Question

A small ring is connected to another larger ring of radius $5\text{m}$. The small ring is given an initial velocity of $10\text{m/s}$ at the bottom of the larger ring. Will the small ring complete a loop around the larger ring? If not, how high will the small ring rise? Assume that there is no friction between the surfaces. [Take $g=10{\text{m/s}}^2$]

• A. The smaller ring rises to a height of $5\text{m}$ above the initial position.
• B. The smaller ring completes a circle around the larger ring.
• C. The smaller ring rises to a height of $7.5\text{m}$ above the initial position.
• D. The smaller ring rises to a height of $10\text{m}$ above the initial position.

Answer 1

The correct option is A The smaller ring rises to a height of $5\text{m}$ above the initial position.


If the smaller ring has to just complete a circle around the larger ring, then, velocity at the topmost point $=0$
$\therefore$ Kinetic energy at the topmost point $=0$
Let $u_1$ be the the minimum speed required at the base to complete vertical circular motion.
Applying the law of conservation of energy between topmost and bottommost point:
$\frac{1}{2}m{u}_1^2=mg\left(2r\right)+0$
$\Rightarrow u_1^2=2\times g\times 2\times 5=200$
$\Rightarrow u_1=\sqrt{200}=10\sqrt{2}\text{m/s}$
$\because u_1>10\text{m/s}$
So, the smaller ring will not complete a loop around the larger ring.
The height to which the smaller ring will rise is given by,
$\frac{1}{2}m{u}^2=mgh\Rightarrow h=\frac{u^2}{2g}=\frac{{10}^2}{2\times 10}$
$\Rightarrow h=\frac{100}{20}=5\text{m}$
The ring will rise to $5\text{m}$ above the base position.