Kenneth Kuttler

300 CHAPTER 14. VECTOR PRODUCTS

Example 14.5.7 A wheel rotates counter clockwise about the vector i+j+k at 60 rev-olutions per minute. This means that if the thumb of your right hand were to point in thedirection of i+j+k your fingers of this hand would wrap in the direction of rotation.Find the angular velocity vector for this wheel. Assume the unit of distance is meters andthe unit of time is minutes.

Let ω = 60× 2π = 120π. This is the number of radians per minute corresponding to60 revolutions per minute. Then the angular velocity vector is 120π√

3(i+j+k) . Note this

gives what you would expect in the case the position vector to the point is perpendicular toi+j+k and at a distance of r. This is because of the geometric description of the crossproduct. The magnitude of the vector is r120π meters per minute and corresponds to thespeed and an exercise with the right hand shows the direction is correct also. However, ifthis body is rigid, this will work for every other point in it, even those for which the positionvector is not perpendicular to the given vector.

Example 14.5.8 A wheel rotates counter clockwise about the vector i+j+k at 60 rev-olutions per minute exactly as in Example 14.5.7. Let {u1,u2,u3} denote an orthogonalright handed system attached to the rotating wheel in which u3 = 1√

3(i+j+k) . Thus

u1 and u2 depend on time but, u1 ×u2 = u3. Find the velocity of the point of the wheellocated at the point 2u1 +3u2 −u3. Note this point is not fixed in space. It is moving.

Since {u1,u2,u3} is a right handed system like i,j,k, everything applies to this sys-tem in the same way as with i,j,k. Thus the cross product is given by

(au1 +bu2 + cu3)× (du1 + eu2 + fu3) =

∣∣∣∣∣∣u1 u2 u3a b cd e f

∣∣∣∣∣∣Therefore, in terms of the given vectors ui, the angular velocity vector is 120πu3. Thevelocity of the given point is∣∣∣∣∣∣

u1 u2 u30 0 120π

2 3 −1

∣∣∣∣∣∣=−360πu1 +240πu2

in meters per minute. Note how this gives the answer in terms of these vectors which arefixed in the body, not in space. Since ui depends on t, this shows the answer in this casedoes also. Of course this is right. Just think of what is going on with the wheel rotating.Those vectors which are fixed in the wheel are moving in space relative to a stationaryobserver. The velocity of a point in the wheel should be constantly changing. However, itsspeed will not change. The speed will be the magnitude of the velocity and this is√

(−360πu1 +240πu2) · (−360πu1 +240πu2)

which from the properties of the dot product equals√(−360π)2 +(240π)2 = 120

√13π

because the ui are given to be orthogonal.

300 CHAPTER 14. VECTOR PRODUCTSExample 14.5.7 A wheel rotates counter clockwise about the vector i+ 3 +k at 60 rev-olutions per minute. This means that if the thumb of your right hand were to point in thedirection of 1+ 3-+k your fingers of this hand would wrap in the direction of rotation.Find the angular velocity vector for this wheel. Assume the unit of distance is meters andthe unit of time is minutes.Let @ = 60 x 27 = 1202. This is the number of radians per minute corresponding to60 revolutions per minute. Then the angular velocity vector is ae (t+j7+k). Note thisgives what you would expect in the case the position vector to the point is perpendicular tot+ 3-+k and at a distance of r. This is because of the geometric description of the crossproduct. The magnitude of the vector is r120” meters per minute and corresponds to thespeed and an exercise with the right hand shows the direction is correct also. However, ifthis body is rigid, this will work for every other point in it, even those for which the positionvector is not perpendicular to the given vector.Example 14.5.8 A wheel rotates counter clockwise about the vector 1+ 9 +k at 60 rev-olutions per minute exactly as in Example 14.5.7. Let {u,,u2,u3} denote an orthogonalright handed system attached to the rotating wheel in which u3 = a (¢+g+k). Thusu, and uz depend on time but, u, X U2 = U3. Find the velocity of the point of the wheellocated at the point 2u,; + 3u2 — U3. Note this point is not fixed in space. It is moving.Since {u1,u2,u3} is a right handed system like 7, 7,k, everything applies to this sys-tem in the same way as with 2,7, k. Thus the cross product is given byUl U2 U3(au, +buz2 + cuz) x (du; teurg+ fu3)=| a bcde fTherefore, in terms of the given vectors u;, the angular velocity vector is 1207u3. Thevelocity of the given point isUz; U2 U30 O 120% | = —3607u; + 2407uU22 3 —1in meters per minute. Note how this gives the answer in terms of these vectors which arefixed in the body, not in space. Since wu; depends on f, this shows the answer in this casedoes also. Of course this is right. Just think of what is going on with the wheel rotating.Those vectors which are fixed in the wheel are moving in space relative to a stationaryobserver. The velocity of a point in the wheel should be constantly changing. However, itsspeed will not change. The speed will be the magnitude of the velocity and this is/(—3607u; + 2402u2) - (—3602u; + 2402u2)which from the properties of the dot product equalsvi (—3602)* + (2402)* = 120V13%because the wu; are given to be orthogonal.