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  • A small mass attached to a string rotates on a frictionless table top.if the tension in the string is increased by pulling the string causing the radius of the circular motion to decrease by a factor of 2 then the kinetic energy of the mass will be?
    Asked by priyanka kumari 4 months ago
     
 
 

Answers (1)

 
  • By , 4 months ago
    No external torque acts on the system consisting of the mass and the string about the centre around which the mass rotates.

    The tension force is increased, but it has zero torque about the centre as it passes through the centre.

    Hence, initial angular momentum about the centre = final angular momentum about the centre.

    So, I? = constant or mr2? = constant.

    Initial angular momentum = mr2?1

    Final angular momentum = mr2?2/4

    Equating, we get - ?1 = ?2/4

    Initial kinetic energy = 1/2 I ?12

    Final kinetic energy = 1/2 I ?22 = 2 I ?12

    Hence final KE/initial KE = 4. The kinetic energy increases 4 times.
 
 
 
 

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