A particle of mass 0300 kg is attached to the 100cm mark of

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 6.00 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. (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,

Solution

a) moment of Inertia of rod baout its center, I = m1*L^2/12

= 0.175*1^2/12

= 0.0145 kg.m^2

moment of inertia of paticle = m2*(L/2)^2

= 0.3*(1/2)^2

= 0.075 kg.m^2

angular momentum = I*w

= (0.0145 + 0.075)*6

= 0.537 kg.m^2/s

b) moment of Inertia of rod bout one end, I = m1*L^2/3

= 0.175*1^2/3

= 0.0583 kg.m^2

moment of inertia of paticle = m2*L^2

= 0.3*1^2

= 0.3 kg.m^2

angular momentum = I*w

= (0.0583 + 0.3)*6

= 2.14 kg.m^2/s