Answer:
5.35 rad/s
Explanation:
From the question, we are toldthat an Identical particles are placed at the 50-cm and 80-cm marks on a meter stick of negligible mass. This rigid body is then mounted so as to rotate freely about a pivot at the 0-cm mark on the meter stick.
Solving this question, the potential energy of the particles must equal to the Kinectic energy i.e
P.E=K.E
Mgh= m½Iω²-------------eqn(*)
Where M= mass of the particles
g= acceleration due to gravity= 9.81m/s^2
ω= angular speed =?
h= height of the particles in the stick on the metre stick= ( 50cm + 80cm)= (0.5m + 0.8m)= 130cm=1.3m
If we substitute the values into eqn(*) we have
m×9.81× (1.3m)= 1/2× m×[ (0.5m)² + [(0.8m)²]× ω²
m(12.74m²/s²)= 1/2× m× (0.25+0.64)× ω
m(12.74m²/s²)= 1/2× m× 0.89× ω²
We can cancel out "m"
12.74= 1/2×0.89 × ω²
12.74×2= 0.89ω²
25.48= 0.89ω²
ω²= 28.629
ω= √28.629
ω=5.35 rad/s
Hence, the angular speed of the meter stick as it swings through its lowest position is 5.35 rad/s
