Amogh Gajare
Last Activity: 9 Years ago
Hello.U
i = -
R
2g
.....................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/s
2.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 = -
R
2g (After solving integration)............let us label it
equation 1But
=
as circumference representing the beads is 1/4th circle.
So when we substitute
value in
equation 1 we get,U
i =
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
+ mRg = mRg(1 –
)
And by work energy theorem W = change in kinetic energy
mRg(1 –
) = ½ m
– 0
So after solving we get
So answer will be square root of whole of the term mentioned above.
