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A solid sphere of mass m and radius R released from the top of an inclined plane such that it rolls down without slipping. If the coefficient of friction is µ and angle of inclination is ø with horizontal, then force of friction acting on the body is

1)   2 square root of G M cubed r end root                                                                2)     1 half square root of G M cubed end root r          

3)    1 half square root of 2 G M cubed end root r                                                           4)       1 half square root of fraction numerator begin display style G M cubed r end style over denominator 2 end fraction end root             

Asked by reachrahul123 14th October 2018, 9:48 PM
Answered by Expert
Answer:
Figure shows the different forces acting on the sphere that is rolling on the inclined plane.
weight mg is resolved into two components, one is perpendicular to inclined plane and
other component  is parallel to inclined plane.
Hence the normal reaction force N = mg cosΦ and the friction force is μN = μ×mg×cosΦ
 
All the expression given in the list of options of answer is proportional to begin mathsize 12px style square root of g m cubed r end root end style.
if we check the units of this expression we are getting begin mathsize 12px style open square brackets m over s squared open parentheses k g close parentheses cubed m close square brackets to the power of bevelled 1 half end exponent space w h i c h space i s space n o t space e q u a l space t o space open square brackets k g m over s squared close square brackets space w h i c h space i s space u n i t space o f space f o r c e end style
I hope g is typed as G, m is typed as M. Similarly there is a mismatch of r and R.
I assume these typing error and considered the correct symbol.
But as explained above there is no matching of units. But I derived the correct expression for frictional force
Answered by Expert 15th October 2018, 4:56 AM
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