A cylinder with rotational inertia I1=2.0kg·m2 rotates clockwise about a vertical axis through its center with angular speed ω1=5.0rad/s. A second cylinder with rotational inertia I2=1.0kg·m2 rotates counterclockwise about the same axis with angular speed ω2=8.0rad/s. If the cylinders couple so they have the same rotational axis what is the angular speed of the combination? What percentage of the original kinetic energy is lost to friction?

A) The angular speed of the combination of the two cylinders is; ω₃ = 0.67 rad/s

B) The percentage of the original kinetic energy lost to friction is;

percentage energy lost = 98.82%

We are given;

Rotational Inertia for first cylinder; I₁ = 2 kg.m²

Angular speed of first cylinder; ω₁ = 5 rad/s

Angular speed of second cylinder; ω₂ = 8 rad/s

Rotational Inertia for second cylinder; I₂ = 1 kg.m²

From conservation of angular momentum, we know that;

Initial angular Momentum() = Final angular Momentum()

Thus;

I₁ω₁ + I₂ω₂ = I₃ω₃

Where;

ω₃ is the angular speed when the two cylinders are combined

I₃ = I₁ + I₂

I₃ = 2 + 1

I₃ = 3 kg.m²

Since the second cylinder rotates in an anticlockwise direction, then its' angular speed will be negative. Thus;

(2 * 5) + (1 * -8) = 3ω₃

10 - 8 = 3ω₃

3ω₃ = 2

ω₃ = 2/3

ω₃ = 0.67 rad/s

B) Let us find initial kinetic energy;

E_i = ¹/₂I₁ω₁² + ¹/₂I₂ω₂²

E_i = ¹/₂((2 * 5²) + (1 * 8²)

E_i = 57 J

Final kinetic energy is;

E_f = ¹/₂I₃ω₃²

E_f = ¹/₂ * 3 * 0.67²

E_f = 0.67335 J

Energy lost = 57 - 0.67335 = 56.32665 J

percentage energy lost = (56.32665/57) * 100%

percentage energy lost = 98.82%

Read more at; link

a) 0.67 rad/sec in the clockwise direction.

b) 98.8% of the kinetic energy is lost.

Explanation:

Let us take clockwise angular speed as +ve

For first cylinder

rotational inertia = 2.0 kg-m^2

angular speed ω = +5.0 rad/s

For second cylinder

rotational inertia = 1.0 kg-m^2

angular speed = -8.0 rad/s

The rotational momentum of a rotating body is given as = ω

where is the rotational inertia

ω is the angular speed

The rotational momenta of the cylinders are:

for first cylinder = ω = 2.0 x 5.0 = 10 kg-m^2 rad/s

for second cylinder = ω = 1.0 x (-8.0) = -8 kg-m^2 rad/s

The total initial angular momentum of this system cylinders before they were coupled together = 10 + (-8) = 2 kg-m^2 rad/s

When they are coupled coupled together, their total rotational inertia = 1.0 + 2.0 = 3 kg-m^2

Their final angular rotational momentum after coupling =

where is their total rotational inertia

= their final angular speed together

Final angular momentum = 3 x = 3

According to the conservation of angular momentum, the initial rotational momentum must be equal to the final rotational momentum

this means that

2 =  3

= final total angular speed of the coupled cylinders = 2/3 = 0.67 rad/s

From the first statement, the direction is clockwise

b) Rotational kinetic energy =

where is the rotational inertia

is the angular speed

The kinetic energy of the cylinders are:

for first cylinder = = = 25 J

for second cylinder = = 32 J

Total initial energy of the system = 25 + 32 = 57 J

The final kinetic energy of the cylinders after coupling =

where

where is the total rotational inertia of the cylinders

is final total angular speed of the coupled cylinders

Final kinetic energy =   = 0.67 J

kinetic energy lost = 57 - 0.67 = 56.33 J

percentage = 56.33/57 x 100% = 98.8%

he . if i may ask what i your question exactly?

explanation:

e

explanation:

there is no change in the electric potential energy as it is in rest on both conditions.