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marysya [2.9K]
3 years ago
7

A record is dropped vertically onto a freely rotating (undriven) turntable. Frictional forces act to bring the record and turnta

ble to a common angular speed. If the rotational inertia of the record is 0.61 times that of the turntable, what percentage of the initial kinetic energy is lost
Physics
1 answer:
alexgriva [62]3 years ago
4 0

Answer:

The loss of initial Kinetic energy = 37.88 %

Explanation:

Given:

Rotational inertia of the turntable = I_t

Rotational inertia (I_r) of the record = 0.61\times I_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  \omega _i\  ,\  \omega_f respectively.

Note :

Angular momentum (L) = Product of moment of inertia  (I)  and angular velocity (\omega) .  

Lets say,

⇒ initial angular momentum = final angular momentum

⇒  L_i=L_f

⇒ (I_t)\times \omega_i = (I_t+I_r)\times \omega_f

⇒ \omega _f=\frac{I_t}{I_t+I_r} \times (\omega_i) ...equation (i)

Now we will find the ratio of the Kinetic energies.

⇒ K_i=\frac{I_t\times \omega_i^2}{2}       ⇒ K_f=\frac{(I_r+I_t)\times \omega_f^2}{2}

Their ratios:

⇒ \frac{K_f}{K_i} =\frac{\frac{(I_t+I_r)\times \omega_f^2}{2} }{\frac{I_t\times \omega_i^2}{2} }    

⇒ \frac{K_f}{K_i} = {\frac{(I_t+I_r)\times \omega_f^2}{2} } \times {\frac{2}{I_t\times \omega_i^2}}

Plugging the values of  \omega _f^2 as \omega _f^2 =(\frac{I_t}{I_t+I_r} \times \omega_i\ )^2 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}

Now,

The Kinetic energy lost in fraction can be written as:

⇒ \frac{K_f-K_i}{K_i}

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)}

Plugging the values of  I_r and I_t .

⇒ \frac{K_f}{K_i} = \frac{-0.61I_t}{0.61I_t+I_t} =\frac{-0.61}{1.61} =-0.3788

To find the percentage we have to multiply it with 100 and here negative means for loss of Kinetic energy.

⇒ \frac{K_f}{K_i} = =-0.3788\times 100= 37.88

So the percentage of the initial Kinetic energy lost is 37.88

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