Question

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?

Answer 1

Answer:

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 $I$ = 2.0 kg-m^2

angular speed ω = +5.0 rad/s

For second cylinder

rotational inertia $I$ = 1.0 kg-m^2

angular speed = -8.0 rad/s

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

where $I$ is the rotational inertia

ω is the angular speed

The rotational momenta of the cylinders are:

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

for second cylinder = $I$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 $I_{t}$ = 1.0 + 2.0 = 3 kg-m^2

Their final angular rotational momentum after coupling = $I_{t}$$w_{f}$

where $I_{t}$ is their total rotational inertia

$w_{f}$ = their final angular speed together

Final angular momentum = 3 x $w_{f}$ = 3$w_{f}$

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

this means that

2 =  3$w_{f}$

$w_{f}$ = 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 = $\frac{1}{2} Iw^{2}$

where $I$ is the rotational inertia

$w$ is the angular speed

The kinetic energy of the cylinders are:

for first cylinder = $\frac{1}{2} Iw^{2}$ = $\frac{1}{2}*2*5^{2}$ = 25 J

for second cylinder = $\frac{1}{2}*1*8^{2}$ = 32 J

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

The final kinetic energy of the cylinders after coupling = $\frac{1}{2}I_{t}w^{2} _{f}$

where

where $I_{t}$ is the total rotational inertia of the cylinders

$w_{f}$ is final total angular speed of the coupled cylinders

Final kinetic energy =  $\frac{1}{2}*3*0.67^{2}$ = 0.67 J

kinetic energy lost = 57 - 0.67 = 56.33 J

percentage = 56.33/57 x 100% = 98.8%