m/s
Therefore, the minimum velocity at which the pail will keep the water in it is 3.13 m/s
NOW, is my assumption that when Tension is zero, the water will fall the correct step in this problem?
This is the only way I could figure to solve it.
Thx
(I hope there i someone out there )  10022005, 05:17 PM
Looks right, so you don't have to solve for any forces in the Y direction?
Just F sub x equals m*a sub x?  10022005, 05:39 PM
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Now, the force was directed toward the center of the circle (pail attached to a string and rotated vertically in a circle).
Now that I think about it, I'm slightly confused about the differences between centripital and radial accelaration. Hmm....I should really do my homework when its assigned haah
Thx. 
 10022005, 06:12 PM
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Oh, and the original equation I setup is:
SUM(F,x)=W+T=ma,r
where W=mgcos(theta), but the angle theta is zero, so mgcos(theta)=mg(1)=mg.
a,r=v^2/r, therefore
SUM(F,x)=g+T=v^2/r (m's cancel)  10022005, 10:31 PM
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haven't the foggiest
but i did stay at a holiday inn express last night.  10022005, 10:47 PM
There's nothing wrong with your approach, you just did it the hard way. The lazy man's way is to consider that so long as the centripetal force is equal to the force of gravity when the bucket is at its apex, water won't fall out. You then derive a formula for the force of gravity on the water, and set it equal to the centripetal force on the water. From there, you solve for the unknown.
m*g = (m*v^2)/r
g = v^2/r
g*r = v^2
v = sqrt(g*r)
Also, radial acceleration is the rate of change of the radial velocity of an object. That is, whether and how much it's spinning slower or faster.  10022005, 10:54 PM
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I like Nabeshin's approach.
Sum the forces in the y direction. Easiest way.  10032005, 12:17 AM
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Originally Posted by Nabeshin
Originally Posted by kwyckemynd00
What I was concerned about was my assumption that Tension = 0 and the validity of the assumption.  10032005, 12:20 AM
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Originally Posted by jmh80  10032005, 12:24 AM
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Originally Posted by Nabeshin
And, what confuses me is that A,r = A,c and I'm not quite sure when to use radial v. centripital.
If you plug in m*v^2/r (A,r) into the above equation, you end up with an imaginary number...  10032005, 12:33 AM
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Oh, and as an FYI, now that I know I've got a few ppl in here familiar with physics, I'm going to be asking for help every now and then.
My teacher SUCKS! I'm basically learning out of the book.
From what I hear from faculty, he should be teaching graduate students and he's working on some research that may get him a nobel (like that means **** after yasser arafat got one, right? ), but the dude just can't teach!  10032005, 12:38 AM
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Originally Posted by kwyckemynd00
smart ppl......  10032005, 12:47 AM
Imagine that you have a pole in the middle of a field. At the top of the pole is a ball which can rotate freely around the axis of the pole. Attached to this pole is a length of string. On the end of this string is a little rocket.
When you turn the rocket on, it will try to go straight, but because it is constantly being pulled by the rope, it ends up traveling in a circle. The force the rope exerts on the rocket, pulling it toward the center of the cirlce, is the centripetal force. Divide this by the mass of the rocket, and you have the centripetal acceleration. This probably isn't news to you.
You can make the rocket go faster or slower using your remote control. If you press the gas, the rocket will accelerate. This acceleration would normally make the rocket fly faster in a straight line, but because it's tied down by a length of rope, it just makes it spin around faster. Therefore, this acceleration is called rotational acceleration. If you multiply this by the mass of the rocket, you get the rotational force.
So, in a nutshell, centripetal force points from the rotating object to the center of rotation, whereas rotational force points out from the direction of the rotating object, and perpendicular to the centripetal force.  10032005, 12:56 AM
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Well, an object has a centripital accelaration pointing toward the center of the circle when it is rotating about said pole, the force is just multiplying by mass (F=ma)...just like you stated.
The centripetal acceleration and centripetal force I'm familiar with.
I'm just confused as to just exactly what the radial accelaration is, because it is opposite the centripital accelaration (same goes for force). And, further, i'm confused as to when to solve for centripital v. radial accelaration (or force).
BTW, i appreciate the help.  10032005, 01:03 AM
It's not really line tension which keeps water in the bucket. It's just an indicator of when water would fall out of the bucket, and while your intuition about this connection is correct, the proof of it is a little tricky.
It's also not a matter of net force in the Y direction being zero, so force summation doesn't really apply.
What you've got is centripetal force pulling the water down, AND gravity pulling the water down. If the water is going to stay in the bucket, the centripetal force has to be pulling the bucket down faster than gravity can. If it pulls the bucket down just as fast as gravity pulls the water down, then the water will just barely stay in the bucket.
Now, if you're swinging the bucket around by your arm, and you swing it such that the centripetal force is less than gravitational force, you'll get wet. When the bucket is over your head, you'll be keeping the bucket from falling with the force of your arm, but the water will get pulled out by gravity.
On the other hand, if you're swinging the bucket around by a string, even if you swing it such that gravitational force outweighs centripetal force, you won't get wet. The string loses tensions because the bucket is now being pulled down by gravity faster than by the centripetal force. This means that the bucket and the water inside it are accelerating to the ground at the same speed.
So, if you have a ball on a string, and you want to figure out at what velocity will the string lose tension, you'll find it to be at the same velocity as that at which the water would leave the bucket if you were swinging it by your arm. (Assuming your arm and the string are the same length.) In both scenarios, what you're really solving for is the velocity at which centripetal force is equal to gravitational force.
Circular motion can get difficult to visualize because the force vectors are always changing, but practice practice practice. If I was in your shoes, I'd do extra rotational problems. It's in your interest to be damn good at them, because you will see them all the time on exams  a good rotational problem is an easy way to test your knowledge of the whole of newtonian kinematics in one fell swoop, so physics profs love to use them.  10032005, 01:12 AM
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Okay, that made sense, thanks!
Now, like all instances of learning a new piece of information, I'm left with only more questions
Now I'm confused on when to use force summation in circular motion...*sigh*
I haven't done **** for homework since I started this class...so, I'm playin a bit of catch up I've just realized I can't half ass physics. Irritating though b/c I slept through calc and chem with great grades, and now i have to apply myself and I feel like I'm "we todd did" to say the least.  10032005, 01:19 AM
Originally Posted by kwyckemynd00
Well, I don't know why your book says that radial acceleration and centripetal acceleration are different things, because every other physics book I've been through (and that would be a lot) considers them synonymous terms.
In fact, if you take apart the words, they mean the same thing. Radial is derived from the latin word "radius," which means the center of a wheel. Centripetal is derived from the latin words "centrum" and "petere", meaning "center" and "to move towards," respectively.
Here's a page from Western Washington University which shows you what I mean:
In order for an object to execute circular motion  even at a constant speed  the object must be accelerating towards the center of rotation. This acceleration is called the centripetal or radial acceleration and has a magnitude of...
But, if your book says that radial acceleration is just the negative of centripetal acceleration, then at least it should be easy to know when you need to use it  never.
BTW, i appreciate the help.  10032005, 01:30 AM
Originally Posted by kwyckemynd00
Think about it. If net force is zero, then there is no acceleration in the system. But in order to have circular motion, you must have a continually accelerating system  without centripetal acceleration, it just doesn't work. And anytime you have acceleration, net force can't be zero.
I haven't done **** for homework since I started this class...so, I'm playin a bit of catch up I've just realized I can't half ass physics. Irritating though b/c I slept through calc and chem with great grades, and now i have to apply myself and I feel like I'm "we todd did" to say the least.
Something that will help is to analyze the physics of your world. What's keeping me in place on this chair? What forces are propelling me forward as I walk? These kinds of little games will help make the laws of physics second nature to you, and they beat reading the textbook by a damn sight.  10032005, 02:07 AM
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Oh, I understand that there has to be acceleration with any circular motion, be definition, there is always a change in velocity, and change in velocity over time is acceleration. So, that's a given.
I thought you mean  for example lets use a falling meteor:
SUM,F,y= F,grav_earth = m,meteor*a < I thought you meant that would be a force summation and that I could not use that relationship with doing circular motion...in which case we just normally would use centripital acceleration.  10032005, 06:47 AM
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Okay.....i've got one other problem that seems easy, but I keep getting the wrong answer!
Objectives:
Determine accelaration of the system.
Find the tension between the two blocks.
Givens:
Coefficient of kinetic friction: 0.100
Force pulling system: 68.0N
Mass of block 1: 12.0 kg
Mass of block 2: 18.0 kg
block 1......T........Block 2
++...........++
..................> F = 68.0 N
++...........++
As you can see, block 1 (12.0kg) is tied together by a rope of negligible mass to block 2 (18.0kg) which is being pulled my a horizontal force of 68.0 N.
To reiterate the objective, find the tension of the rope and the acceleration of the system.
If anyone has anymore help here, it would be appreciated. its a pain in the ass because it looks so easy, but I keep coming up with the exact same WRONG answer.
I get T = 36.6 N and Ax = 2.07m/s^2 < wrong  10032005, 12:59 PM
I don't have time for an in depth reply now (gotta get to school myself), but here's the Cliff Notes. First, find the acceleration of the whole system, which is just net mass divided by net force. Be sure to account for the frictional force, which will decrease the netforce.
Once you have the acceleration of the whole system, you then figure out what the net force you would need for the block that's being towed by the other block. That net force plus the frictional force (remember, the frictional force is opposing the force pulling the blocks) will be the force of tension in the line between the two blocks.  10032005, 07:05 PM
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Thx guys, I appreciate it very much. I"ll have to take a look at this in a little bit.
So far, everyone I've had help from has had a very different approach to solving the problem than I've been taught...makes things more interesting  10042005, 09:50 PM
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Okay guys, thx. I reviewed this right now, and I was able to put the info you gave me together to make it work in a system similar to how I was taught, and how my instructor wants to see it on the test.
SumForces,x,system=Force,syste mFriction,System=m,system*a,sys tem; Force = 68N, m,system = m1+m2 = 12+18=30
a=ForceFriction/m = 68N â€“ meau*mass*gravity/mass = (68N â€“ 0.100*30.0kg*9.8m/2^2)/30kg= ~ 1.29m/s^2
Now, lets focus on one box specifically to find T: (see box 1, m = 12.0kg)
Fx = T â€“ Fk = ma (a = a,system, normal force = mass*gravity (m*g))
T = ma + Fk
T= ma + meau*g*m
T = 12.0*1.29 + 0.1*9.8*12.0 = 27.2 N
Thx again guys.  10052005, 09:29 AM
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