Question

Course Contents ... Exam3 Due in: od 00:29:46 Timer Notes Evaluate Feedback Print Info A thin 1-m-long uniform rod with a total mass of 750 g is suspended vertically at the upper end. The moment of inertia of the rod in respect to the center of mass is ml-/12 (where L is the total length of the rod). A 9-9 bullet is shot to the lower end of the rod and embeds there. The bullet speed before impact is 580 m/s. Calculate the amount of energy transferred to the heat during the collision. Submit Answer Tries 0/99 Communication Blocked Send Feedback

Answer 1

let
M = 750 g = 0.75 kg
L = 1 m
m = 9 g = 0.009 kg
v = 580 m/s

let w is the angular speed of the rod after the collision.

using parallel axis theorem, moment of inertia of the rod about one axis, Ip = Icm + M*(L/2)^2

= M*L^2/12 + M*L^2/4

= M*L^2/3

Apply conservation of angular momentum

Lf =Li

(M*L^2/3 + m*L^2)*w = m*v*L

w = m*v*L/(M*L^2/3 + m*L^2)

= 0.009*580*1/(0.75*1^2/3 + 0.009*1^2)

= 20.15 rad/s

KEi = (1/2)*m*v^2

= (1/2)*0.009*580^2

= 1514 J

KEf = (1/2)*I*w^2

= (1/2)*(M*L^2/3 + m*L^2)*w^2

= (1/2)*(0.75*1^2/3 + 0.009*1^2)*20.15^2

= 53 J

The amount of energy transfered to the heat during the collision = KEi - KEf

= 1514 - 53

= 1461 J <<<<<<<<<<<<<-----------------Answer