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

Answer:

a The kinetic energy is  $KE = 0.0543 J$

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

Explanation:

From the question we are told that

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

   The mass of the rod $m = 140 = \frac{140}{1000} = 0.140 \ kg$  

   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

                   $I = \frac{mL^2}{3}$  

Substituting values

               $I = \frac{(0.140) (1.4)^2}{3}$

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

Generally the  kinetic energy rod is mathematically represented as

             $KE = \frac{1}{2} Iw^2$

                    $KE = \frac{1}{2} (0.0915) (1.09)^2$

                           $KE = 0.0543 J$

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

                   $KE = PE = mgh$

                        $0.0543 = mgh$

                             $h = \frac{0.0543}{9.8 * 0.140}$

                                $h = 1.372 \ m$    

                 

Answer 2

Answer:

a) Kr = 0.0543 J

b) Δy = 0.0396 m

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