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. If this body is released from rest in a horizontal position, what is the angular speed of the meter stick as it swings through its lowest position

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= m½Iω²eqn(*)

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

you get 2.3=1.21*\frac{a}2 or 4.6=1.21a or a=3.802 (b)

explanation: