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

The loss of initial Kinetic energy = 37.88 %

Explanation:

Given:

Rotational inertia of the turntable =

Rotational inertia () of the record =

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   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.

⇒       ⇒

Their ratios:

⇒    

⇒

Plugging the values of   as from equation (i) in the ratios of the Kinetic energies.

⇒

Now,

The Kinetic energy lost in fraction can be written as:

⇒

Now re-arranging the terms.

Plugging the values of   and .

⇒

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

⇒

So the percentage of the initial Kinetic energy lost is 37.88

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