A block of mass M lies on a frictionless table. A string is attached to the mass

Question

A block of mass M lies on a frictionless table. A string is attached to the mass and passes through a small hole in the middle of the table, so that it hangs vertically. The block is set to rotating; once it is rotating with a constant angular velocity, the string is slowly pulled straight downward. As the string is pulled,

Select one:

A. the kinetic energy of the block decreases

B. none of these answers are correct

C. the angular momentum of the block remains constant

D. the angular momentum of the block decreases

E. the kinetic energy of the block remains constant

Explanation / Answer

We know that thw kinetic energy K.E of rotation is given as

K.E =1/2 I omega2 ....(1)

Here, I is the moment of ineria and omega is the angular velocity. Since the shape of the block remaons same so I=mr^2 on pulling down the string r will increase and hence, I and it is given that the angular velocity is constant so omega is also same. Hence, we can conclude that the K. E of rotation increases.

The translation K.E is given as

K.E_t =1/2 m v^2

As the string is pulled down than the v and the r will change and they will change in such a manner so that their ration remains the same leading to constant value of omega. As the string is pulled down so r will increase that means v has to decrease which employs that K.E _t should decrease. Hence, we can conclude that any of the statement relating to K.E is not true.

Angular momentum L is

L=I omega

L= mr^2 omega

On pulling down the string r increase which in turn increase I and omega is constant which is equal to v/r so L can also be given as

L=mvr

m remains the same. v and r changes by same amount in order to keep omega to be constant. Hence, we can say that the angular momentum remains the same

So we can say that option C thatthe angular momentum remains same is correct.

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