Question

A particle of mass 0.300 kg is attached to the 100 cm mark of a meter stick of mass 0.200 kg. The meter stick rotates on a horizontal, frictionless table with an angular speed of 4.00 rad/s.]

(a) Calculate the angular momentum of the system when the stick is pivoted about an axis perpendicular to the table through the 75.0 cm mark.

(b) What is the angular momentum when the stick is pivoted about an axis perpendicular to the table through the 0 cm mark?

Answer 1

formula for angular momentum is

L = I w

the total angualr momentum of the system when the stick is pivoted about an axis perpendicular to the table through the 75.0 cm mark is

L total = I_ toal w

=( I_ stick + I particle) w

=( ( mL^2/12 + M( L- L')^2)w

= ( 0.2( 1)^2/12 +0.3 ( 1-0.75)^2) 4 rad/s

=0.141 kg m^2/s

(b)

the total angualr momentum of the system  when the stick is pivoted about an axis perpendicular to the table through the 0 cm mark is

L total = I_ toal w

=( I_ stick + I particle) w

=( ( mL^2/3 + M( L- 0)^2)w

= ( 0.2( 1)^2/3 +0.3 ( 1-0)^2) 4 rad/s

=1.46 kg m^2/s