Question

Problem 7.62 15 of 15 Constants A mallet consists of a uniform cylindrical head of mass 2.75 kg and a diameter 0.0800 m mounted on a uniform cylindrical handle of mass 0.550 kg and length 0.240 m, as shown in (Figure 1). Part A If this mallet is tossed, spinning into the air, how far above the bottom of the handle is the point that will follow a parabolic trajectory? Express your answer to three significant figures and include the appropriate units. Figure 1 of 1 ? d. Value Units Submit Recuestas 24.0 cm 8.00 cm < Return to Assignment Provide Feedback

Answer 1

[Answer] : Given information, For cylindrical heud muss mi 2.75 kg diameter d = 0.08 m For cylindrical handle muss m2 = 0.55 kg length l = 0.240 m The center of mass will follow a parabolic trajectory. parto purtat Co-ordinate of Centre of muss m, 4 + m2 2. Xcm m+m2 where, of mass of part I Chead) 2= centre o et dhe H! 2 0.24 + 0.08 2 0,98 = 0.28 m * = Centre of mass of purt 2 Chundle) 11 ut 2 0.240 2 1 0.12 m

Now, we have. m = 2.75 kg x=0.28 m m2 = = 0.55 kg ų= 0.12 m substitute the values. : Zem (2.75) (0.28) + (0.55) (0.12) 2.75 +0.55 i. Xem 0.836 3.3 1, | An = 0.253 m The point 0.253 m above the bottom of the handle will follow parabolic trujectory.