Question

A particle has an angular velocity vector ω whose magnitude is the angular speed of the particle and the direction represents the axis about which the particle is instantaneously rotating. The particle’s linear velocity v can be written as v = ω × r where r is the particle’s position vector.

(a) If ω is constant, show that the particle’s linear acceleration a = dv/dt can be written as a = ω(ω · r) − |ω| 2 r
(b) Suppose a particle is constrained in the x−y plane. In cylindrical coordinates r = Rˆs and is perpendicular to ω = ωˆz and ω = dφ/dt is the constant rate of rotation. Use the result from part a) to find the acceleration of this body. The result should look familiar. Draw the situation and label the position vector r, the velocity vector v and the acceleration vector a in the drawing.

Answer 1

a) a = w (w. r) - r w² , b) a = - r d²φ /dt²

Step-by-step explanation:

For this exercise we derive the velocity of the particle

v = w x r

a = dv / dt = d/dt (w x r)

a = dw/dt x r + w x dr/dt

a = α x r + w x v

a = α x r + w x (w x r)

In the first term alpha is in the same direction, so its vector product zero

The second term, we develop it using

A x B x C = B (A .C) - C (A. B)

we substitute

a = 0 + w (w .r) - r (w .w)

a = w (w. r) - r w²

b) x- y plane

r = R s^

w = w k^

w = dφ / dt

a = w (w. R) - R w²

w² = (dφ / dt )²

a = w (w. R) - r d²φi / dt²

since the motion is in the x -y plane and the angular velocity is in the z axis, the scalar product of two perpendicular vectors zero (w. R) = 0

a = - r d²φ /dt²