Question

A uniform rod with a mass m and length L has one end attached to a pivot. The rod swings around on a frictionless horizontal table with angular speed wo. A ball with mass m (so same mass as the rod) is placed on the table a distance d from the pivot. The ball is made of clay so when the rod strikes it, the ball sticks to the rod (i.e., inelastic collision). If the final (post-collision) angular velocity of the rod + clay- ball combo is one half that of the rod's initial angular velocity (wo), then what is the ratio of the lengths d/L? Note that the figure to the right is not drawn to scale. [The moment of inertial of a rod about one end is 1 3 mL 2 where L is the length of the rod.] Please remember to explain the process of getting the answer and show work. Wo m L d Pivot

Answer 1

It is given ง m of inertia of rool The mass of the ball im The moment about one end Ir = 1/2 me? Now there is no tarque acting hence by conservation principle of angular momentum of then Pivot Z=0 Total Total angular momentum before inelastic collision 5)=( angular ular momentum of rod andelay ball Aftercollision I, 10. = Iz W2 1 o inertia of rod I, - - Moment I = 1 m 2² Wir Angular awo velocity befort collision ودي - ركرا I= Angular momentum of rodandballccombo)about pivot point Md² I2 = 1 / 3 M²+ W2 = Angular velocity after inelastic collision wys هلیا 2 Now putting these all value in egr 0 C 1922. Wowie ( M22+ (5 m2 + matt). wat CamScanner

~ 하자 2 2 2. 을 me wo = Mg+)'를 좋고 d² 를 떻 d² 은 - 3 하고 11 12 3 13 () - () - 1 CS Scanned with CamScanner