Question

A bead of mass m kept at the top of a smooth hemispherical wedge of mass M and radius R, is gently pushed towards right. As a result, the wedge slides due left. Find the a. speed of the wedge
b. magnitude of velocity of the bead relative to the wedge

Answer 1

Applying work-energy theorem
$W_{gravity}=\Delta K^{{}^{\prime }}+\Delta K_c$ (i)
ΔK′=ΔK′= change in KE w.r.t. centre of mass from and ΔKC=ΔKC= change in kinetic energy of centre of mass w.r.t. ground.
Let v is the velocities of ball with respect to wedge and V is the velocity of wedge.
Linear momentum will be conserved in the horizontal direction.
Hence

$MV=m(vcos\theta -V)$

$V=\frac{mvcos\theta }{(m+M)}$ (ii)

The velocity of centre of mass will be zero in horizontal direction but the velocity of centre of mass will change in vertical direction.
${{(v}_{cm})}_y=\frac{mvsin\theta }{(m+M)}$ (iii)
Now applying Work Energy theorem

$-mgR(1-cos\theta )=\frac{1}{2}\mu v^2+\frac{1}{2}(m+M){{{(v}_{cm})}_y}^2$

−mgR(1−cosθ)=12μv2+12(m+M)vcm,y2
Here,

$\mu =\frac{mM}{m+M}$

from here we get,

$v=\sqrt{\frac{2gR(1-cos\theta )(m+M)}{(M+m{sin}^2\theta )}}$

and from equation (ii)

$V=\sqrt{\frac{2gR{m}^2(1-cos\theta ){cos}^2\theta }{(m+M)(M+m{sin}^2\theta )}}$ (Ans.)