Awooden rod of negligible mass and length 84.0 cm is pivoted about a horizontal axis through its center. a white rat with mass 0.600 kg clings to one end of the stick, and a mouse with mass 0.200 kg clings to the other end. the system is released from rest with the rod horizontal. if the animals can manage to hold on, what are their speeds as the rod swings through a vertical position?

v = 2.029 m/s

Explanation:

Given

L = 84.0 cm   ⇒  R = 0.5*L = 0.5*84 cm = 42 cm = 0.42 m

m₁ = 0.600 kg

m₂ = 0.200 kg

g = 9.8 m/s²

u₁ = u₂ = 0 m/s

v₁ = ?

v₂ = ?

Due to gravity, the bar oscillates and becomes vertical. The mass that occupies the lower position is the one with the highest torque. The one that reduces the potential energy (the system tends to the position of minimum energy). This is achieved if the mass that goes down is 0.6kg (that goes down 42cm) and the one that goes up is 0.2kg (goes up 42cm).

In this system mechanical energy is conserved, so we can match its value in the horizontal position with the one in the vertical.

then

Ei = Ki + Ui = 0.5*(m₁+m₂)*(0)² + (m₁+m₂)*9.8*(0) = 0 J

Ef = Kf + Uf

⇒ Kf = 0.5*(m₁+m₂)*v² = 0.5*(0.6+0.2)*v² = 0.4*v²

⇒ Uf = m₁*g*h₁ + m₂*g*h₂ = 0.6*9.8*(-0.42) + 0.2*9.8*0.42 = - 1.6464

⇒ Ef = Kf + Uf = 0.4*v² - 1.6464

Since

0 = 0.4*v² - 1.6464  ⇒  v = 2.029 m/s

v is the same value due to the wooden rod is pivoted about a horizontal axis through its center and the masses are on opposite ends.

v₁ = v₂ = v    ⇒  ω₁*R₁ = ω₂*R₂  ⇒ ω₁*R = ω₂*R  ⇒ ω₁ = ω₂ = ω

⇒ v = ω*R

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.