Question

A particle of mass 0.300 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

Answer 1

mass1 0.300 kg

mass2 0.175 kg

L = 1m

(a)
I = 1/12 m1L^2 + m2 0.5^2
I = 1/12 0.175*1^2 + 0.300*0.5^2
I = 0.08958 kg m2

L = Iw
L = 0.08958*2
L = 0.1791 kg m2 /s
L = 0.1791 J-sec

(b)
I = 1/3 m1L2 + m2 R2
I = 1/3 0.175*1^2 + 0.300*1^2
I = 0.3583 kg m2

L = Iw
L = 0.3583*2
L = 0.7166 kg m2 /s
L = 0.766 J-sec