# Here I go

A bead of mass 'm' is kept at the top of a smooth hemispherical wedge of mass M and radius R . The bead Is gently pushed towards right . As a result the wedge slides due left . For the angular displacemt $$\theta$$ of the bead , If the speed of the wedge can be represented as (where a,b,c are constants) find a+b+c

$$\sqrt(\frac{am^{2}gR(1-bcos\theta)(cos\theta)^{2}}{(m+M)(M+mc(sin\theta)^{2}})$$

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