Relative Motion using Rotatin Axis

[joemama69]

Homework Statement

While the bridge is closing with a constant rotation of .5 rad/s, a man runs along the roadway such that when d=10ft, he is running outward from the center at 5 ft/s with an acceleration of 2 ft/s^2, both measured relative to the roadway. Determine his velocity and acceleration at this instant.

Homework Equations

V(a) = V(o) + Omega X r(a/o) + v(a/o)
A(a) = A(o) + Omega' X r(a/o) + Omega X (Omega X r(a/o)) + 2Omega X V(a/o) + A(a/o)

The Attempt at a Solution

V(o) = 0, Omega = .5k, Omega' = 0, r(a/o) = 10j, V(a/o) = -5j, A(a/o) = -2j

Lets do velocity first

Omega X r(a/o) = -5i

V(a) = 0 + -5i + -5j = -5i - 5j ft/s


Now Acceleration

Omega' X r(a/o) = 0
Omega X (Omega X r(a/o)) = 2.5j
2Omega X v(a/o) = 5j

A(a) = 0 + 0 + 2.5j + 5j + -2j = 5.5j ft/s^2


Shouldn't the velocity and acceleration be in the same direction.

 

[Chi Meson]

Shouldn't the velocity and acceleration be in the same direction.

No, they rarely are in the same direction. Think about circular motion: what's the direction of the tangential velocity, and what is the direction of centripetal acceleration? (the names give away the answer)

 

[nlightner]

Yes, the velocity and acceleration should be in the same direction. However, in this case, the acceleration is not solely due to the rotation of the bridge, but also the man's own acceleration relative to the roadway. Therefore, the direction of the acceleration will be slightly different from the direction of the velocity. Your calculations are correct, and the small difference in direction is due to the man's acceleration.