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2 rods of mass m and length L are connected by a massless pin at point O such that they are free to rotate about O as well as point O can move in a horizontal plane , as shown. Two balls of mass m moving with a velocity u strike the rods inelastically . What is the angular velocity of the rods just after the collission. ____________ O ____________ O m O m

2 rods of mass m and length L are connected by a massless pin at point O such that they are  free to rotate about O as well as point O can move in a horizontal plane , as shown. Two balls of mass m moving with a velocity u strike the rods inelastically . What is the angular velocity of the rods just after the collission.


____________ O ____________


O m                                 O m  

Grade:12

2 Answers

mohit yadav
54 Points
7 years ago
54
SOURAV MISHRA
37 Points
7 years ago

THE SITUATION IS VERY SYMMETRIC.

THE POINT IN THIS QUESTION IS THE POINT O IS JUST A CONNECTION BETWEEN THE TWO RODS. IT IS NOT A FIXED POINT OR A HINGE. SO DO NOT ASSUME PURE ROTATION ABOUT O. IN FACT THE POINT O WILL MOVE BACKWARDS JUST AFTER THE COLLISION.

THE QUESTION CAN BE SOLVED THROUGH ANGULAR MOMENTUM CONSERVATION PRINCIPLE. WE MAY TAKE ONLY ONE OF THE RODS AS HE SITUATION WILL BE SAME FOR THE OTHER ROD AS WELL DUE TO THE SYMMETRY IN THE PROBLEM.

THEREFORE AFTER COLLISION A ROD-PARTICLE SYSTEM WILL ROTATE ABOUT THEIR COMMON CENTER OF MASS WHICH WILL BE AT A DISTANCE OF L/4 FOM THE END OF THE ROD.

THE SITUATION JUST AFTER THE COLLISION WILL BE PURE ROTAION ABOUT THE TWO CENTERS OF MASS ALONG WITH THE TRANSLATION OF THE ENTIRE SYSTEM. THE POINT O WILL MOVE BACKWARDS FROM ITS INITIAL POSITION.

SO WE CAN WRITE 

muL/4 = (5mL2/24)w

WHERE MOMENT OF INERTIA OF A ROD-PARTICLE SYSTEM ABOUT THEIR CENTER OF MASS = (5mL2/24)

AND w = ANGULAR VELOCITY OF A ROD-PARTICLE SYSTEM.

THIS GIVES w = 6u/5L.

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