Physics-
General
Easy

Question

The acceleration of the center of mass of a uniform solid disc rolling down an inclined plane of angle alpha is

  1. g sin alpha    
  2. 2/3 g sin alpha    
  3. 1/2 g sin alpha    
  4. 1/3 g sin alpha    

hintHint:

a equals fraction numerator g blank s i n blank alpha over denominator 1 plus fraction numerator K subscript 2 end subscript over denominator R to the power of 2 end exponent end fraction end fraction this is the formula to be used here.

The correct answer is: 2/3 g sin alpha


    The acceleration of the body which is rolling down an inclined plane of angle alpha is

    a equals fraction numerator g blank s i n blank alpha over denominator 1 plus fraction numerator K subscript 2 end subscript over denominator R to the power of 2 end exponent end fraction end fraction
    where K equalsradius of gyration,
    R equalsradius of body.
    Now, here the body is a uniform solid disc.
    So, fraction numerator K to the power of 2 end exponent over denominator R to the power of 2 end exponent end fraction equals fraction numerator 1 over denominator 2 end fraction
    therefore a equals fraction numerator g blank s i n blank alpha over denominator 1 plus fraction numerator 1 over denominator 2 end fraction end fraction
    or a equals fraction numerator g blank s i n alpha over denominator 3 divided by 2 end fraction
    or a equals fraction numerator 2 g sin invisible function application alpha over denominator 3 end fraction

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