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Angular velo.

  • Thread starter ixbethxi
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  • #1
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find an expression for the balls angular velocity in terms of R,y,and g


mr*omega^2=mg
so i had omega=sqrt(g/R) and im saying that the angular velocity does not depend on its height. am i right?
 

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  • #2
mukundpa
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what is the problem???
 
  • #3
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oo its basically the picture.. "a small ball rolls arond a horizontal circle at height y inside a frictionless hemispherical bown of raidus R as shown in my picutre

find an expresion for the balls angular velo in terms of r,y,and g
 
  • #4
mukundpa
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What is your work?
 
  • #5
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my work is what i showed above, i'm trying to see if thats right, but the picture is still pending aproval, im sure u'll understand as soon as its approved.
 
  • #6
mukundpa
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As I guess the problem is of centripetal force.
Resolve the normal reaction of the bowl in horizontal and vertical direction.
the vertical component balances the weight and the horizontal component provides the required centripetal force. The ball is rotating in a horizontal circle of radius Rsinq(theeta)
q is the angle the radius of the bowl to the ball makes with verticle.
q is cos^-1(R-y / R)
 
  • #7
mukundpa
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The angular velocity depends on the height at which you want to revolve the ball. because with height the angle q will change and vertical component of the normal reaction will change accordingly.
 
  • #8
mukundpa
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you have equated centripetal force with the weight of the ball. How it is possible. The weight of the ball is acting vertically and the centripetal force is required horizontally.
 
  • #9
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honestly i tried understanding what you said, maybe im just slow. i understand your last comment about how making them equal is completely wrong, yes that makes total sense. but if its possible can u explain to me the part about the q and the cos...you completely lost me. sorry
 
  • #10
mukundpa
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My guess for the problem was correct.
Draw a sideface diagram and indicate the forces acting on the ball.

indicate the distances R and h in the fig. and try to resolve the force in horizontal and vertical directions.


will be back after four hours.
 
  • #11
VietDao29
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Okay, let's call O is the center of thecircle at the top of the hemisphere.
A is where the object is.
C is the center of the circle the object is moving arround.
[itex]\alpha[/itex] is the angle OAC. You notice, when the object travels arround the circle, [itex]\alpha = const[/itex].
Since it's frictionless, there are 2 forces acting on the object : P, N.
In Oy - axis:
[tex]P = N\sin{\alpha}[/tex]
The ball travels at the constant speed, and arround a circle (circular motion).
Therefore you also have:
[tex]N\cos{\alpha} = m \frac{v ^ 2}{r}[/tex]
[tex]\Leftrightarrow v = \sqrt{\frac{N \cos \alpha r}{m}} = \sqrt{\frac{P \cos \alpha r}{\sin \alpha m}} = \sqrt{g r \cot \alpha}[/tex]
r is the radius of the small circle the object is travelling arround.
All you have to do now is to calculate the r, and cot(alpha) with respect to R, and y.
Is that clear enough?
------
EDIT:
You can find the angular velocity by:
[tex]\omega = \frac{v}{r} = \sqrt{\frac{g \cot \alpha}{r}}[/tex]
Viet Dao,
 
Last edited:
  • #12
here is my solution to this question

i second VietDao29's idea!
here is my solution to this question!

my english is poor,so i don't discribe how i do it!

i think you can get a idea about it and understand the equations i make!

i want to make friends with you all
it is helpful to improve my english!
ronggangsky@163.com thanks for reading
 

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