If the equation for the displacement of a particle moving on a circular path is given by θ =2t3+0.5, where θ is in radian and t in second, then the angular velocity of the particle after 2 s from its start is

Question

If the equation for the displacement of a particle moving on a circular path is given by θ =2t3+0.5, where θ is in radian and t in second, then the angular velocity of the particle after 2 s from its start is (A) 8 rad/s (B) 12 rad/s (C) 24 rad/s (D) 36 rad/s

Tardigrade Admin · Accepted Answer

Since, θ =2t3+0.5, the angular velocity of the particle =(dθ /dt)=6t2, At t=2 text s, angular velocity =6× 22=24 text rad/s

Q. If the equation for the displacement of a particle moving on a circular path is given by $θ = 2 t^{3} + 0.5,$ where $θ$ is in radian and t in second, then the angular velocity of the particle after 2 s from its start is

8 rad/s

12 rad/s

24 rad/s

36 rad/s

Since, $θ = 2 t^{3} + 0.5,$ the angular velocity of the particle $= \frac{d θ}{d t} = 6 t^{2},$ At $t = 2 s,$ angular velocity $= 6 \times 22 = 24 r a d / s$

Q. If the equation for the displacement of a particle moving on a circular path is given by θ=2t3+0.5, where θ is in radian and t in second, then the angular velocity of the particle after 2 s from its start is