Question

A frictionless track $ABCDE$ ends in a circular loop of radius $R$. A body slides down the track from point $A$ which is at height $h=5\text{cm}$. Find the maximum value of $R$ for which the body completes the loop successfully.

• A. $2\text{cm}$
• B. $4\text{cm}$
• C. $2.5\text{cm}$
• D. $5\text{cm}$

Answer 1

The correct option is A $2\text{cm}$

Applying mechanical energy conservation between points $A$ and $B$,
$mg\left(5\right)+0=0+\frac{1}{2}m{v}^2$
(Velocity at $A=0$ and taking P.E at $B=0$)
$\Rightarrow$ Velocity of body at point $B$
$v=\sqrt{10g}$ ... (1)

F.B.D of block when body at the highest point $D$


Hence, $mg+N=\frac{m{v}^{\prime 2}}{R}$

For just completing the circle(i.e. $R=R_{Max}$) $N=0$, so the required centripetal force is provided by the gravitaional force:
$\Rightarrow mg=\frac{m{v}^{\prime 2}}{R_{Max}}$
$\Rightarrow v^{\prime 2}=g{R}_{Max}$ ... (2)

Applying mechanical energy conservation between points $B$ and $D$:
$0+\frac{1}{2}m{v}^2=mg\left(2R_{Max}\right)+\frac{1}{2}m{v}^{\prime 2}$
$\Rightarrow \frac{v^2}{2}=2g{R}_{Max}+\frac{1}{2}g{R}_{Max}$ (from (2))
$\Rightarrow \frac{10g}{2}=\frac{5g{R}_{Max}}{2}$ ($v^2=10g$ from (1))
$\Rightarrow R_{max}=2\text{cm}$