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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)

Anjishnu Adhikari , 9 Years ago

Grade 12

1 Answers

Amogh Gajare

Last Activity: 9 Years ago

Hello.U_i = - $\lambda$ R²g $\int^\frac{\pi }{2}_0 sin\theta d\theta$ .....................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 U_i = - $\lambda$ R²g (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,

U_i = $\frac{-2MR^2 g}{\pi }$

R is also height here.So the final potential energy U_f = -mgR (Negative sign because of opposite direction.)

So work done by gravity = U_i – U_f

$\frac{-2MR^2 g}{\pi }$ + mRg = mRg(1 – $\frac{2}{\pi}$ )

And by work energy theorem W = change in kinetic energy

mRg(1 – $\frac{2}{\pi}$ ) = ½ m $v^2$ – 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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