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QJEEAdv21Ph2dff: JEE Advanced 2021 Physics - Class 11 JEEAdv21Ph2dff Solution: A horizontal force F is applied at the center of mass of a…

A horizontal force F is applied at the center of mass of a cylindrical object of mass m and radius R, perpendicular to its axis as shown in the figure. The coefficient of friction between the object and the ground is μ\mu . The center of mass of the object has an acceleration a. The acceleration due to gravity is g. Given that the object rolls without slipping, which of the following statement(s) is(are) correct?

JEE Advanced 2021 Paper 1 Online Physics - Rotational Motion Question 34 English

Multiple Choice Options

  • A For the same F, the value of a does not depend on whether the cylinder is solid or hollow
  • B For a solid cylinder, the maximum possible value of a is 2 μ\mu g
  • C The magnitude of the frictional force on the object due to the ground is always μ\mu mg
  • D For a thin-walled hollow cylinder, a=F2ma = {F \over {2m}}

Step-by-Step Verified Solution

JEE Advanced 2021 Paper 1 Online Physics - Rotational Motion Question 34 English Explanation
F −- f = mac

fR = IC α\alpha

aC −- α\alpha R = 0

F−ICαR=macF - {I_C}{\alpha \over R} = m{a_c}

aC=FICR2+m{a_C} = {F \over {{{{I_C}} \over {{R^2}}} + m}}

f=ICαR=ICR2,aC=ICR2F[ICR2+m]f = {{{I_C}\alpha } \over R} = {{{I_C}} \over {{R^2}}},{a_C} = {{{I_C}} \over {{R^2}}}{F \over {\left[ {{{{I_C}} \over {{R^2}}} + m} \right]}}

f=F(m+ICR2)f = {F \over {\left( {m + {{{I_C}} \over {{R^2}}}} \right)}}

Thin walled hollow cylinder,

IC = mR2

aC=F2m{a_C} = {F \over {2m}}

fR=ICα=ICaCRfR = {I_C}\alpha = {{{I_C}{a_C}} \over R}

f=ICaCR2≤μmgf = {{{I_C}{a_C}} \over {{R^2}}} \le \mu mg

aC≤μmgR2IC{a_C} \le {{\mu mg{R^2}} \over {{I_C}}}

For solid cylinder, IC=mR22{I_C} = {{m{R^2}} \over 2}

aC≤2μg{a_C} \le 2\mu g

(aC)max⁡=2μg{({a_C})_{\max }} = 2\mu g

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