Answer:
a The kinetic energy is [tex]KE = 0.0543 J[/tex]
b The height of the center of mass above that position is [tex]h = 1.372 \ m[/tex]
Explanation:
From the question we are told that
The length of the rod is [tex]L = 1.4m[/tex]
The mass of the rod [tex]m = 140 = \frac{140}{1000} = 0.140 \ kg[/tex]
The angular speed at the lowest point is [tex]w = 1.09 \ rad/s[/tex]
Generally moment of inertia of the rod about an axis that passes through its one end is
[tex]I = \frac{mL^2}{3}[/tex]
Substituting values
[tex]I = \frac{(0.140) (1.4)^2}{3}[/tex]
[tex]I = 0.0915 \ kg \cdot m^2[/tex]
Generally the kinetic energy rod is mathematically represented as
[tex]KE = \frac{1}{2} Iw^2[/tex]
[tex]KE = \frac{1}{2} (0.0915) (1.09)^2[/tex]
[tex]KE = 0.0543 J[/tex]
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
[tex]KE = PE = mgh[/tex]
[tex]0.0543 = mgh[/tex]
[tex]h = \frac{0.0543}{9.8 * 0.140}[/tex]
[tex]h = 1.372 \ m[/tex]
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
