Problem 8.8. A rod of length L and mass m is held in a vertical position on a rough horizontal surface (Fig. P.8.8). The rod is then slightly displaced from the vertical unstable equilibrium position (Fig. P.8.8a) and let go. The rod will rotate clockwise around point A without slip (Fig. P.8.8b), but eventually the slipping must take place before the horizontal position is reached (Fig. P.8.8c). Determine the angular velocity of the rod as a function of its angle of inclination with the vertical axis before the slip takes place. At which value of theta does the slip takes place?

ω = √((3g/L)*(1 - Cos θ))

Step-by-step explanation:

We need to apply the Principle of Conservation of mechanical energy as follows

Ei = Ef    ⇒ Ki + Ui = Kf + Uf

In the vertical position

ωi = 0 ⇒ Ki = 0

yi = L/2 ⇒ Ui = m*g*L/2

We can get the rotational inertia I using the formula

I = m*L²/3

then

Kf = I*ω²/2 = (m*L²/3)*ω²/2 = m*L²*ω²/6

Now, we obtain the potential energy Uf as follows

Uf = m*g*y

where

y = (L/2)*Cos θ

⇒ Uf = m*g*(L/2)*Cos θ

Now, we have

Ui = Kf + Uf

⇒ m*g*L/2 = (m*L²*ω²/6) + (m*g*(L/2)*Cos θ)

⇒ ω² = (3g/L)*(1 - Cos θ)

⇒ ω = √((3g/L)*(1 - Cos θ))

We need to know the coefficient of static friction in order to get the value of theta where slip takes place.

∠cbe and ∠deb are same-side interior angles.

i took the test