Answer:
The loss of initial Kinetic energy = 37.88 %
Explanation:
Given:
Rotational inertia of the turntable = ![I_t](https://tex.z-dn.net/?f=I_t)
Rotational inertia (
) of the record = ![0.61\times I_t](https://tex.z-dn.net/?f=0.61%5Ctimes%20I_t)
According to the question:
<em>Frictional forces act to bring the record and turntable to a common angular speed.</em>
So,angular momentum will be conserved as it is an inelastic collision.
Considering the initial and final angular velocity of the turn table as
respectively.
Note :
Angular momentum
= Product of moment of inertia
and angular velocity
.
Lets say,
⇒ initial angular momentum = final angular momentum
⇒
⇒
⇒
...equation (i)
Now we will find the ratio of the Kinetic energies.
⇒
⇒ ![K_f=\frac{(I_r+I_t)\times \omega_f^2}{2}](https://tex.z-dn.net/?f=K_f%3D%5Cfrac%7B%28I_r%2BI_t%29%5Ctimes%20%5Comega_f%5E2%7D%7B2%7D)
Their ratios:
⇒
⇒ ![\frac{K_f}{K_i} = {\frac{(I_t+I_r)\times \omega_f^2}{2} } \times {\frac{2}{I_t\times \omega_i^2}}](https://tex.z-dn.net/?f=%5Cfrac%7BK_f%7D%7BK_i%7D%20%3D%20%7B%5Cfrac%7B%28I_t%2BI_r%29%5Ctimes%20%5Comega_f%5E2%7D%7B2%7D%20%7D%20%5Ctimes%20%7B%5Cfrac%7B2%7D%7BI_t%5Ctimes%20%5Comega_i%5E2%7D%7D)
Plugging the values of
as
from equation (i) in the ratios of the Kinetic energies.
⇒ ![\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}](https://tex.z-dn.net/?f=%5Cfrac%7BK_f%7D%7BK_i%7D%20%3D%5Cfrac%7B%28I_t%2BI_r%29%5Ctimes%20%5Cfrac%7B%28I_t%29%5E2%7D%7B%28I_t%2BI_r%29%5E2%7D%20%5Ctimes%20%5Comega_i%5E2%7D%7BI_t%5Ctimes%20%5Comega_i%5E2%7D%20%3D%5Cfrac%7B%28I_t%29%5E2%7D%7B%28I_t%2BI_r%29%7D%5Ctimes%20%5Cfrac%7B1%7D%7BI_t%7D%3D%5Cfrac%7BI_t%7D%7BI_t%2BI_r%7D)
Now,
The Kinetic energy lost in fraction can be written as:
⇒
Now re-arranging the terms.
![\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)}](https://tex.z-dn.net/?f=%5Cfrac%7BK_f-K_i%7D%7BK_i%7D%20%20%3D%28%5Cfrac%7BK_f%7D%7BK_i%7D%20-1%29%3D%20%5Cfrac%7BI_t%7D%7BI_t%2BI_r%7D%20-1%3D%5Cfrac%7BI_t-I_t-I_r%7D%7BI_t%2BI_r%7D%20%3D%5Cfrac%7B-I_r%7D%7B%28I_t%2BI_r%29%7D)
Plugging the values of
and
.
⇒ ![\frac{K_f}{K_i} = \frac{-0.61I_t}{0.61I_t+I_t} =\frac{-0.61}{1.61} =-0.3788](https://tex.z-dn.net/?f=%5Cfrac%7BK_f%7D%7BK_i%7D%20%3D%20%5Cfrac%7B-0.61I_t%7D%7B0.61I_t%2BI_t%7D%20%3D%5Cfrac%7B-0.61%7D%7B1.61%7D%20%3D-0.3788)
To find the percentage we have to multiply it with
and here negative means for loss of Kinetic energy.
⇒ ![\frac{K_f}{K_i} = =-0.3788\times 100= 37.88](https://tex.z-dn.net/?f=%5Cfrac%7BK_f%7D%7BK_i%7D%20%3D%20%3D-0.3788%5Ctimes%20100%3D%2037.88)
So the percentage of the initial Kinetic energy lost is 37.88