Question

A particle is kept at rest at the top of a sphere of diameter $42m$. When disturbed slightly, it slides down. At what height $h$ from the bottom, the particle will leave the sphere.

• A. $14m$
• B. $28m$
• C. $35m$
• D. $7m$

Answer 1

Answer: (C)

$35m$

$x=r-rcos\theta mgcos\theta -m{a}_c=0cos\theta =\frac{a_c}{g}{a}_c=centripetalaccelerationcos\theta =\frac{a_c}{g}=\frac{v^2}{rg}\frac{1}{2}m{v}^2=mg(r-rcos\theta )\frac{1}{2}m\times mgcos\theta =mg(1-cos\theta )cos\theta =\frac{2}{3}cos\theta =\frac{r-x}{r}=\frac{2}{3}3r-3x=2rx=\frac{r}{3}heightfrombottom2r-\frac{r}{3}=\frac{5r}{3}=\frac{5\left(21\right)}{3}=35m$