Question

A thin rod of length 0.75 m and mass 0.42 kg is suspended

freely from one end. It is pulled to one side and then allowed to swing
like a pendulum, passing through its lowest position with angular
speed 4.0 rad/s. Neglecting friction and air resistance, find (a) the
rod's kinetic energy at its lowest position and (b) how far above that
position the center of mass rises.

Answer 1

a) K = 0.63 J, b) h = 0.153 m

Step-by-step explanation:

a) In this exercise we have a physical pendulum since the rod is a material object, the angular velocity is

w² =
`(m g d)/(I)`

where d is the distance from the pivot point to the center of mass and I is the moment of inertia.

The rod is a homogeneous body so its center of mass is at the geometric center of the rod.

d = L / 2

the moment of inertia of the rod is the moment of a rod supported at one end

I = ⅓ m L²

we substitute

w =
`\sqrt{(mgL)/(2) \ (1)/((1)/(3) mL^2) }`

w =
`\sqrt{(3)/(2) \ (g)/(L) }`

w =
`\sqrt{ (3)/(2) \ (9.8)/(0.75) }`

w = 4.427 rad / s

an oscillatory system is described by the expression

θ = θ₀ cos (wt + Φ)

the angular velocity is

w = dθ /dt

w = - θ₀ w sin (wt + Ф)

In this exercise, the kinetic energy is requested in the lowest position, in this position the energy is maximum. For this expression to be maximum, the sine function must be equal to ±1

In the exercise it is indicated that at the lowest point the angular velocity is

w = 4.0 rad / s

the kinetic energy is

K = ½ I w²

K = ½ (⅓ m L²) w²

K = 1/6 m L² w²

K = 1/6 0.42 0.75² 4.0²

K = 0.63 J

b) for this part let's use conservation of energy

starting point. Lowest point

Em₀ = K = ½ I w²

final point. Highest point

Em_f = U = m g h

energy is conserved

Em₀ = Em_f

½ I w² = m g h

½ (⅓ m L²) w² = m g h

h = 1/6 L² w² / g

h = 1/6 0.75² 4.0² / 9.8

h = 0.153 m