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STD 11 - 6. system of particles and rotational motion — NEET STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceNEETSTD 11 - 6. system of particles and rotational motion4 Marks
Question
A uniform solid cylindrical roller of mass $'m\ '$ is being pulled on a horizontal surface with force $F$ parallel to the surface and applied at its centre. If the acceleration of the cylinder is $'a\ '$ and it is rolling without slipping then the value of $'F\ '$ is 
✓

Answer

From figure,$ma = F - f ...(i)$
And, torque $\tau = I\alpha $
$\frac{{m{R^2}}}{2}\alpha = fR\,$
$\frac{{m{R^2}}}{2} - \frac{a}{R} = fR\,\,\left[ {\alpha = \frac{a}{R}} \right]$
$\frac{{ma}}{2} = f ,...\left( {ii} \right)$
Put this value in equation $(i),$
$ma = F - \frac{{ma}}{2}\,\,or\,\,F = \frac{{3ma}}{2}$

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