JEE Main202228 Jun 2022Evening ShiftPhysicsMotion in Two DimensionsActual
A ball is spun with angular acceleration α = 6 t 2 - 2 t where t is in second and α is in rad s - 2 . At t = 0 , the ball has angular velocity of 10 rad s - 1 and angular position of 4 rad . The most appropriate expression for the angular position of the ball is
Options
- A3 2 t 4 - t 2 + 10 t
- Bt 4 2 - t 3 3 + 10 t + 4
- C2 t 4 3 - t 3 6 + 10 t + 12
- D2 t 4 - t 3 2 + 5 t + 4
Correct answer
B. t 4 2 - t 3 3 + 10 t + 4
Step-by-step solution
Given: α = 6 t 2 - 2 t Using relation α = d ω d t = 6 t 2 - 2 t Integrating the above, ∫ 10 ω d ω = ∫ 0 t 6 t 2 - 2 t d t ⇒ ω - 10 = 2 t 3 - t 2 Now, ω = d θ d t = 10 + 2 t 3 - t 2 ∫ 4 θ d θ = ∫ 0 t 10 + 2 t 3 - t 2 d t Integrating the above relation, θ - 4 = 10 t + t 4 2 - t 3 3 Thus, angular position of the ball is θ = t 4 2 - t 3 3 + 10 t + 4 .