Tethebal| problem, radial acceleration, angle of ball orbit

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The discussion revolves around calculating the angle that a string makes with the vertical when a 1-kilogram ball is whirled in a circular motion tied to a meter-long string at a frequency of 1.6 Hz. Participants clarify that the tension in the string can be resolved into vertical and horizontal components, leading to the equations Tcos(x) = mg and Tsin(x) = centripetal force. The mass of the ball cancels out in the equations, simplifying the calculations. The centripetal force is expressed as mω²r, where ω is the angular velocity. Formatting suggestions for clearer presentation of equations are also provided.
izforgoat
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Homework Statement



This is a concept/procedure question. I am playing teth3rbalI with a 1 kilogran ball and it is tied to a meterlong string. When I hit the ball it circles the pole at 1.6 Hz. Find the anglethat the string makes with thepole.

Homework Equations


a_{r}=v^{2}/r
a_{r}=(4\pi^{2}r)/T^{2}

T=1/f
f_{r}=ma_{r}

The Attempt at a Solution



So I found the period for one cycle but that's basically it and I'm stuck without a radius or at least a velocity to go by. What am I overlooking?
 

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The whirling string has to make an angle, say 'x', with the vertical in the downward direction. It lies on a cone, whose semi-vertical angle is 'x'. If 'T' is the tension in the string, then mg=T*cos(x). The centipetal force is T*sin(x). The mass m travels in a horizontal plane.

Now you try for a while.
 
hmmm

k so wow, i' not getting very far at all on this.

since Tcos(x) = mg then Tcos(x) = 9.8 N and since T sin(x) = centripedalforce then Tsin= 1kg((4\pi^{2}r)/(.625^{2}))

I'm slightly nervous proceeding since my next move would be to do a pythagoras of 9.8^{2} + ((4\pi^{2}r)/TimePeriod^{2})^{2} = T^{2} which doesn't seem right nor does it make me any more confident.
 
Last edited:
kk think i got it but does the mass cancel out?
 
It'd be neater if you didn't write the values of g etc in every step. Use symbols and plug in the values later.

Yes, the mass cancels out.

The centripetal force is m*w^2*r= T*sin x.

tan x = Tsin x/Tcos x = mw^2*r/mg =rw^2/g.

You know the value of r, and w is given.
 
izforgoat said:
kk think i got it but does the mass cancel out?

If you look at Shooting Star's post this should become apparent.

It would also help if you formatted your posts in a more readable way. For example [ tex] [ /tex] tags should be used for LaTeX set on its own, whereas you should use [ itex] [ /itex] tags for inline maths. Furthermore, it would help if you either did all your equations in tex, or none of them, since that would prevent strange mixed formatting.

Just a few thoughts, anyway.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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