Question

When the disks collide and stick together, their temperature rises. Calculate the increase in internal energy of the disks, assuming that the process is so fast that there is insufficient time for there to be much transfer of energy to the ice due to a temperature difference. (Also ignore the small amount of energy radiated away as sound produced in the collisions between the disks.)

Answer 1

Answer:

   ΔT = $\frac{\Delta K}{(m_1+m_2) c_e }$    

Explanation:

This is an interesting problem, no data is given, so the result is a general expression.

Suppose that the disks are initially rotating with angular velocity w₁ and w₂, as well as that they have radii r₁ and r₂ and masses m₁ and m₂

we start the problem finding odl final angular velocity of the discs together, for this we define a system formed by the two discs, in this case the torques during the collision are internal and the angular momentum is conserved

       

initial instant. Just before the crash

           L₀ = L₁ + L₂

with

           L₁ =  I₁ w₁

the moment of inertia of a disc with an axis passing through its center is

           I₁ = ½ m₁ r₁²

we substitute

           I₀ = ½ m₁ r₁² w₁ + ½ m₂ r₂² w₂

final instant. Right after the crash

         L_f = I w

     

in angular momentum it is a scalar quantity, so it is additive

         I = I₁ + I₂

angular momentum is conserved

         L₀ = L_f

         I₁ w₁ + I₂ w₂ = I w

         w =  $\frac{ I_1 w_1 + I_2 w_2 }{I}$            (1)

We already have the angular velocities of the system, let's find the kinetic energy of it

initial

          K₀ = K₁ + K₂ = ½ I₁ w₁² + ½ I₂ w₂²

final

          K_f = K = ½ I w²

the variation of the kinetic energy is the loss in the increase of the temperature of the system, they indicate us that we neglect the other possible losses

         ΔK = K_f -K₀

         ΔK = ½ I w² - (½ I₁ w₁² + ½ I₂ w₂²)           (2)

In this chaos we know all the values ​​for which the numerical value of ΔK can be calculated, the symbolic substitution gives expressions with complicated

Now if all this variation of energy turns into heat

         Q = ΔK

         m_{total} c_e ΔT = ΔK

where the specific heat of the bear discs must be known, suppose they are of the same material

         ΔT = $\frac{\Delta K}{(m_1+m_2) c_e }$               (3)

to make a special case, we suppose some data

the discs have the same mass and radius, disc 2 is initially at rest and the discs are made of bronze that has c_e = 380 J / kg ºC

we look for the angular velocity

          I₁ = I₂ = I₀

          I = 2 I₀

         

we substitute in 1

           w = $\frac{I_o w_1 + I_o 0 }{2I_o}$ I₀ w₁ + I₀ 0 / 2Io

           w = w₁ /2

we look for the variation of the kinetic energy with 2

       

             ΔK = ½ (2I₀) (w₁ /2)² - (½ I₀ w₁² + ½ I₀ 0)

             ΔK = ¼ I₀ w₁² -½ I₀ w₁²

             ΔK = - ¼ I₀ w₁²

the negative sign indicates that the kinetic energy decreases

We look for the change in Temperature with the expression 3

              ΔT = $\frac{ \Delta K}{(m_1 +m_2) c_e}$ΔK / (m1 + m2) ce

              ΔT = $\frac{ \frac{1}{4} I_o w_1^2 }{ 2m c_e}$

              ΔT =  $\frac{1}{8} \frac{ (\frac{1}{2} m r_1^2 ) w_1^2 }{ m c_e}$

              ΔT = $\frac{1}{16} r_1^2 w_1^2 / c_e$

in this expression all the terms are contained

Answer 2

The increase in internal energy of the disks will be $\rm \triangle E= mc\frac{\triangle k }{(m_1+m_2)c_e}$.

What is internal energy?

The energy contained within a thermodynamic system is known as its internal energy. It's the amount of energy required to build or prepare a system in any given internal state.

The given data in the problem is;

$\rm \omega_1$  is the angular velocity of  disk 1

$\rm \omega_2$ is the  angular velocity of disk 2

r₁ is the radius of  disk 1

r₂ is the radius of  disk 2

m₁ is the mass of disk 1

m₂ is the mass of disk 2

Momentum before the collision;

$\rm L_1 = I_1 \omega_1$

The moment of inertia of disc 1

$\rm i_1 = \frac{1}{2} m_1r_1^2$

The momentum gets conserved;

$\rm L_0 = L_f \\\\ I_1 \omega_1 + I_2\omega_2 = I \omega \\\\ \rm \omega= \frac{I_1 \omega_1 + I_2\omega_2}{I}$

The change in the kinetic energy is;

$\traingle KE= K_f - K_0 \\\\ \traingle KE= \frac{1}{2} I \omega^2-(\frac{1}{2} I_1\omega_1^2 + (\frac{1}{2} I_2\omega_2^2 )$

The change in the energy gets converted into heat;

$\rm Q= \triangle k \\\\\ m_{total } c_e dt = \triangle k$

The change in the temperature is

$\triangle T= \frac{\triangle k }{(m_1+m_2)c_e}$

The internal energy change is found by;

$\rm \triangle E = mc_v dt$

$\rm \triangle E= mc\frac{\triangle k }{(m_1+m_2)c_e}$

Hence the increase in internal energy of the disks will be $\rm \triangle E= mc\frac{\triangle k }{(m_1+m_2)c_e}$.

To learn more about the internal energy refer to the link;

https://brainly.com/question/11278589