Question

A small block slides with velocity $0.5\sqrt{g}r$ on the horizontal friction less surface as shown in the figure. The block leaves the surface at point $C$. The angle $\theta$ in the figure is:

• A. ${\cos }^{-1}\left(4/9\right)$
• B. ${\cos }^{-1}\left(3/4\right)$
• C. ${\cos }^{-1}\left(1/2\right)$
• D. $noneoftheabove$

Answer 1

The correct option is B ${\cos }^{-1}\left(3/4\right)$

Vertical displacement, $s=r(1-\cos \theta )$

Apply kinematic equation,

$V^2=u^2+2as=[{\left(0.5\sqrt{gr}\right)}^2+2gr(1-\cos \theta \left)\right]$

Kinetic energy = Potential Energy

$\frac{1}{2}m{v}^2=mgr(1-\cos \theta )$

$v^2=2gr(1-\cos \theta ).........\left(1\right)$

Normal reaction on block

$N=mg\cos \theta -\frac{m{v}^2}{r}$

$0=mg\cos \theta -\frac{m{v}^2}{r}$

$0=mg\cos \theta -\frac{m{\left(0.5\sqrt{gr}\right)}^2+2gr(1-\cos \theta )}{r}$

$0=mg\cos \theta -2.25mg+2mg\cos \theta$

$\cos \theta =\frac{3}{4}$

$\theta ={\cos }^{-1}\left(\frac{3}{4}\right)$

Hence, at angle ${\cos }^{-1}\left(\frac{3}{4}\right)$ , block will leave the surface.