Question

7. A child stands at the centre of a turn table with his two arms out stretched. The turn table
is set rotating with an angular speed of 40 rpm. How much is the angular speed of the child
change if he folds his hands back and thereby reduces his moment of inertia to 2/3 times
the initial value? Assume that the turntable rotates without friction.​

Answer 1

Answer:

(a) Since no external force acts on the boy, the angular momentum is a constant.

Thus I  

1

​  

ω  

1

​  

=I  

2

​  

ω  

2

​  

 

⟹ω  

2

​  

=  

I  

2

​  

 

I  

1

​  

 

​  

ω  

1

​  

=  

2

5

​  

×40rev/min=100revmin

(b) Initial kinetic energy=E  

1

​  

=  

2

1

​  

I  

1

​  

ω  

1

2

​  

 

Final kinetic energy=E  

2

​  

=  

2

1

​  

I  

2

​  

ω  

2

2

​  

 

Thus  

E  

1

​  

 

E  

2

​  

 

​  

=  

I  

1

​  

 

I  

2

​  

 

​  

×  

ω  

1

2

​  

 

ω  

2

2

​  

 

​  

=2.5

The increase in the rotational kinetic energy is attributed to the internal energy of the boy.