Respuesta :
Answer:
The loss of initial Kinetic energy = 37.88 %
Explanation:
Given:
Rotational inertia of the turntable = [tex]I_t[/tex]
Rotational inertia ([tex]I_r[/tex]) of the record = [tex]0.61\times I_t[/tex]
According to the question:
Frictional forces act to bring the record and turntable to a common angular speed.
So,angular momentum will be conserved as it is an inelastic collision.
Considering the initial and final angular velocity of the turn table as [tex]\omega _i\ ,\ \omega_f[/tex] respectively.
Note :
Angular momentum [tex](L)[/tex] = Product of moment of inertia [tex](I)[/tex] and angular velocity [tex](\omega)[/tex] .
Lets say,
⇒ initial angular momentum = final angular momentum
⇒ [tex]L_i=L_f[/tex]
⇒ [tex](I_t)\times \omega_i = (I_t+I_r)\times \omega_f[/tex]
⇒ [tex]\omega _f=\frac{I_t}{I_t+I_r} \times (\omega_i)[/tex] ...equation (i)
Now we will find the ratio of the Kinetic energies.
⇒ [tex]K_i=\frac{I_t\times \omega_i^2}{2}[/tex] ⇒ [tex]K_f=\frac{(I_r+I_t)\times \omega_f^2}{2}[/tex]
Their ratios:
⇒ [tex]\frac{K_f}{K_i} =\frac{\frac{(I_t+I_r)\times \omega_f^2}{2} }{\frac{I_t\times \omega_i^2}{2} }[/tex]
⇒ [tex]\frac{K_f}{K_i} = {\frac{(I_t+I_r)\times \omega_f^2}{2} } \times {\frac{2}{I_t\times \omega_i^2}}[/tex]
Plugging the values of [tex]\omega _f^2[/tex] as [tex]\omega _f^2 =(\frac{I_t}{I_t+I_r} \times \omega_i\ )^2[/tex] from equation (i) in the ratios of the Kinetic energies.
⇒ [tex]\frac{K_f}{K_i} =\frac{(I_t+I_r)\times \frac{(I_t)^2}{(I_t+I_r)^2} \times \omega_i^2}{I_t\times \omega_i^2} =\frac{(I_t)^2}{(I_t+I_r)}\times \frac{1}{I_t}=\frac{I_t}{I_t+I_r}[/tex]
Now,
The Kinetic energy lost in fraction can be written as:
⇒ [tex]\frac{K_f-K_i}{K_i}[/tex]
Now re-arranging the terms.
[tex]\frac{K_f-K_i}{K_i} =(\frac{K_f}{K_i} -1)= \frac{I_t}{I_t+I_r} -1=\frac{I_t-I_t-I_r}{I_t+I_r} =\frac{-I_r}{(I_t+I_r)}[/tex]
Plugging the values of [tex]I_r[/tex] and [tex]I_t[/tex] .
⇒ [tex]\frac{K_f}{K_i} = \frac{-0.61I_t}{0.61I_t+I_t} =\frac{-0.61}{1.61} =-0.3788[/tex]
To find the percentage we have to multiply it with [tex]100[/tex] and here negative means for loss of Kinetic energy.
⇒ [tex]\frac{K_f}{K_i} = =-0.3788\times 100= 37.88[/tex]
So the percentage of the initial Kinetic energy lost is 37.88
