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Laws of Motion

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Physics Part I

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Physics

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Class 10 Class 12
A thin circular loop of radius rotates about its vertical diameter with an angular frequency ωShow that a small bead on the wire loop remains at its lowermost point for straight omega space less-than or slanted equal to space square root of straight g over straight R end rootWhat is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for ω square root of fraction numerator 2 straight g over denominator straight R end fraction end root? Neglect friction.

Let the radius vector joining the bead with the centre make an angle θ, with the vertical downward direction.


             

Here,

OP = R = Radius of the circle

  N = Normal reaction

The respective vertical and horizontal equations of forces can be written as, 

mg = N Cosθ                         ... (i) 

mlω2 = Sinθ                       … (ii)

In ΔOPQ, we have

Sin θ = l / R 

R Sinθ                              … (iii)

Substituting equation (iii) in equation (ii), we get
 
m(R Sinθω2 = N Sinθ 

mR ω2 = N                            ... (iv) 

Substituting equation (iv) in equation (i), we get

mg = mR ω2 Cosθ 

Cosθ = g / Rω2                        ...(v) 

Since cosθ ≤ 1, the bead will remain at its lowermost point for g / Rω2 ≤ 1.

i.e., for                      ω ≤ (g / R)1/2  

For ω = (2g / R)1/2 

rightwards double arrow      
ω2 = 2g / R                  ...(vi) 

On equating equations (v) and (vi), we get

  fraction numerator 2 straight g over denominator straight R end fraction = g / RCos θ  

 Cos θ = 1 / 2 

∴ θ = Cos-1(0.5)  =  600 , is the angle made by the radius vector joining the centre to the bead with the vertical downward direction. 
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