Sol. (i) (Scalar-geometric). The center O is chosen as the reference point for the relative-velocity equation since its motion is given. We therefore write

v A = v O + v A/O

where the relative-velocity term is observed from the translating axes x-y attached to O. The angular velocity of AO is the same as that of the wheel ω = v 0 /r = 3/0.3 = 10 rad/s. Thus, from Eq. 5/5 we have

[v AO = r O ] v A/O = 0.2(10) = 2 m/s

Which is normal to AO as shown. The vector sum v A is shown on the diagram and may be calculated from the law of cosines. Thus,

v A 2 = 3 2 + 2 2 + 2(3) (2) cos 60º = 19 (m/s) 2 v A

= 4.36 m/s Ans.

The contact point C momentarily has zero velocity and can be used alternatively as the reference point, in which case, the relative-velocity equation becomes

v A = v C + v A/C = v A/C where

v A/C = = v 0 = (3) = 4.36 m/s v A = v A/C = 4.36 m/s

The distance = 436 mm is calculated separately. We see that v A is normal to AC since A is momentarily rotating about point C.

(ii) (Vector) : We will now use Eq. and write

v A = v O + v A/O = v O + ω × r 0 .

Where ω = –10 k rad/s

r 0 = 0.2 (–i cos 30º + j sin 30º) = –0.1732i+0.1j m

v O = 3i m/s

We now solve the vector equation

v A =3i+ =3i + 1.732j + 1.0i

= 4i + 1.732j m/s Ans.

The magnitude v A = = = 4.36 m/s and direction agree with previous solution

Helpful Hints :

(i) Be sure to visualize v A/O as the velocity which A appears to have in its circular motion relative to O.

(ii) The vectors may also be laid off to scale graphically and the magnitude and direction of v A measured directly from the diagram.

(iii) The velocity of any point on the wheel is easily determined by using the contact point C as the reference point. You should construct the velocity vectors for a number of points on the wheel for practice.

(iv) The vector ω is directed into the paper by the right-hand rule, whereas the positive z-direction is out from the paper; hence, the minus sign.