JEE Main 2014 · 19 Apr · Shift 1 · Physics Q4

← JEE Main 2014 · 19 Apr · Shift 1

PhysicsSingle correct+4 / -1Question 4 of 90

Consider a cylinder of mass M resting on a rough horizontal rug that is pulled out from under it with acceleration 'a' perpendicular to the axis of the cylinder. What is $F_{\text{friction}}$ at point P ? It is assumed that the cylinder does not slip.

A cylinder with centre O rests on a rug at the contact point P. The rug is pulled to the left with acceleration a; the cylinder's centre moves left with velocity v while it rotates with angular velocity ω; the friction force at P is marked.

\dfrac{Ma}2

\dfrac{Ma}3

Answer and worked solution

Official answer: D

Friction $f$ (towards the rug's motion) is the only horizontal force: $f=MA$ for the centre. About the centre, $fR=\frac12MR^2\alpha$ .

No slipping at P: the contact point moves with the rug, $A+R\alpha=a$ . So $A+2A=a$ , giving $A=\frac a3$ and $f=\dfrac{Ma}3$ .

Topics:Rolling motion Torque and rotational dynamics

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JEE Main 2014 · 19 Apr · Shift 1, Physics question 4