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
- 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 g .
- C. The magnitude of the frictional force on the object due to the ground is always μ m g .
- 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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