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A small ball rolls around a horizontal circle at height y inside a frictionless hemispherical bowl of radius R, as shown in the attached picture

Posted on 2014-10-13
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Ball on a hemispherical bowl
What is the angular speed (w) in rpms if R is .30m and the ball is halfway up?

I found an expression for the ball's angular velocity (w) in terms of R, g, and y   (y is the ball's height, as shown in the picture)
and the expression is: w = sqrt(g/(R-y))

And I don't know if this i relevant, but i found an expression for the ball's minimum speed for moving in a circle:
 w = [sqrt(Rg)]/R


So far, I have tried using the first equation with the given information to get w = sqrt(9.8/(.3-.15)) = 8.082903769m/s
Then, considering that the circumference is 2piR = .9424777961m = 1rev, I divided w by .9424777961 to get the number of revolutions in 1 second, which i got to be 8.576227262 revs/sec. Then, I multiplied that by 60 to get revolutions per minute which is 514.57 rpms.

The answer is supposed to be rounded to two significant figures, so i entered 510, but it is still wrong. Does anybody know what to do here? I also don't know how to derive the first equation, but my homework website (MasteringPhysics) says that that is what it is.
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Question by:Logan Latosz
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by:phoffric
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Thanks for letting us know this is homework. My physics is a bit rusty, so please help me understand how did you get your units of m/s when you wrote: w = sqrt(9.8/(.3-.15)) = 8.082903769m/s.

Where did you find the expression? In general, when studying basic physics, you aren't supposed to find expressions; instead you are expected to derive expressions from expressions already established. If you do not understand where the "found" expression comes from, then it is easy to use it incorrectly.
I found an expression for the ball's angular velocity (w) in terms of R, g, and y   (y is the ball's height, as shown in the picture)
 and the expression is: w = sqrt(g/(R-y))
I was wondering what the radius of the circle travelled by the ball is at height y. Not sure if that plays in a role in your problem.
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by:Logan Latosz
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I got my units because angular velocity is a speed and speed is a distance over a time interval, so meters per second.
Unless.... should w = sqrt(g/(R-y)) be in radians per second because it is angular velocity?

It depends what expression you're talking about, I found the first expression by getting the problem wrong enough times and it gave me the correct answer (so i didn't actually find it lol). I found the second expression by using two equations from previous experience:
Minimum speed needed to travel in a circle without succumbing to the force of gravity:
v = sqrt(Rg)  where R is radius and g is acceleration of gravity

And angular velocity (w):
w = v/R

So I substituted the min speed equation into the v in the angular velocity equation to get w = [sqrt(Rg)]/R

The radius of the ball's trajectory at height y is unknown, but perhaps it could be possible to figure out. Looking at the picture and thinking about normal force, could the radius be the sine of the angle formed by the normal force?
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phoffric earned 500 total points
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I remember this much: "meters per second" is velocity, not angular velocity. I think angular velocity units is radians per second. Yes, I just checked - it is.
http://en.wikipedia.org/wiki/Angular_velocity
it gave me the correct answer
At this point, you have to take a step back and try to derive the correct answer before continuing, I believe. Physics is very hard and if you skip a small step in understanding, it will not be long before nothing makes any sense.

You probably have learned by now how to check for the units.
angular velocity is a speed and speed is a distance over a time interval, so meters per second.

And angular velocity (w):  w = v/R
What are the units of v? And units of R are what? Then do division of the units, and what do you get?
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by:Logan Latosz
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Oh, okay. Units of v are m/s and units of R are m... In v/R, the meters will cancel leaving 1/seconds. I remember my professor saying that rad/s are strange in that they are used when you have 1/s. Besides, rad/s would only make sense as it is a distance over time AND radians deal with circles. Okay...

So now, that we found that the angular speed is 8.0829 rad/s, I would have to divide that by 2*pi (because there are 2pi radians in the circumference of a circle) to get revolutions per SECOND. Then, multiply that by 60 to get revolutions per MINUTE, right?

I think I'm starting to understand this :D  thank you for all your help so far!
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by:Logan Latosz
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Helped with mostly the understanding, rather than just giving an answer. Which is awesome!
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by:Logan Latosz
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Thank you so much sir! I've been working on that problem for about 4 hours prior to you helping me!
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by:phoffric
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I guess you got the right numbers for the online exercise, right?

But, it is important that you be able to derive the expressions yourself.
Have you determined whether you need the radius of the ball's trajectory at height, y? If you need it, and don't know how to get it (and you should know how even if you don't need it at the moment), then please don't hesitate to ask another question. (This is just a math geometry problem - not physics.)

If you need to figure out how to derive the formula where the online exercise just gave it to you, then I advise that you take a step back to an area that you are comfortable with and start deriving the formula. If you get stuck on that, then ask another physics question. (It's OK if you ask two questions at the same time. Don't forget to mention that this is for study purposes. Otherwise, you may get answers that are correct, but incomprehensible at least in the long run.)
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by:phoffric
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Make sure you understand this lecture given in the first video.
Uniform Circular Motion
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by:Logan Latosz
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Thank you for the link to the video! I watched it and it actually did reinforce my understanding of the concept
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by:phoffric
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Glad the video helped.
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by:phoffric
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The link I gave you cut off in less than 10 minutes on two browsers. In case that happened to you, here is a better link.
http://ocw.mit.edu/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/video-lectures/lecture-5/

Also, you can now view many more lectures. They are very entertaining.
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