Question

A thin rod of length 1.4 m and mass 140 g 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 1.09 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) Kr = 0.0543 J

b) Δy = 0.0396 m

Step-by-step explanation:

a) Given

L = 1.4 m

m = 140 g = 0.14 kg

ω = 1.09 rad/s

Kr = ?

We have to get the rotational inertia as follows

I = Icm + m*d²

⇒ I = (m*L²/12) + (m*(L/2)²)

⇒ I = (0.14 kg*(1.4 m)²/12) + (0.14 kg*(1.4 m/2)²)

⇒ I = 0.09146 kg*m²

Then, we apply the formula

Kr = 0.5*I*ω²

⇒ Kr = 0.5*(0.09146 kg*m²)*(1.09 rad/s)²

⇒ Kr = 0.0543 J

b) We apply the following principle

Ei = Ef

Where the initial point is the lowest position and the final point is at the maximum height that its center of mass can achieve, then we have

Ki + Ui = Kf + Uf

we know that ωf = 0 ⇒ Kf = 0

⇒ Ki + Ui = Uf

⇒ Uf - Ui = Ki

⇒ m*g*yf - m*g*yi = Ki

⇒ m*g*(yf - yi) = Ki

⇒ m*g*Δy = Ki

⇒ Δy = Ki/(m*g)

where

Ki = Kr = 0.0543 J

g = 9.81 m/s²

⇒ Δy = (0.0543 J)/(0.14 kg*9.81 m/s²)

⇒ Δy = 0.0396 m

Answer 2

a The kinetic energy is
`KE = 0.0543 J`

b The height of the center of mass above that position is
`h = 1.372 \ m`

Step-by-step explanation:

From the question we are told that

The length of the rod is
`L = 1.4m`

The mass of the rod
`m = 140 = (140)/(1000) = 0.140 \ kg`

The angular speed at the lowest point is
`w = 1.09 \ rad/s`

Generally moment of inertia of the rod about an axis that passes through its one end is

`I = (mL^2)/(3)`

Substituting values

`I = ((0.140) (1.4)^2)/(3)`

`I = 0.0915 \ kg \cdot m^2`

Generally the kinetic energy rod is mathematically represented as

`KE = (1)/(2) Iw^2`

`KE = (1)/(2) (0.0915) (1.09)^2`

`KE = 0.0543 J`

From the law of conservation of energy

The kinetic energy of the rod during motion = The potential energy of the rod at the highest point

Therefore

`KE = PE = mgh`

`0.0543 = mgh`

`h = (0.0543)/(9.8 * 0.140)`

`h = 1.372 \ m`