A smooth sphere of radius R moves along a str | vedclass

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(D) The sphere moves with acceleration $a = g$ towards the left,so the particle experiences a pseudo force $F_p = ma = mg$ towards the right.Let the p

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$\sqrt{2Rg(1 + \sin 	heta - \cos 	heta)}$

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(D) The sphere moves with acceleration $a = g$ towards the left,so the particle experiences a pseudo force $F_p = ma = mg$ towards the right.Let the particle be at the top of the sphere at $t = 0$. As it slides to an angle $	heta$ from the vertical,the vertical displacement is $h = R(1 - \cos 	heta)$ and the horizontal displacement is $d = R \sin 	heta$.Applying the work-energy theorem relative to the sphere:$W_g + W_{pseudo} = \Delta K.E.$Work done by gravity: $W_g = mg \cdot h = mgR(1 - \cos 	heta)$.Work done by pseudo force: $W_{pseudo} = F_p \cdot d = (mg)(R \sin 	heta)$.Since the initial kinetic energy is zero,the final kinetic energy is $\frac{1}{2}mv^2$:$mgR(1 - \cos 	heta) + mgR \sin 	heta = \frac{1}{2}mv^2$.Dividing by $m$ and multiplying by $2$:$2gR(1 - \cos 	heta + \sin 	heta) = v^2$.Therefore,$v = \sqrt{2gR(1 + \sin 	heta - \cos 	heta)}$.

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