0

Guest

Courses New

A bob of mass m , suspended by a string of length l1 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 l2, 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 l1/l2 is ?

A bob of mass m , suspended by a string of length l1 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 l2, 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 l1/l2 is ?

Grade:11

2 Answers

Saurabh Koranglekar
askIITians Faculty 10335 Points
2 years ago
576-543_s.PNG
Vikas TU
14149 Points
2 years ago
For a bob at the top position in the circle, the forces are centripetal force, weight and tension.
         T = m v² / r - m g   --- (1)
String remains tight if  T > 0.  Hence,  v² >= r g     --- (2)
Let velocity given to the bob at its lowest position in the circle = u.
    1/2 m u² = 1/2  m v² + m g (2 r)
=> u² = v² + 4 g r
=> u >= √(5 rg)        --- (3)    using (2)
Now for the Top bob tied to string of radius L2:
Velocity given at it lowest position =  v2 >= √(5 L2 g)    -- (4)
The collision of two bobs of the same mass m is elastic, and the bob tied to string of length L2, was at rest before collision, we apply conservation of linear momentum and energy. We get
     v1 = v2 >= √(5 L2 g)    --- (5)
Velocity of bob tied to string of length L1 is same as that of the other bob after collision. Then after collision, the first bob's velocity is 0.
Let the velocity of first bob at its lowest position = v0.  Apply conservation of energy:
           1/2 m v0² = 1/2 m v1² + m g (2 L1)
            vo² = v1² + 4 g L1     
                  = 5 g L2 + 4 g L1   --- (6)
    As the first bob was given just enough velocity v0 at its lower position to make a full circle,
             v0² >= 5 L1 g     using (3)
                     = 5 g L2 + 4 g L1           from (6)
Solve these two to get   ans.
Hence,  L1/ L2 = 5
 

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

Other Related Questions on Engineering Entrance Exams

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More