Question

A stick is resting on a concrete step with 1/5 of its length hanging over the edge. A single ladybug lands on the end of the stick hanging over the edge, and the stick begins to tip. A moment later, a second, identical ladybug lands on the other end of the stick, which results in the stick coming momentarily to rest 62.1° from the horizontal. If the mass of each bug is 3.09 times the mass of the stick and the stick is 16.3 cm long, what is the magnitude of the angular acceleration of the stick?

Answer 1

angular acceleration=torque/moment of inertia

distance of first lady bug from the edge when the stick is momentarily at rest=(1/5)*length*cos(angle)

=0.2*0.163*cos(62.1)=0.015255 m

distance of center of mass of the stick from the edge=(0.5*length-0.2*length)*cos(angle)=0.022882 m

distance of the second lady bug from the edge=(4/5)*length*cos(angle)=0.061018 m

let mass of the stick be m kg and mass of the bugs are 3.09*m

then total torque=3.09*m*0.061018+m*0.022882-3.09*m*0.015255=0.16429*m N.m

moment of inertia about the edge:

moment of inertia of the stick about its center of mass=(1/12)*mass*length^2=(1/12)*m*0.163^2

using parallel axis theorem, moment of inertia about the edge=moment of inertia about the center of mass+mass*(distance of edge from center of mass)^2

=(1/12)*m*0.163^2+m*((0.5-0.2)*0.163)^2

=0.0046053*m kg.m^2

moment of inertia of the first ladybug=mass*distance^2=3.09*m*(0.2*0.163)^2=0.0032839*m kg.m^2

moment of inertia of the second ladybug=mass*distance^2=3.09*m*(0.8*0.163)^2=0.052543*m kg.m^2

so total moment of inertia=0.0046053*m+0.0032839*m +0.052543*m =0.060432*m kg.m^2

then angular acceleration=torque/moment of inertia=0.16429*m/(0.060432*m )=2.7186 rad/s^2