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

`t - t_1 = (-\omega_o + √(\omega_o^2 + 8\alpha(2\omega_o t + \alpha t^2)))/(4\alpha)`

Step-by-step explanation:

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

`\theta_a = \theta_o + \omega_o t + (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 + (\omega_o)/(2) (t - t_1) + (1)/(2)(2\alpha) (t - t_1)^2`

now since both positions are same

`\theta_a = \theta_b`

`\omega_o t + (1)/(2)\alpha t^2 = (\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 = (-\omega_o + √(\omega_o^2 + 8\alpha(2\omega_o t + \alpha t^2)))/(4\alpha)`