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.
       kg·m2/s

(b) What is the angular momentum when the stick is pivoted about anaxis perpendicular to the table through the 0 cm mark?
    kg·m2/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.
       kg·m2/s

(b) What is the angular momentum when the stick is pivoted about anaxis perpendicular to the table through the 0 cm mark?
    kg·m2/s

Explanation / Answer

(a) Moment of inertia about the vertical axis thru the 25 cmmark,                  I = m1r12 +m2.r22                    = 0.45 * 0.752 + 0.125 * 0.052                    = 0.25343 kg.m2       Therefore angularmomentum,                 L = I.
                   = 0.25343 * 6.0
   = 1.51875 kgm2 / s                    (b) Moment of inertia about the vertical axis throughthe 0 cm mark,                  I = m1r12 +m2.r22                    = 0.45 * 1.02 + 0.125 * 0.502                    = 0.48125 kg.m2       Therefore angularmomentum,                 L = I. = 0.48125 * 6.0 = 2.8875 kgm2 / s

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