Question

A massless rod of length L has a small mass m fastened at its center and another mass m fastened at one end. On the opposite end from the mass m, the rod is hinged with a frictionless hinge. The rod is released from rest from an initial horizontal position; then it swings down. What is the angular velocity as the rod swings through its lowest (vertical) position?

Answer 1

Answer:

onservation of energy

U top = K bottom

(m + m)*g*L = 1/2*I*?^2 where I = m*(L/2)^2 + m*L^2 = 1.25*m*L^2

So 2m*g*L = 1/2*1.25*m*L^2*?^2

So ? = sqrt(3*g*/(1.25*L) ) = sqrt(12g/5L)

Answer 2

The angular velocity as the rod swings through its lowest (vertical) position is; √(8g/3L)

What is the angular velocity?

Let the length from the left end of the rod to the center of gravity be x.

Thus, from equilibrium we have;

3m * x = 2m(L/2) + mL

divide through by m to get;

3x = 2L

x = ²/₃L

Moment of inertia is;

I = 2m(L/2)² + m(L)²

I = mL²/2 + mL²

I = ³/₂mL²

Potential energy is;

P.E = 3mg(²/₃L)

P.E = 2mgL

Formula for kinetic energy is;

K.E = ¹/₂Iω²

K.E = ¹/₂ * ³/₂mL² * ω²

From conservation of energy;

P.E = K.E

Thus;

2mgL = ¹/₂ * ³/₂mL² * ω²

ω = √(8g/3L)

Read more about angular velocity at; https://brainly.com/question/9408577