Question

012 (part 1 of 3) 10.0 points A uniform rod, supported and pivoted at its midpoint, but initially at rest, has a mass of 88 g and a length 8 cm. A piece of clay with mass 44 g and velocity 2.8 m/s hits the very top of the rod, gets stuck and causes the clay- rod system to spin about the pivot point o at the center of the rod in a horizontal plane. Viewed from above the scheme is 2.8 m/s 44 g HO 88 g (a) before (b) during (c) after With respect to the pivot point o, what is the magnitude of the initial angular mo- mentum Li of the clay-rod system? After the collisions the clay-rod system has an angular velocity w about the pivot.
I did part 1 and got 0.0049, which is correct. But I don't know how to do part 2 and 3. Please help and thanks!

Answer 1

12 Mputty = M = 44 = 01044 Kg. Mooca = 2m = 88 = 0.088 Kg l= 8 cm =) 0.08 m. V =2.8 m/s Conservation of angular momentum L-8XP = M J x & Text = de = 16 2; = 118 x 11 =mov. mve 13) = (0.00419) (2.8 m/s) (ovog m)/2 12; = 0.0049 Kg m/s.) If Irodt I day. = 72 2me + m ( 2 ) 2 F eme? ) 5 (0.oggkg) ('08 m)? If = 0.0002346 Kg . m² 16 0.0002346 kg.m!

14 16 = 1; If = mul Islef-mue Wp = b (2.8 m/s) 0.08m solve this answer en get an sods.