Angular velocity using Newton's law

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The discussion focuses on calculating the minimum and maximum angular velocity (ω) for a mass on a rotating conical dish, using Newton's laws. Participants emphasize the importance of correctly identifying the components of forces acting on the object, particularly the centripetal force mω²r, and how to resolve these forces along the chosen axes. A free body diagram is recommended to visualize the forces, and clarification is provided on the angles involved in the equations. The correct angles for the forces are identified as 30° for mg and 60° for mω²r, leading to successful calculations for ωmax and ωmin. Understanding the direction of forces and their components is crucial for solving the problem accurately.
Huumah
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Homework Statement



Small object of mass m is placed on the inner surface of the conical dish which can rotate with angular speed ω.

For μ=0.3 calculate the minumum and the maximum ω such that the object is remains at rest. Solve the problem using the indicated axes x-y.

Do the following steps:
(a) Show all forces acting on the object and and the acceleration
(b) Write Newton’s second law along both axes and solve.
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Homework Equations


F=ma=m*ω2*r

The Attempt at a Solution



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I have plugged in numbers and got answers for ωmax and ωmin but I don't get the right answer.
I have solved it right when I choose axis the ordinary X-Y axis but I need to know how to solve with using the other axis mentioned in the problem

The right answer is
ωmax=7.214 rad/s and ωmin =3.405 rad/s
 
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Your equation for both the X and Y direction is wrong.
You have taken the wrong component of mg and mω2r along the X direction. In the RHS, it won't be completely mω2r, it will be some component of this force. Make a free body diagram.
 
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In my free body diagram i have the fricton coefficent on both sides because it can go both upwards and downwards.

I don't really get what I should do on the right hand side for mω2*r

I have been trying for a few hours now and I am completely lost now.
 
Huumah said:
In my free body diagram i have the fricton coefficent on both sides because it can go both upwards and downwards.
Correct!
I don't really get what I should do on the right hand side for mω2*r
The force mω2r doesn't act in the X direction, so you will have to take its component along the X direction and solve.
See the image below, that should help you.
5bxqfc.png
 
Thanks very much.

I did not know in what direction the ω2 force was heading. Tried to read it in my physics book but english is not my first language so i didn't understand it.

I feel like I'm getting closer but there is something wrong in my 2 equations because I can't solve them ( because there is something wrong with them)

Can you give my a hint, Sir?
r5m4p.png
 
Huumah said:
I did not know in what direction the ω2 force was heading. Tried to read it in my physics book but english is not my first language so i didn't understand it.
ω2r is the centripetal acceleration. Don't think of mω2r as a force, but as mass*acceleration. (View things from the usual inertial frame.)

I feel like I'm getting closer but there is something wrong in my 2 equations because I can't solve them ( because there is something wrong with them)
Why can't you solve them? Eliminate N. (I think you have one of your angles wrong.)
 
I always got an error when i tried to solve it using wolfram alpha.

I just probably had a silly typo because it worked when I typed it in again.

I got it right when i changed the angle to 30° in the formula for y axis.

But I still don't get why it is not 60° in the y-axis formula.

Anyway thank you both so much for the help. Such a lovely community.
 
Huumah said:
I got it right when i changed the angle to 30° in the formula for y axis.

But I still don't get why it is not 60° in the y-axis formula.
The angle between mg and your y-axis is 30 degrees; the angle between mω2r and your y-axis is 60 degrees. And sin(30)=cos(60) & sin(60)=cos(30).
 

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