Consider a thin 4 m rod pivotedat one end whose mass is 7 kg. A uniformdensity 7
ID: 1756570 • Letter: C
Question
Consider a thin 4 m rod pivotedat one end
whose mass is 7 kg. A uniformdensity 7 kg
spherical object (whose radiusis 0.53 m) is
attached to the end of the rod.The moment
of inertia of the rod about an endis
Note: The length Cmass in the figurerepre-
sents the location of thecenter-of-mass of the
rod plus mass system.
Determine the position of thecenter ofmass
from the pivot point;i.e., findCmass . Answer
Note: The rod is initially at rest at 38? below
the horizontal.
What is the angular accelerationof the rod
immediately after it isreleased? Answer in
units of rad/s2.
Explanation / Answer
Given that the mass of rod is m = 7 kgLength ofrod is L = 4 m
massof spherical object is m = 7 kg Radius of object is r = 0.53 m --------------------------------------------------------------------------- The momnet of inertia of rod is Irod =(1/3)mL2 The momnet of inertia of sphere isIsphere =(2/5)mr2 The momnet of inertia of total system is I =Isphere + Irod + m*( L + r)2 = (2/5)mr2 + (1/3)mL2 + m*( L + r)2 = -------- kg.m2 The center of mass of the system aboutpivot is X = [ m*L/2 + m*(L+r) ] / 2m = [ L/4 + (L+r)/2 ] = ----------- m ( from pivot ) So the total weightof the system acts at center of mass so Apply torque about pivot thentorque is = 2m*g*X*mg sin I= 2m*g*X*mg sin =2m*g*X*mg sin / I =----------rad/s2 Radius of object is r = 0.53 m --------------------------------------------------------------------------- The momnet of inertia of rod is Irod =(1/3)mL2 The momnet of inertia of sphere isIsphere =(2/5)mr2 The momnet of inertia of total system is I =Isphere + Irod + m*( L + r)2 = (2/5)mr2 + (1/3)mL2 + m*( L + r)2 = -------- kg.m2 = (2/5)mr2 + (1/3)mL2 + m*( L + r)2 = -------- kg.m2 The center of mass of the system aboutpivot is X = [ m*L/2 + m*(L+r) ] / 2m = [ L/4 + (L+r)/2 ] = ----------- m ( from pivot ) So the total weightof the system acts at center of mass so Apply torque about pivot thentorque is = 2m*g*X*mg sin I= 2m*g*X*mg sin =2m*g*X*mg sin / I =----------rad/s2 = [ L/4 + (L+r)/2 ] = ----------- m ( from pivot ) So the total weightof the system acts at center of mass so Apply torque about pivot thentorque is = 2m*g*X*mg sin I= 2m*g*X*mg sin =2m*g*X*mg sin / I =----------rad/s2