Ameterstick is initially standing vertically on the floor. if the meterstick falls over, with what angular velocity will it hit the floor? assume that the end in contact with the floor experiences no friction and slips freely.

L = length of the meter stick = 1 m

h = height of center of mass of stick from bottom end on the floor = L/2 = 1/2 = 0.5 m

m = mass of the meter stick

I = moment of inertia of the meter stick about the bottom end

w = angular velocity as it hits the floor

moment of inertia of the meter stick about the bottom end is given as

I = m L²/3

using conservation of energy

rotational kinetic energy of meter stick as it hits the floor = potential energy when it is vertical

(0.5) I w² = m g h

(0.5) (m L²/3) w² = m g h

( L²) w² =  6g h

( 1²) w² =  6 (9.8) (0.5)

w = 5.4 rad/s

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg