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

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.