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JEE Advanced Physics Rotational Motion 2021 JEE Advanced 2021 (Paper 1)

JEE Advanced Physics Question (2021) — Solution

Question

A horizontal force  F  is applied at the centre 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 . The centre 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?

Options

  1. A. For the same  F , the value of  a  does not depend on whether the cylinder is solid or hollow.
  2. B. For a solid cylinder, the maximum possible value of  a  is  2 μ g .
  3. C. The magnitude of the frictional force on the object due to the ground is always  μ m g .
  4. D. For a thin-walled hollow cylinder,  a = F 2 m .

Answer

D. For a thin-walled hollow cylinder,  a = F 2 m .

Step-by-step solution

Let moment of inertia  = I For pure rolling,  a = α R and  F − f = m a               . . . 1 and  f R = I α or  f R = I a R or  f = I a R 2 Therefore,  F − I a R 2 = m a ∴       F = m + I R 2 a ⇒  For the same value of  F , value of a depends on  I . ⇒  For a solid cylinder I = m R 2 2  (to calculate maximum  a ) f = μ m g ∴   μ m g = m R 2 2 × a R 2 Therefore,  a = 2 μ g ⇒ f  is dependent on applied force. ⇒  For thin-walled hollow cylinder I = m R 2 , F = m + m R 2 R 2 a = 2 m a ∴     a = F 2 m  

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