A wheel of radius 0.5 m rolls without slipping on a horizontal surface, starting from rest. If the linear distance traveled by its center of mass in the first second is m, the number of revolutions made by the wheel during the 3^ rd second of its motion is :

Step-by-step solution

Let the linear acceleration of the center of mass be a . The linear distance covered in the first second ( t = 1 s) is: s = ut + 1 2 at^2 = 0 + 1 2 a(1)^2 a = 2 m/s ^2 For rolling without slipping, the angular acceleration is: = a R = 2 0.5 = 4 rad/s ^2 The angular displacement during the 3^ rd second ( n = 3 ) is: = ₀ + 2 (2n - 1) = 0 + 4 2 (2(3) - 1) = 2 5 = 10 rad The number of revolutions made during the 3^ rd second is: N = 2 = 10 2 = 5 Answer: 5