Topic: Engineering Mechanics
A point moves according to the relation x = 8t² cos(ωt), y = -4t³ sin(ωt), where x and y are in meters, and t in seconds. If angular velocity ω = 8π rad/sec, the velocity vector of the point at t = 0.5 sec is given by
- A. 8 + j4π
- B. 6 + j20π
- C. 8 - j4π
- D. -20π – j6
Answer: C. 8 - j4π
Explanation: Differentiating x and y with respect to t gives velocity components u and v. At t=0.5s, ωt = 4π. Since sin(4π)=0 and cos(4π)=1, the velocity vector V = [16t cos(ωt) - 8t²ω sin(ωt)]i + [-12t² sin(ωt) - 4t³ω cos(ωt)]j simplifies to V = (16*0.5*1 - 0)i + (0 - 4*(0.5)³*8π*1)j = 8i - 4πj.