JEE Main & Advanced Physics NLM, Friction, Circular Motion Sticking of Person With Wall of Rotor

Sticking of Person With Wall of Rotor A person with a mass m stands in contact against the wall of a cylindrical drum (rotor). The coefficient of friction between the wall and the clothing is m. If Rotor starts rotating about its axis, then person thrown away from the centre due to centrifugal force at a particular speed w, the person stuck to the wall even the floor is removed, because friction force balances its weight in this condition. From the figure. Friction force (F) = weight of person (mg) \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mu R = mg\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mu \,{{F}_{c}}=mg\]  [Here, \[{{\operatorname{F}}_{c}}= centrifugal force\]] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mu m\omega _{\min }^{2}r=mg\] \[\therefore \,\,\,\,\,\,\,\,\,{{\omega }_{\min }}=\sqrt{\frac{g}{\mu \,r}}\] Sample Problems (Miscellaneous) Problem 29. A motorcycle is travelling on a curved track of radius 500m if the coefficient of friction between road and tyres is 0.5. The speed avoiding skidding will be [MH CET (Med.) 2001]             (a) 50 m/s          (b) 75 m/s          (c) 25 m/s          (d) 35 m/s Solution: (a) \[v=\sqrt{\mu rg}=\sqrt{0.5\times 500\times 10}=50\,m/s.\] Problem 30. A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is 0.2. The weight of the block is [AIEEE 2003]                                     (a) 2 N                                      (b) 20 N             (c) 50 N                                     (d) 100 N Solution: (a) For equilibrium                         Weight (W) = Force of friction (F)                           \[W=\mu R=0.2\times 10=2\,N\] Problem 31. A body of mass 2 kg is kept by pressing to a vertical wall by a force of 100 N.  The friction between wall and body is 0.3. Then the frictional force is equal to [Orissa JEE 2003]                         (a) 6 N              (b) 20 N             (c) 600 N           (d) 700 N Solution: (b) For the given condition Static friction = Applied force = Weight of body \[= 2 \times  10 = 20 N.\] Problem 32. A fireman of mass 60kg slides down a pole. He is pressing the pole with a force of 600 N. The coefficient of friction between the hands and the pole is 0.5, with what acceleration will the fireman slide down \[\left( g = 10 m/{{s}^{2}} \right)\] [Pb. PMT 2002]             (a) \[1 m/{{s}^{2}}\]    (b) \[2.5 m/{{s}^{2}}\] (c) \[10 m/{{s}^{2}}\]   (d) \[5 m/{{s}^{2}}\]             Solution: (d) \[\operatorname{Friction} =mR = 0.5 \times  600 = 300 N, \,Weight = 600 N\]             \[\therefore \,\,\,\,\,\,\,a=5\,m/{{s}^{2}}\] Problem 33. The system shown in the figure is in equilibrium. The maximum value of W, so that the maximum value of static frictional force on 100 kg body is 450 N, will be                                     (a) 100 N                                   (b) 250 N             (c) 450 N                                   (d) 1000 N Solution: (c) For vertical equilibrium \[{{T}_{1}}\sin {{45}^{o}}=W\]                                     \[\therefore \,\,\,\,\,{{T}_{1}}=\frac{W}{\sin {{45}^{o}}}\]             For horizontal equilibrium \[{{T}_{2}}={{T}_{1}}\cos {{45}^{o}}\]\[=\frac{W}{\sin {{45}^{o}}}\cos {{45}^{o}}=W\]             and for the critical condition \[{{T}_{2}}=\text{ }F\]             \[\therefore \,\,\,\,\,\,\,\,\operatorname{W} = {{T}_{2}}= F = 450 N~~~~\]