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.

the functions have the same rate of change i think. i narrowed it down to this answer and "the rate of change of tour 1 is greater" but i got it wrong using that answer so im assuming its the answer above. (this is a question i also answered from the iready site if anyone needs a reference)

use distributive property on all like terms and change your signs if you have a double negative or if you have positive/negative