A stick is resting on a concrete step with 1/6 of its length hanging over the ed
ID: 1355314 • Letter: A
Question
A stick is resting on a concrete step with 1/6 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 46.5° from the horizontal. If the mass of each bug is 3.60 times the mass of the stick and the stick is 17.5 cm long, what is the magnitude of the angular acceleration of the stick at the instant shown?
Explanation / Answer
momentum of the inertia of 1st bug=mr^2
=m*(L/6)^2
=3.60*(1/36)*L^2
=0.1 mL^2
momentum of the inertia of 2nd bug=mr^2
=3.60*(5L/6)^2
=3.60*(25/36)*L^2
=2.5 mL^2
moment of inertia of stick = (1/12) m L^2 + md^2
=(1/12)m*L^2+m*(1.5L/6)^2
=(1/12+2.25/36)m*L^2
=0.145 mL^2
total moment of inertia =0.1 mL^2+2.5 mL^2+0.145 mL^2
=2.7458 mL^2
torque from first bug=-mgr cos46.5
=0.1720*mg*L
torque from second bug = mgr cos46.5
=3.60*mg*(5L/6) cos46.5
=2.065 *mg*L
torque from weight of stick=mgr cos46.5
=(1.5L/6)*mg*cos46.5
=0.172*mg*L
total torque = (0.1720+2.065-0.172) mg *L
=2.065*mg*L
angular acceleration =total torque/total momentum of inertia
=2.065*mg*L/2.7458 mL^2
=0.7520 g/L
=0.7520*9.8/0.175
=42.115 rad/s^2
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