Question

Consider two particles A and B. The angular position of particle A, with constant angular acceleration, depends on time according to $\theta_A(t)$=θ₀+ω₀t+12αt². At time t=t₁, particle B, which also undergoes constant angular acceleration, has twice the angular acceleration, half the angular velocity, and the same angular position that particle A had at time t=0.How long after the time t₁ does the angular velocity of B have to be to equal A's?

Answer 1

Answer:

$t - t_1 = \frac{-\omega_o + \sqrt{\omega_o^2 + 8\alpha(2\omega_o t + \alpha t^2)}}{4\alpha}$

Explanation:

After time "t" the angular position of A is given as

$\theta_a = \theta_o + \omega_o t + \frac{1}{2}\alpha t^2$

now we know that B start motion after time t1

so its angular position is also same as that of position of A after same time "t"

so we have

$\theta_b = \theta_o + \frac{\omega_o}{2} (t - t_1) + \frac{1}{2}(2\alpha) (t - t_1)^2$

now since both positions are same

$\theta_a = \theta_b$

$\omega_o t + \frac{1}{2}\alpha t^2 = \frac{\omega_o}{2}(t - t_1} + \alpha(t - t_1)^2$

$2\omega_o t + \alpha t^2 = \omega_o(t - t_1) + 2\alpha(t - t_1)^2$

$t - t_1 = \frac{-\omega_o + \sqrt{\omega_o^2 + 8\alpha(2\omega_o t + \alpha t^2)}}{4\alpha}$