Rigid rods of negligible mass lying along the y axis connect three particles. Th
ID: 777544 • Letter: R
Question
Rigid rods of negligible mass lying along the y axis connect three particles. The system rotates about the x axis with an angular speed of 3.20 rad/s. (The center of mass of mass m1 = 7.40 kg is at location y1 = 5.55 m, the center of mass of m2 = 3.70 kg is at y2 = 3.70 m, and the center of mass of m3 = 5.55 kg is at y3 = 7.40 m.)
(a) Find the moment of inertia about the x axis. kg · m2
(b) Find the total rotational kinetic energy evaluated from 1 2 I2. J
(c) Find the tangential speed of each particle. v1 = m/s v2 = m/s v3 = m/s
(d) Find the total kinetic energy evaluated from 1 2 mivi2 . J
(e) Compare the answers for kinetic energy in parts (b) and (d). (If your answers from (b) and (d) differ by less than 7%, then answer that they are equal; otherwise, answer which one is greater.) 1 2 mivi2 > 1 2 I2 1 2 mivi2 = 1 2 I2 1 2 mivi2 < 1 2 I2
my.Explanation / Answer
(a)
the moment of inertia about the x axis I = m1*y1^2 + m2*y2^2 + m3*y3^3
the moment of inertia about the x axis I = (7.4*5.55^2) + (3.7*3.7^2) + (5.55*7.4^2) = 582.5 J
------------------------------------
(b)
the total rotational kinetic energy evaluated from 1/2*I*w^2 = (1/2)*582.5*3.2^2 = 2982.4 J
---------------------------------
(c)
v1 = y1*w = 5.55*3.2 = 17.76 m/s
v2 = y2*w = 3.7*3.2 = 11.84 J
v3 = y3*w = 7.4*3.2 = 23.68 J
================================
(d)
total kinetic energy = (1/2)*m1*v1^2 + (1/2)*m2*v2^2 + (1/2)*m3*v3^2
total kinetic energy = (1/2)*7.4*17.76^2 + (1/2)*3.7*11.84^2+(1/2)*5.55*23.68^2 = 15.78.34 J
total kinetic energy = 2982.45 J
=============================
(e)
(1/2)*mi*vi^2 = (1/2)*I*w^2
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.