Physics-
General
Easy

Question

A metal rod of length L and mass m is pivoted at one end. A thin disk of mass M and radius R left parenthesis less than L right parenthesis is attached at its centre to the free end of the rod. Consider two ways the disc is attached case A- the disc is not free to rotate about its centre and case B – the disc is free to rotate about its centre. The rod-disc system performs SHM in vertical plane after being released from the same displaced position. Which of the following statement(s) is/are true?

  1. Restoring torque in case A=Restoring torque in case B    
  2. Restoring torque in case A<Restoring torque in case B    
  3. Angular frequency for case A<Angular frequency for case B    
  4. Angular frequency for case A<Angular frequency for case B    

The correct answer is: Angular frequency for case A<Angular frequency for case B


    tau subscript A end subscript equals tau subscript B end subscript equals left parenthesis m g fraction numerator L over denominator 2 end fraction sin invisible function application theta plus M g L sin invisible function application theta right parenthesis
    = Restoring torque about point O.
    In case A, moment of inertia will be more. Hence, angular acceleration left parenthesis alpha equals tau divided by I right parenthesis will be less. Therefore angular frequency will be less. Note Question is difficult because this type of SHM is rarely.

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