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A necklace made of beads is kept on a half of a semicircular surface as shown in the attached image(see the attached image).What will be its velocity when it become horizontal, falling down ?(neglect friction )(given gravity =g ,radius of the semicircular half =r)

A necklace made of beads is kept on a half of a semicircular surface as shown in the attached image(see the attached image).What will be its velocity when it become horizontal, falling down ?(neglect friction )(given gravity =g ,radius of the semicircular half =r)

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Grade:12

1 Answers

Amogh Gajare
60 Points
8 years ago
Hello.Ui = -\lambdaR2g\int^\frac{\pi }{2}_0 sin\theta d\theta.....................that’s the formula which is used in these kind of problems.Lambda represents mass per unit length,R is radius,g is 10 m/s2.Upper limit is taken as 90 degree as quarter cicle is given in question and lower limit is obviously 0 degree as it is becoming horizontal finally.
Therefore Ui =  -\lambdaR2g (After solving integration)............let us label it equation 1
But \lambda = \frac{Mass}{\pi R/2} as circumference representing the beads is 1/4th circle.
So when we substitute \lambda value in equation 1 we get,
Ui = \frac{-2MR^2 g}{\pi }
 
R is also height here.So the final potential energy Uf = -mgR (Negative sign because of opposite direction.)
So work done by gravity = Ui – Uf
                             \frac{-2MR^2 g}{\pi } + mRg = mRg(1 – \frac{2}{\pi})
 
And by work energy theorem W = change in kinetic energy
                                         mRg(1 – \frac{2}{\pi}) = ½ mv^2 – 0
 
So after solving we get v^2 = 2Rg(1 - \frac{2}{\pi})
So answer will be square root of whole of the term mentioned above.
 
 
 
 

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