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Problem setup-Merry Go Round.10 ft radius,1 rev/sec (54 mph at perimet…

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Problem setup-Merry Go Round.10 ft...
Problem setup-Merry Go Round.10 ft radius,1 rev/sec (54 mph at perimeter) steel ball 8oz groove in flr straight from perimeter to center?
How much impulse force pushed in a short instant once, on a steel ball while resting at perimeter (A), and how long does it take to reach center(B)? No need to calc friction. The steel ball will start off fast and slow down due to the opposite centripetal force as the MGR is spinning. The goal is to find out how much force is just enough to get it to the center with just a little extra force left to hit.
2. How long does it take to reach the center (A -> B) to travel the 10 ft radius? You can imagine sitting at the perimeter of the MGR as it is spinning 1 rev/sec, you'd be experiencing a lot of outward force. Then you'd try to roll the steel ball to someone in the center of the MGR. The groove on the floor of the MGR keeps the ball in line as it rolls straight to the center. Would the ball get there before one revolution?
Submitted: 2 years ago.Category: Math Homework
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Answered in 17 hours by:
6/29/2015
Math Tutor or Teacher: Ingo U, Software Engineer replied 2 years ago
Ingo U
Ingo U, Software Engineer
Category: Math Homework
Satisfied Customers: 796
Experience: Over 25 years experience in software development. Expert in Microsoft .Net, C#, C++, VB, SQL
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Hi, I imagine you need to see all the steps? And do you typically convert everything to metric units first, since most physics is studied using metric? What units do you need for the answer, N or lb ?
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Math Tutor or Teacher: Ingo U, Software Engineer replied 2 years ago
Also, can you specify the length of the "short instant" for the impulse force? Or can we just pick something, like 1ms. It does enter into the equation for force in the end.
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Customer reply replied 2 years ago

English units are easier for me at this time. And the short instant can be 1 ms. That sounds as good as anything. It sounds like you are familiar with the challenge, which is nice. No one i know can figure this out due to a few things, not least of which is the decreasing resisting centripetal force on the ball the closer it gets to the center. Thank you for your help!

Math Tutor or Teacher: Ingo U, Software Engineer replied 2 years ago
Yes I realized the non-constant force/acceleration after giving it some thought... I'm afraid that's a bit beyond my scope. Here's a link that might help though:http://physics.stackexchange.com/questions/15587/how-to-get-distance-when-acceleration-is-not-constantBasically I think first you have to solve the initial/final value problem where acceleration is a function of distance travelled, ultimately resulting in the answer to "what initial velocity does the ball need right after moment of impact". The rest is just solving for the amount of (average) force needed (for a short time, i.e. 1ms) to change the momentum of the ball such that it has the desired velocity.But the first part is the hard one. I will opt out in the hope that someone else here can help you.Regards,Ingo U
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Customer reply replied 2 years ago

Ingo,

I understand and thanks for trying. I saw the link and some related problems, but don't see how to get to my answer specifically there. Is there a way you can forward this onto someone else in your network that might be able to solve it?

Thanks again :-)

Pat

Math Tutor or Teacher: Ingo U, Software Engineer replied 2 years ago
Actually I found another link that appears to answer the exact same question:https://answers.yahoo.com/question/index?qid=20150512220924AAzVHqPI don't quite follow it but it appears to be well received.
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Customer reply replied 2 years ago

Thanks but that answer has a serious flaw. It is actually an answer to my question I submitted. They don't understand the opposite centripetal force acting on the ball as exemplified by him saying that the ball must go slowly to the center. If it goes slowly of course it will be quickly whirled back off the MGR before it makes any progress towards the center. I think they have part of it, but they didn't want to go any further. Can you please send on to someone in your network that might know more advanced physics? Maybe they can work off that partial yahoo answers input. I know it can't be too hard for the right person. In principle it is a very straight forward and simple qquesti I n with very few variablea....If one knew how to structure the equations of course. Thanks again!

Pat

Customer reply replied 2 years ago

Ingo,

any luck finding someone else you work with to take this on? I would much rather have someone in your company finish rather than have to start over somewhere else. But I need this info and would like to know if you are still trying to find someone on your end.

Thanks,

Pat

Math Tutor or Teacher: Ashok Kumar, Electrical Engineer replied 2 years ago
Ashok Kumar
Ashok Kumar, Electrical Engineer
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Satisfied Customers: 3,150
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Do you still need answer?
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Customer reply replied 2 years ago

Yes please.

Math Tutor or Teacher: Ashok Kumar, Electrical Engineer replied 2 years ago
Kindly indicate time in hours.
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Customer reply replied 2 years ago

Actually I should ask first if you are going to be able to get help. Do you think you can get help? If so, yes, please do, otherwise what is the process for me to go to another expert? Thanks again!

Pat

Math Tutor or Teacher: Ashok Kumar, Electrical Engineer replied 2 years ago
I hope I can solve it. I just wanted to know time in hours by which you would need answer. If time is short, I will opt out and some other expert may help you. If there is enough time, I will try to reply. I want time in hours as I am in totally different time zone.
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Customer reply replied 2 years ago

How about 48 hours?

Math Tutor or Teacher: Ashok Kumar, Electrical Engineer replied 2 years ago
Thanks. I will try it and let you know if I can solve it.
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Math Tutor or Teacher: Ashok Kumar, Electrical Engineer replied 2 years ago
Sorry. I tried but I could not solve it and , hence, will prefer to opt out.
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Customer reply replied 2 years ago

How do others in your network get alerted to this question? Can you forward it onto anyone? Thanks!

Pat

Math Tutor or Teacher: Ashok Kumar, Electrical Engineer replied 2 years ago
Now this question is open to all eक्ष्perts. Anybody interested will try to reply and you will be intimated by mail.
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Customer reply replied 2 years ago

Thanks Ashok :-)

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