A particle of mass 0.500 kg is attached to the 100-cm mark of a meterstick of ma

Question

A particle of mass 0.500 kg is attached to the 100-cm mark of a meterstick of mass 0.175 kg. The meterstick rotates on the surface of a frictionless, horizontal table with an angular speed of 2.00 rad/s.

(a) Calculate the angular momentum of the system when the stick is pivoted about an axis perpendicular to the table through the 50.0-cm mark.
kg · m2/s

(b) Calculate the angular momentum of the system when the stick is pivoted about an axis perpendicular to the table through the 0-cm mark.
kg · m2/s

Explanation / Answer

a)moment of inertia of the meter stick about an axis passing through its center=(1/12)*mass*length^2

=(1/12)*0.175*1^2=0.0146 kg.m^2

moment of inertia of the particle=mass * distance from axis^2

=0.5*0.5^2=0.125 kg.m^2

net moment of inertia=0.1396 kg.m^2

then angular momentum=moment of inertia*angular speed
=0.1396*2=0.2792 kg.m^2/s

b)moment of inertia of the meter stick about an axis passing through its edge=(1/3)*mass*length^2

=(1/3)*0.175*1^2=0.0583 kg.m^2

moment of inertia of the particle=mass * distance from axis^2

=0.5*1^2=0.5 kg.m^2

net moment of inertia=0.5583 kg.m^2

then angular momentum=moment of inertia*angular speed
=0.5583*2=1.1166 kg.m^2/s

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