Question

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

• A. $\frac{mRcos\theta }{M+m}$ to the right
• B. $\frac{mRsin\theta }{M+m}$ to the right
• C. $\frac{mRcos\theta }{M+m}$ to the left
• D. $\frac{mRsin\theta }{M+m}$ to the left

Answer 1

The correct option is D $\frac{mRsin\theta }{M+m}$ to the left

$m_1\Delta {\overrightarrow{x}}_1+m_2\Delta {\overrightarrow{x}}_2=0$
$m(Rsin\theta -x)-Mx=0$
$x=\frac{mRsin\theta }{m+M}$