**Problem Description:**Two 4.50 kg balls are attached to the ends of a thin rod of length 72.0 cm and negligible mass. The rod is free to rotate in a vertical plane without friction about a horizontal axis through its center. With the rod initially horizontal, a 45.0 g wad of wet putty drops onto one of the balls, hitting it with a speed of 4.51 m/s and then sticking to it.**Questions:**(a) What is the angular speed of the system just after the putty wad hits?(b) What is the ratio of the kinetic energy of the system after the collision to that of the putty wad just before?(c) Through what angle (degrees) will the system rotate before it momentarily stops?**Diagram Explanation:**The diagram shows a horizontal rod with a ball on each end. A "Putty wad" is shown above one ball, with an arrow indicating it is falling towards the ball. The center of the rod is marked as the "Rotation axis."**Answers:**(a) Angular speed of the system: **0.0623 rad/s**(b) Ratio of kinetic energies: **0.00498**(c) Angle of rotation: **269 degrees**

Solution: Given Values, Mass of Ball (M)=4.50 Kg Mass of wet putty (m)=45.0 g = 0.045 Kg Length of the rod (L)=72 cm=0.72 ma) The angular speed of the system just after the putty wad hits is calculated as: Firstly, Let us calculate the Moment of inertia of the system after the collision is given as: Moment of inertia (I)=(2M+m)(L/2)2Moment of inertia (I)=(2×4.50 Kg+0.045 Kg)(0.72 m/2)2 Moment of inertia (I) = 1.1722 Kg.m2 By Applying the Law of conservation of angular momentum before and after the collision is: mv(L/2)= IωAngular velocity (ω)=mv(L/2)IAngular velocity (ω)=0.045 Kg×4.51 m/s×(0.72 m/2)1.1722 Kg.m2 =0.0623 rad/sThus,The angular speed of the system just after the putty wad hits is 0.0623 rad/s b) The Ratio of the Kinetic Energy system after the collision to that of the putty wad just before is: Here, Let us take, Initial Kinetic energy is (Ki)=12mv2Final Kinetic energy is (Kf)=12Iω2Then the Ratio is given as:KfKi=12Iω212mv2 =Iω2mv2 From the above a bit, we haveω=mvL/2ISo, substitute in the above equation: = ImvL/2I2mv2 =mL24I =mL24(2M+m)(L/2)2 since, I=(2M+m)(L/2)2 =m(2M+m)KfKi=0.045 Kg24.50 Kg+0.045 Kg=0.0049 Thus, The Ratio of the Kinetic Energy system after the collision to that of the putty wad just before is 0.0049 c) The Angle will the system rotate before it momentarily stops is given as: From the Law of conservation of energy, Ki+Ui=Kf+Uf12Iω2+mgL2=0+mgL2(1-cosθ)Therefore,cosθ=12Iω2mgL2 =121.1722 Kg.m20.062320.045 Kg×9.8 m/s20.72 m2cosθ = 0.17θ =cos-10.17=80.210Thus,The Angle will the system rotate before it momentarily stops is 80.210

Answered: Two 4.50 kg balls are attached to the…