Chapter 11, Problem 065 Two 0.510 kg balls are attached to the ends of a thin ro
ID: 1554325 • Letter: C
Question
Chapter 11, Problem 065
Two 0.510 kg balls are attached to the ends of a thin rod of length 53.0 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 (the figure), a 97.0 g wad of wet putty drops onto one of the balls, hitting it with a speed of 3.62 m/s and then sticking to it. (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 (deg) will the system rotate before it momentarily stops?
Chapter 11, Problem 065
Two 0.510 kg balls are attached to the ends of a thin rod of length 53.0 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 (the figure), a 97.0 g wad of wet putty drops onto one of the balls, hitting it with a speed of 3.62 m/s and then sticking to it. (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 (deg) will the system rotate before it momentarily stops?
Your answer is partially correct. Try again.Explanation / Answer
a)
m = mass of each ball at the ends of rod = 0.510 kg
L = length of rod = 53 cm = 0.53 m
r = distance of each ball from center of rod = L/2 = 0.53/2
initial Moment of inertia of rod + balls is given as
Ii = 2 mr2 = 2 (0.510) (0.53/2)2 = 0.072 kgm2
wi = initial angular velocity of ball+rod = 0 rad/s
M = mass of putty = 97 g = 0.097 kg
v = speed of putty , before collision = 3.62 m/s
If = final moment of inertia of system = 2 mr2 + Mr2 = 0.072 + (0.097) (0.53/2)2 = 0.0788 kgm2
wf = angular velocity of system just after colllision
using conservation of angular momentum
Ii wi + Mvr = If wf
(0.072) (0) + (0.097) (3.62) (0.53/2) = (0.0788) wf
wf = 1.18 rad/s
number = 1.18
Unit = rad/s
b)
KEi = kinetic energy of putty = (0.5) M v2 = (0.5) (0.097) (3.62)2 = 0.636 J
KEf = final kinetic energy of system = (0.5) If wf2 = (0.5) (0.0788) (1.18)2 = 0.055 J
Ratio = KEf /KEi =0.055/0.636 = 0.0865
Number = 0.0865
c)
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.