Consider a thin 4 m rod pivotedat one end whose mass is 7 kg. A uniformdensity 7

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

The acceleration of gravityg =9.8m/s2 .

The moment ofinertia of the rod

plus masssystem with respect to the pivot

Note: The lengthCmass in the figure repre-

sents the location of the center-of-mass ofthe

rod plus mass system.

Determine the position of the centerofmass

from the pivot point; i.e., find Cmass . Answer

inunits of m.

Explanation / Answer

       The length of the rod isL = 4 m         mass of rod andsphere is m = 7 kg   -----------------------------------------------------------------------------------        The center of mass of thesystem about pivot 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          

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