The ratio of the accelerations for a solid sphere(mass 'm' and radius 'r') rolling down an incline of angle 'theta' without slipping and slipping down the incline without rolling is:a)7:5b)5:7c)2:3d)2:5The above question was asked in AIPMT 2014.

Asked by Renuka Ganesh | 6th May, 2014, 09:49: PM

Expert Answer:

For slipping motion on an inclined plane the acceleration for a solid sphere making an angle θ  is given by
 
 straight a subscript slipping space end subscript space equals space straight g space sin space straight theta
For rolling motion of a sphere without slipping :
 
The acceleration of a sphere of mass m, radius r and moment of inertia I is 
straight a subscript rolling space equals fraction numerator straight g space sinθ over denominator 1 plus begin display style straight I over mr squared end style end fraction
For a uniform sphere we know that straight I over mr squared equals 2 over 5
Substituting this  we get that the ratio straight a subscript rolling over straight a subscript slipping equals space fraction numerator straight g space sin space straight theta over denominator begin display style fraction numerator straight g space sin space straight theta over denominator 1 plus begin display style straight I over mr squared end style end fraction end style end fraction space equals fraction numerator straight g space sin space straight theta over denominator begin display style fraction numerator straight g space sin space straight theta over denominator 1 plus begin display style 2 over 5 end style end fraction end style end fraction equals space 5 over 7
Hence the ratio is 5:7 
Option b) is the correct answer.

Answered by Jyothi Nair | 7th May, 2014, 10:32: AM

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