A solid sphere of mass 1   kg and radius 1   m rolls without slipping on a fixed inclined plane with an angle of inclination θ = 30 ° from the horizontal. Two forces of magnitude 1   N each, parallel to the incline, act on the sphere, both at distance r = 0 . 5  

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A solid sphere of mass 1 &#160; kg and radius 1 &#160; m rolls without slipping on a fixed inclined plane with an angle of inclination &#952; = 30 &#176; from the horizontal. Two forces of magnitude 1 &#160; N  each, parallel to the incline, act on the sphere, both at distance r = 0 . 5 &#160; m from the center of the sphere, as shown in the figure. The acceleration of the sphere down the plane is m &#160; s - 2 . (Take g = 10 &#160; m &#160; s - 2 ) If the numerical value has more than two decimal places, truncate/round-off the value to TWO decimal places.

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Correct answer: A. A

The forces acting on the solid sphere is shown below. Here,  N  is the normal force acting on sphere and weight  m g  is acting downwards.    Taking torque about contact point. &#964; &#8594; = m g R sin 30 &#8855; + 1 &#215; 1 &#8857; = 10 &#215; 1 &#215; 1 2 - 1       (Taking &#8855;  as positive) Then, we have  &#8658; 5 - 1 = I sphere&#160;about&#160;tangent &#945; Using parallel axis theorem, &#964; = 2 5 m R 2 + m R 2 &#945; = 7 5 m R 2 &#945; &#8658; &#945; = 20 7 rad &#160; s - 2 So, acceleration of sphere down the plane is  a cm = &#945; R = 20 7 &#160; m &#160; s - 2 a cm = 2 . 86 &#160; m &#160; s - 2

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