Question

A small solid sphere of radius $r$ rolls down an incline without slipping which ends into a vertical loop of radius $R$. Find the height above the base so that it just loops the loop

• A. $\frac{5}{2}R$
• B. $\frac{5}{2}(R-r)$
• C. $\frac{25}{10}(R-r)$
• D. $\frac{27}{10}R-\frac{17r}{10}$

Answer 1

Answer: (D)

$\frac{27}{10}R-\frac{17r}{10}$

Let $v_1$ and $w$ be the velocity and angular velocity of the sphere at B respectively.

As the is no friction. so conservation of energy is valid.

$P.E_A+K.E_A=P.E_B+K.E_B$

$mgh+0=mg(2R-r)+\frac{1}{2}m{v}_1^2+\frac{1}{2}I{w}^2$

$mgh=mg(2R-r)+\frac{1}{2}m{v}_1^2+\frac{1}{2}\times \frac{2}{5}m{r}^2{w}^2$

$mgh-mg(2R-r)=\frac{1}{2}m{v}_1^2+\frac{1}{5}m{v}_1^2$ $[v_1=rw]$

$⟹mg(h+r-2R)=\frac{7}{10}m{v}_1^2$

Circular motion equation: $\frac{m{v}_1^2}{R-r}=mg$

$⟹mg(h+r-2R)=\frac{7}{10}mg(R-r)$

$⟹h=\frac{27}{10}R-\frac{17}{10}r$