Question

An object of mass m slides down on the surface of a hemispherical bowl of radius R from one end to the bottom of bowl as shown in fig. If there is no friction between object and the bowl then what will be the speed of object at the bottom of bowl? ​

Answer 1

Answer:

Conservation & energy Loss in P.E = gian in K.E. mgv = 1 2mv2 v = √(2gr)

Conservation & energy Loss in P.E = gian in K.E. mgv = 1 2mv2 v = √(2gr) More

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Answer 2

As given in the question an object of mass m slides down the hemisphere bowl

Radius of the bowl =r

As per the question no friction is present between the object and the bowl

By using law of conservation of energy

Loss in potential energy = gain in kinetic energy

PE initial + PE final. = KE initial. + KE final

$mgh \: + 0 \: = 0 + \frac{1m {v}^{2} }{2}$

$mgr \: = \frac{1m {v}^{2} }{2}$

$v \: = \sqrt{2gr}$