JEE Main 2015 · 10 Apr · Shift 1 · Physics Q5

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PhysicsSingle correct+4 / -1Question 5 of 90

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 :

2 ma

\dfrac32

\dfrac53

Answer and worked solution

Official answer: C

$F-f=ma$ , and $fR=\frac12mR^2\alpha$ with $\alpha=\frac aR$ , so $f=\frac12ma$ . Then $F=\dfrac32ma$ .

Topics:Rolling motion

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JEE Main 2015 · 10 Apr · Shift 1, Physics question 5