A particle of mass 0.450 kg is attached tothe 100 cm mark of a meter stick of ma

Question

A particle of mass 0.450 kg is attached tothe 100 cm mark of a meter stick of mass 0.125 kg. The meter stick rotates on a horizontal,frictionless table with an angular speed of 6.00 rad/s. (a) Calculate the angular momentum of thesystem when the stick is pivoted about an axis perpendicular to thetable through the 25.0 cm mark.
(b) What is the angular momentum when the stick is pivoted about anaxis perpendicular to the table through the 0 cm mark?
(a) Calculate the angular momentum of thesystem when the stick is pivoted about an axis perpendicular to thetable through the 25.0 cm mark.
(b) What is the angular momentum when the stick is pivoted about anaxis perpendicular to the table through the 0 cm mark?

Explanation / Answer

a. The moment of inertia in this case as, I=MR2+mr2=0.45*0.252+0.125*0.752 The angular momentum is P=I=I*6 b. In this case the moment of inertia is I=0.45*0.52+0.125*12 P=I Hence we get by it. P=I Hence we get by it.

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