A uniform rod of mass m and length l is hinged at upper end rod is free to rotatein vertical plane a ball of mass m moving horizontally with velocity v sticks to it
Answers
KE of the ball converts into rotational energy of the whole (ball + rod) system
KE of ball = 1/2mv²
rot energy of system = 1/2Iw² = 1/2 (ml²/3 + ml²)w²
here I is used of rod(ml²/3) + ball attached to its end (ml²)
equating the energies
1/2mv² = 1/2Iw²
1/2mv² = 1/2 (ml²/3 + ml²)w²
solving it we get
w² = 3v²/4l²
Answer: 3v² = 4l²w²
Explanation: To answer this question, we need to understand the relationship between Kinetic Energy and Rotational Energy.
As given in the question, the ball is moving horizontal with a velocity v so we know that the kinetic energy (KE) of the ball converted into the rotational energy (RE) of the system of Ball & Rod.
We know that KE = ½ mv² and that RE = ½ Iw² where I is the moment of Inertia of the system and w is the angular velocity.
RE of the Rod = ½ ml² × ⅓w² as the moment of Intertia of a Rod along its one end is ⅓ml².
RE of the Ball = ½ ml² w² as the moment of Inertia of a ball along its center is ml².
RE of the system = ½ w²(⅓ml² + ml²)
= ½ w² (4ml²/3)
= ⅔ ml² w²
Since we know that the Kinetic Energy converted into Rotational Energy,
KE of Ball = RE of the System
½mv² = ⅔ ml² w²
3v² = 4l² w²