Angular Velocity of Rotating Mass

AI Thread Summary
A solid sphere rotates uniformly in a horizontal circle, suspended by a cord at an angle theta from the vertical. The discussion involves analyzing the forces acting on the mass, including tension, gravity, and centripetal force. The relationship between these forces leads to the equation tan(theta) = (v^2)/(rg), where r = Lsin(theta). Participants clarify that the horizontal component of tension equals the centripetal force, and they derive angular velocity using the relationship v = ωr. Ultimately, the correct expression for angular velocity is confirmed to involve taking the square root of the derived equation.
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1. Homework Statement

A solid sphere of mass m rotates uniformly in a horizontal circle suspended from a fixed
point on a cord of length L and negligible mass. The cord
makes an angle theta from the vertical. What is the angular
velocity?

2. Homework Equations

Torque= (Rotational Inertia)x(angular acceleration)


3. The Attempt at a Solution

Tried to mess around with a few things but ultimately got no-where so any help would be greatly appreciated.
 
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This is the Diagram for the problem
 

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Welcome to Physics Forums.

I would suggest that you draw a free body diagram, identifying all the forces acting on the mass.
 
thanks for the welcome,

so there would be a tension force through the cord, a gravitational force acting straight down and a centripetal force. am i missing anything?
 
20930997 said:
thanks for the welcome,

so there would be a tension force through the cord, a gravitational force acting straight down and a centripetal force. am i missing anything?
No, that's spot on. So, what can you say about the horizontal and vertical components of the forces acting on the mass?
 
the vertical components sum to zero.
does the horizontal component of the tension= the centripetal force?
 
20930997 said:
the vertical components sum to zero.
does the horizontal component of the tension= the centripetal force?
Correct!
 
ok so i got to
tan(theta)=(v^2)/rg where r= Lsin(theta)

i was thinking solve for v and then use Angular velocity=v(r)

is this heading in the right direction?
 
20930997 said:
ok so i got to
tan(theta)=(v^2)/rg where r= Lsin(theta)

i was thinking solve for v and then use Angular velocity=v(r)

is this heading in the right direction?
Looks okay to me.
 
  • #10
ok so i got angular velocity= g/Lcos(theta)

but the answers tell me that it should be the sqrt of that.
im a bit lost as to where i have made my mistake :S
 
  • #11
oh wait i forgot to square both sides of the equation!

thanks for your help!
 
  • #12
20930997 said:
ok so i got angular velocity= g/Lcos(theta)

but the answers tell me that it should be the sqrt of that.
im a bit lost as to where i have made my mistake :S
So you have

\frac{v^2}{r} = g\tan\theta

And you know that v = \omega r. Thus,

\omega^2 r = g\tan\theta

Do you follow and can you take it from here?
 
  • #13
20930997 said:
oh wait i forgot to square both sides of the equation!

thanks for your help!
Ahh! Never-mind!

No problem, it was a pleasure!
 
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