Question

A block is released on the convex surface of a hemispherical wedge as shown in figure. Determine the displacement of the wedge when the block reaches the angular position $\theta$.

Answer 1

Let the displacement is $x$.

The displacement is given as,

$m_1\Delta {\overrightarrow{x}}_1+m_2\Delta {\overrightarrow{x}}_2=0$

$m\left(R\sin \theta -x\right)+Mx=0$

$x=\frac{mR\sin \theta }{m+M}$

Thus, the displacement of the wedge is $\frac{mR\sin \theta }{m+M}$.