Question 336: Work, Energy and Power & Circular Motion

Question

A bob of mass m, suspended by a string of length l
1
is given a minimum velocity required to complete a full circle in the vertical plane. At the highest point, it collides elastically with another bob of mass m suspended by a string of length l
2
, which is initially at rest. Both the strings are mass-less and inextensible. If the second bob, after collision acquires the minimum speed required to complete a full circle in the vertical plane, the ratio $${{{l_1}} \over {{l_2}}}$$ is

Accepted Answer

To determine the minimum velocity required for the bob of mass $ m $ to complete a full circle in the vertical plane, we start with the condition at the bottommost point:
$ u = \sqrt{5 g l_1} 	ext{ (at point A)} $
At the highest point B, using the conservation of energy, we get:
$ \frac{1}{2} m u_1^2 = m g (2 l_1) + \frac{1}{2} m v_1^2 $
Simplifying, we have:
$ v_1^2 = u_1^2 - 4 g l_1 $
Substituting $ u_1 = \sqrt{5 g l_1} $:
$ v_1^2 = 5 g l_1 - 4 g l_1 $
$ v_1 = \sqrt{g l_1} $
When the bob at point B elastically collides with another identical mass that is initially at rest, the velocities swap due to the nature of elastic collisions. Thus, the second bob attains the velocity:
$ u_2 = v_1 = \sqrt{g l_1} $
For the second bob to complete its circular path, the minimum velocity requirement is:
$ u_2 = \sqrt{5 g l_2} = \sqrt{g l_1} $
$	herefore \frac{l_1}{l_2}=5$

Work, Energy and Power & Circular Motion

Integer / Numeric Answer Type