Relative motion: bug on a disc

=a²Seeettt+a²S² r eeennn

=a²

β^x 1

2β ₂

+x²+1 2ln

2β^x+

1+4β^2x2

eeettt

+ 2β^a2 (1+4β^2x2)^3/2

β^x

1 2β

2

+x²+1 2ln

2β^x+

1+4β^{2x2 2}

eeennn

Then, the velocity of the bug relative to a fixed observer vvv_p/op/op/ois, vvv_p/op/op/o=vvv_ooo/o/o/o+vvv_p/op/op/o+ωωωaxaxax×rrr_p/op/op/o

Now, we consider each of the terms in the above. The term vvv_ooo/o/o/o represents the velocity oforelative to the fixed observer. However, though theoaxes are rotating, the hub is not actually moving andois always right on top ofo. Therefore, vvv_ooo/o/o/o=0.

Next, we consider vvv_p/op/op/o. This represents the velocity of the bug relative to the disc, which we know; the bug’s walking speed is provided in the problem descrip- tion, and the direction is always in a straight line along what we have defined as eee_xxx. Therefore, vvvppp///ooo=vbugeee_xxx.

The angular velocity vector of the moving axes,ωωω^axaxaxis also known. The moving axes are stuck to the disc, and the angular velocity of the disc is provided in the problem description. The magnitude of the angular velocity of the disc is given asω, and the direction is always along eeezzz. The position vector of the bug relative to the moving reference frame has a magnitude ofr(as specified by the problem statement), and, according to our definition, is always in the eee_xxxdirection, so rrr_p/op/op/o= reee_xxx. Then,

vvv_p/op/op/o=vvv_ooo/o/o/o+vvv_p/op/op/o+ωωωaxaxax×rrr_p/op/op/o

=0+vbugeee_xxx+ω^r(eeezzz×eeex_xx) vvv_p/op/op/o=v_bugeee_xxx+ω^reeey_yy

Since we can only describe the velocity of the bug at a single instant in time, we might as well ensure that we defineosuch that the moving axes eee_xxxand eee_yyylie right on top of the fixed axes eeexxxand eeeyyyat that moment in time. Then, at that moment only, eee_xxx=eeexxxand eee_yyy=eeeyyy, and

v_p/o

vv_p/op/o=v_bugeeexxx+ω^reeeyyy

We were asked to determine the velocity relative to the fixed observer, so the results cannot be presented in terms of some arbitrary moving unit vectors which we created for our own convenience. Because of how we definedoando, the conversion from the moving frame to the fixed one was very simple. Had we stucko to the disc somewhere other than right on top ofo, or had eee_xxxpointed in a direction other than directly at the bug, there would have been a lot of sines and cosines involved in converting from eee_xxxand eee_yyyto eeexxxand eeeyyy, but the result (expressed in the universal, fixed coordinates) would have been exactly the same. This equivalence shall be demonstrated later.

The acceleration of the bug relative to a fixed observer may be expressed as,

aaa_p/op/op/o=aaa_ooo/o/o/o+aaa_p/op/op/o+2ωωωaxaxax×vvv_p/op/op/o

+αααaxaxax×rrr_p/op/op/o+ωωωaxaxax×ωωωaxaxax×rrr_p/op/op/o

Again, we will look at each of these terms individually. First, aaa_ooo/o/o/orepresents the acceleration of the moving originorelative to the fixed reference frameo. Sinceo andoalways lie one on top of the other, aaa_ooo/o/o/o=0.

The term aaa_p/op/op/orepresents the acceleration of the bug relative to the moving frame (which is stuck to the disc). Since the bug is walking with constant velocity along the surface of the disc, aaa_p/op/op/o=0 as well.

In the term 2ωωωaxaxax×vvvppp///ooo, we have already determined thatωωωax=ω^eeezzz, and that vp/o

vp/o

vp/o =vbugeee_xxx.

The angular acceleration of the o reference frame,ααα^axaxax, is known to be zero because the rotational velocity of the reference frame is constant. Therefore,αααaxaxax× r_p/o

r_p/o

r_p/o=0.

Re-assembling these terms,

a_p/o

aa_p/op/o=0+0+2ω^vbug(eeezzz×eeex_xx) +0+ω^eeezzz×

ω^r(eeezzz×eeex_xx)

=2ω^vbugeee_yyy+ω^eeezzz×ω^reeey_yy

=2ω^vbugeee_yyy+ω^2r(eeezzz×eeey_yy) a_p/o

aa_p/op/o=2ω^vbugeee_yyy−ω^2reeex_xx

If we again consider the instant when eee_xxx=eeexxxand eee_yyy=eeeyyy, we can give the result, ap/o

aapp//oo=2ω^vbugeeeyyy−ω^2reeexxx

The first term is the Coriolis acceleration, which disappears ifvbug→0. The second term is the centripetal acceleration, and is pointing in the−eeexxxdirection (toward the hub).

Alternatively, we may also have determined the velocity and acceleration of the bug as a function of position (and indirectly, then, as a function of time). Rather than assuming that eee_xxxis aligned with eeexxxat the ‘moment of interest’, we may equally solve the problem when eee_xxxis oriented at some arbitrary angleθrelative to the fixed coordinate system, as illustrated in Figure 2.19.

Once again we will define a fixed coordinate systemowith its origin at the centre of the disc. We shall define eeezzzas being normal to the disc, pointing upwards; eeexxxand ey

ey

eywill be fixed in space and oriented as illustrated. We will define a moving coordi- nate systemowith its origin always coincident witho, but with its unit vectors eee_xxx and eee_yyystuck to the surface of the disc and oriented so that eee_xxxalways points toward the bug, as before. Once again eeezzzis always parallel with eee_zzz.

The velocity of the bug relative to a fixed observer is then,

Fig. 2.19 Example 4b

vvv_p/op/op/o=vvv_ooo/o/o/o+vvv_p/op/op/o+ωωωaxaxax×rrr_p/op/op/o

Once again, since the centre of the disc is not moving, vvv_ooo/o/o/o=0. The eee_xxxunit vector always points toward the bug, so rrr_p/op/op/o =reee_xxx. Similarly, sinceois stuck to the disc, vvv_p/op/op/o is just the velocity of the bug relative to the surface of the disc, so vvv_p/op/op/o = vbugeee_xxx. Substituting these expressions into the velocity equation, then,

vvv_p/op/op/o=0+vbugeee_xxx+ω^eeezzz×reee_xxx

(Again, since o is stuck to the disc,ωωωaxaxax is simply the angular velocity ω ^{of the} disc). We are now left with an equation with mixed ‘prime’ and ‘non-prime’ unit vectors; in order to carry out the addition and cross product, all of the unit vectors must be in the same coordinate system. We will therefore resolve everything into the ocoordinate system (this choice was arbitrary; we could equally resolve everything into theosystem). Since eeezzzis always parallel with eee_zzz, these two are interchangeable, and so

vp/o

vvpp//oo=vbugeee_xxx+ω^eeez_zz×reee_xxx

=vbugeee_xxx+ω^reeez_zz×eeex_xx v_p/o

vv_p/op/o=v_bugeee_xxx+ω^reeey_yy

Now, to express this result from the point of view of a fixed observer, we must resolve the moving unit vectors into components parallel to the fixed unit vectors.

From the geometry, we see that

e_x

ee_xx=|eee_xxx|cos(θ)eeexxx+|eee_xxx|sin(θ)eeeyyy

e_y

ee_yy=−|eeey_yy|sin(θ)eeexxx+|eeey_yy|cos(θ)eeeyyy

e_z

ee_zz=|eee_zzz|eeezzz

Where it should be noted again that unit vectors have unit magnitude by definition, so this result reduces to

eee_xxx=cos(θ)eeexxx+sin(θ)eeeyyy

eee_yyy=−sin(θ)eeexxx+cos(θ)eeeyyy

e_z ee_zz=eeezzz

Substituting this into our earlier result,

v_p/o

vv_p/op/o=vbugeee_xxx+ω^reeey_yy

=vbug

cos(θ)eeexxx+sin(θ)eeeyyy

+ω^r−sin(θ)eeexxx+cos(θ)eeeyyy

v_p/o

vv_p/op/o=

v_bugcos(θ)−ω^rsin(θ) eeexxx+

v_bugsin(θ) +ω^rcos(θ) ey

ey

ey

Alternatively, we could have first resolved our expression for vvvppp///ooointo compo- nents along eeexxxand eeeyyyand then carried out the cross-product:

v_p/o

vv_p/op/o=v_bugeee_xxx+ω^eeez_zz×reee_xxx v_p/o

vv_p/op/o=vbug

cos(θ)eeexxx+sin(θ)eeeyyy

+ω^eeezzz×r

cos(θ)eeexxx+sin(θ)eeeyyy

(2.19) Distributing the cross-product,

v_p/o v_p/o v_p/o=vbug

cos(θ)eeexxx+sin(θ)eeeyyy

+ω^rcos(θ) ez

eezz×eeexxx

+ωrsin(θ) ez

eezz×eeeyyy

v_p/o

v_p/o

v_p/o=v_bug

cos(θ)eeexxx+sin(θ)eeeyyy

+ω^rcos(θ) ey

eeyy

+ω^rsin(θ)

−eeexxx

Gathering terms,

vp/o

vvpp//oo=

vbugcos(θ)−ωrsin(θ) eeexxx+

vbugsin(θ) +ωrcos(θ) ey

ey

ey

which is exactly the same result as arrived at earlier.

Next, for the acceleration of the bug,

a_p/o

aa_p/op/o=aaa_ooo/o/o/o+aaa_p/op/op/o+2ωωωaxaxax×vvv_p/op/op/o

+αααaxaxax×rrrppp///ooo+ωωωaxaxax×ωωωaxaxax×rrrppp///ooo

Since the centre of the disc is not moving, aaa_ooo/o/o/o=0. Also, the bug is running with constant speed relative to the disc, so aaa_p/op/op/o=0. The disc itself is rotating with con- stant speed, and since our moving axes are stuck to the disc,αααaxaxax=0 as well. Then,

ap/o

ap/o

ap/o=0+0+2ωωωaxaxax×vvvppp///ooo+0+ωωωaxaxax×ωωωaxaxax×rrrppp///ooo

ap/o

ap/o

ap/o=2ω^eeezzz×vbugeee_xxx+ω^eeezzz×ω^eeezzz×reee_xxx

Substituting eeezzz=eee_zzz, we can ensure that all of our unit vectors are in the same coor- dinate system so that the vector addition and multiplication may be carried out.

aaappp///ooo=2ω^vbug

eee_zzz×eee_xxx

+ω^eeez_zz×ω^reeez_zz×eeex_xx

=2ω^vbugeee_yyy+ω^eeez_zz×ω^reeey_yy

=2ω^vbugeee_yyy+ω^2reeez_zz×eee_yyy aaa_p/op/op/o=2ω^vbugeee_yyy−ω^2reeex_xx

Again, this must be expressed in components along the fixed unit vectors; substitut- ing, then,

aaa_p/op/op/o=2ω^vbugeee_yyy−ω^2reeex_xx

=2ω^vbug

−sin(θ)eeexxx+cos(θ)eeeyyy

−ω^2rcos(θ)eeexxx+sin(θ)eeeyyy

aaa_p/op/op/o=

−2ω^vbugsin(θ)−ω^2rcos(θ) eeexxx+

2ω^vbugcos(θ)−ω^2rsin(θ) ey

eeyy

Notice that for the special case ofθ=0, the earlier results (from when we assumed that the moving unit vectors were parallel with the fixed unit vectors at the moment of interest) are recovered.

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