A ball is spun with angular acceleration α = 6t² - 2t where t is in second and α is in rads⁻². At t = 0, the ball has angular velocity of 10 rads⁻¹ and angular position of 4 rad. The most appropriate expression for the angular position of the ball is :

Solution

Physics sir · 3 steps

Integrate α twice, and use the two starting values to fix the two constants.

1. First integration. Since α = dω/dt, integrate 6t² - 2t to get ω = 2t³ - t² + C₁. At t = 0, ω = 10 rad s⁻¹, so C₁ = 10.
2. Second integration. Since ω = dθ/dt, integrate 2t³ - t² + 10 to get θ = t⁴/2 - t³/3 + 10t + C₂. At t = 0, θ = 4 rad, so C₂ = 4.

   θ = t⁴/2 - t³/3 + 10t + 4

3. So. The angular position is option (b). As a check, it gives θ(0) = 4 and dθ/dt at t = 0 equal to 10, matching both starting values.

The idea to keep. Going from α to θ means two integrations and two constants. The initial angular velocity fixes the first and the initial angular position fixes the second. Never drop either one.

Answer from the NTA final answer key. Solution by PrepWiser.