99 Percentile Qs Bank for JEE MainPhysicsLaws of Motion
A uniform cylinder of radius 1 ~m , mass 1 ~kg spins about its axis with an angular velocity 20 rad / s . At certain moment, the cylinder is placed into a corner as shown in the figure. The coefficient of friction between the horizontal wall and the cylinder is , whereas the vertical wall is frictionless. If the number of rounds made by the cylinder is 5 before it stops, then the value of is (acceleration due to grav
Options
- A3
- B2
- C1
- D0.4
Correct answer
C. 1
Step-by-step solution
Free body diagram of cylinder is We have following equations, aligned N₂ & =m g N₂ & =N₁ aligned Torque about centre of cylinder is N₂ R= m R^2 2 From Eqs. (i) and (iii), we get m g R= m R^2 2 = 2 g R Now using, ^2= ₀^2-2 We have, 0=(20)^2-2 ( 2 10 1 ) Here, =5 2 So, (20)^2=40 10 ( ) = 1