Crank-slider mechanism

=θ¨^Reeex_xx−θ˙^2Reeez_zz×eee_xxx aaa_p/op/op/o=θ¨^Reeex_xx−θ˙^2Reeey_yy

Resolving eee_xxxand eee_yyyinto components parallel to the fixed unit vectors eeexxxand eeeyyy,

a_p/o

aa_p/op/o=θ¨^Rcos(θ)eeexxx−sin(θ)eeeyyy

−θ˙^2Rsin(θ)eeexxx+cos(θ)eeeyyy

ap/o

aapp//oo=θ¨^Rcos(θ)−θ˙^2Rsin(θ) ex

ex

ex−θ¨^Rsin(θ) +θ˙^2Rcos(θ) eeeyyy

Once again, since the motion of the slider can only track the horizontal component of the motion of the pin,

aaabbb=θ¨Rcos(θ)−θ˙^2Rsin(θ) ex

ex

ex

Note, of course, that the same result may have been arrived at by directly differenti- ating the velocity, as

ab

ab

ab=˙v˙v˙v_bbb

= d dt

θ˙^Rcos(θ)eeexxx

a_b

a_b

a_b=θ¨^Rcos(θ)eeexxx+θ˙^R˙cos(θ)eeexxx−θ˙^2Rsin(θ)eeexxx+θ˙^Rcos(θ)˙e˙e˙exxx

where we have applied the product rule. SinceRis constant (the radius of the crank does not change with time), ˙R=0. Also, because the fixed unit vector eeexxxdoes not change direction, ˙e˙e˙ex_xx=0 as well. Then,

aaabbb=θ¨^Rcos(θ)eeexxx+0−θ˙^2Rsin(θ)eeexxx+0 aaabbb=θ¨^Rcos(θ)−θ˙^2Rsin(θ)

ex

eexx

as expected.

at an angle ofθ relative to the vertical as shown. The connecting rod has a length L. Determine the angular velocity and angular acceleration of the connecting rod as functions ofθ^.

Fig. 2.28 Crank-slider mechanism

SOLUTION

This is an example of a typical, simple mechanism, and is actually a fairly good representation of a reciprocating engine. This problem is very similar to the earlier bug-on-a-ladder example: there are constraints on both ends of the connecting rod C, though one end of the connecting rod is moving in circles rather than in a straight line. While the geometry may be fairly simple, the algebra involved in solving this problem symbolically can get extremely tedious.

To begin with, let us define the point p as the pin joining the connecting rod and the crank-wheel, and let us define the pointqas the pin joining the connecting rod to the slider. In order to determine the angular velocity and angular acceleration of the connecting rod, we will need to apply the relative motion equations using a coordinate system stuck to the rod. Since all the points of interest are stuck to the rod, vvv_p/op/op/o =aaa_p/op/op/o =0 for anypoint of interest on the rod relative toanymoving originostuck to the rod. The only unknowns in the relative velocity or acceleration equation will therefore be

(1) the velocity (or acceleration) ofo;

(2) the velocity (or acceleration) of the point of interest

(3)ωωωaxaxax, and (4)αααaxaxax.

Since the problem is two-dimensional, each of the relative velocity and relative ac- celeration vector equations each represent a system of two equations (one for eeexxxand one for eeeyyy). Therefore, we need to find two of these four unknowns before we can solve the relative motion equations for the connecting rod.

Because the pointspandqare constrained (pcan only move in circles of radius Rabout the centre of the crank-wheel, whileq can only move along a horizontal line), it should be reasonably easy to find vvvppp///ooo and vvvqqq///ooo (and, similarly, aaappp///ooo and aaaqqq///ooo). If we use one of these as our ‘point of interest’ on the connecting rod and the other as the moving origin, this will eliminate unknowns (1) and (2). Let us begin, then, by examining the crank-wheel in order to determine vvv_p/op/op/oand aaa_p/op/op/o.

Fig. 2.29 Crank-slider mechanism- the crank-wheel

Let us define a moving coordinate system which is stuck to the wheel, with the originoon the centre of the wheel. The moving unit vector eee_yyyrotates along with the wheel and is always pointing fromotowardp. The unit vector eee_xxxis perpendicular to eee_yyyand points downward and to the right at the instant of interest, while eee_zzzalways points out of the page. We will also define a fixed coordinate systemosuch that eeexxx

points to the right, eeeyyypoints up and eeezzzpoints out of the page, as illustrated. Since they are not given, we will take magnitudes of the angular velocity and acceleration of the wheel as ˙θ^{and ¨}θ, respectively, and treat these as unknowns for which we will have to solve later. From the relative velocity equation,

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

Since the moving originois stuck to the centre of the wheel and the centre of the wheel is not moving, vvv_ooo/o/o/o=0. Also, since both the moving coordinate system and the pointpare fixed to the same rigid wheel, vvv_p/op/op/o =0. Substituting,

v_p/o

vv_p/op/o=0+0+ (−θ˙)eee_zzz×Reee_yyy

=−θ˙^Reeez_zz×eeey_yy v_p/o

vv_p/op/o=θ˙^Reeex_xx

However, at the moment of interest,

e_x

ee_xx=cos(θ)eeexxx−sin(θ)eeeyyy

e_y

ee_yy=sin(θ)eeexxx+cos(θ)eeeyyy

e_z e_z e_z=eeezzz

Substituting,

v_p/o v_p/o

v_p/o=θ˙^Reeex_xx

=θ˙^Rcos(θ)eeexxx−sin(θ)eeeyyy

v_p/o

v_p/o

v_p/o=θ˙^Rcos(θ)eeexxx−θ˙^Rsin(θ)eeeyyy

Similarly, for the acceleration,

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, the pin at the centre of the wheel is not moving and both the coordinate systemoand the pointpare stuck to the same wheel, so aaa_ooo/o/o/o=aaa_p/op/op/o=0.

a_p/o

aa_p/op/o=0+0+2ωωωaxaxax×0+αααaxaxax×rrr_p/op/op/o+ωωωaxaxax×ωωωaxaxax×rrr_p/op/op/o

= (−θ¨)eeez_zz×Reee_yyy+ (−θ˙)eeez_zz×(−θ˙)eeez_zz×Reee_yyy

=−θ¨^Reeez_zz×eeey_yy

−θ˙^eeez_zz× −θ˙^Reeez_zz×eeey_yy

=θ¨^Reeex_xx−θ˙^eeez_zz×θ˙^Reeex_xx

=θ¨^Reeex_xx−θ˙^2Reeez_zz×eee_xxx a_p/o

aa_p/op/o=θ¨^Reeex_xx−θ˙^2Reeey_yy

Converting back into the fixed coordinates,

aaa_p/op/op/o=θ¨^Rcos(θ)eeexxx−sin(θ)eeeyyy

−θ˙^2Rsin(θ)eeexxx+cos(θ)eeeyyy

aaappp///ooo=θ¨^Rcos(θ)−θ˙^2Rsin(θ)

ex

ex

ex+

−θ¨^Rsin(θ)−θ˙^2Rcos(θ) ey

eeyy

We now have expressions for the velocity and acceleration of the pointprelative to a fixed observer, in terms of the angular velocity and acceleration of the crank wheel (which remain unknown). Next, let us turn our attention to the pointq. Since qis fixed to the sliderB, and since the velocity and acceleration of the sliderBis known, we may simply write,

v_q/o vv_q/oq/o=v_beeexxx

a_q/o aa_q/oq/o=abeeexxx

Finally, we direct our attention to the connecting rod. We will redefine the co- ordinate systemosuch that the moving origin is stuck to the connecting rod at the pointq, and the moving unit vector eee_yyypoints along the connecting rod, always to- ward the pointp. The unit vector eee_xxxis perpendicular to eee_yyy, pointing upwards and to the right, while eee_zzzagain points out of the page. The fixed coordinate system remains unchanged. For convenience, we shall define the angleθc as the angle subtended between the connecting rod and the horizontal. Note that we could equally have placedoon top of pointpand selectedqas our point of interest; the result would be identical.

Fig. 2.30 Crank-slider mechanism- the connecting rod

Applying the relative velocity equation, v_p/o

v_p/o

v_p/o=vvv_ooo///ooo+vvv_ppp///ooo+ωωω^axaxax×rrr_ppp///ooo

Since the moving origin o is coincident with the pointq, vvv_ooo/o/o/o=vvv_q/oq/oq/o, which we have already determined. Also, since the coordinate systemoand the pointqare both stuck to the same rigid bar, vvv_p/op/op/o=0. Substituting,

v_p/o v_p/o

v_p/o=vbeeexxx+0+ω^eeez_zz×Leee_yyy

=vbeeexxx+ω^Leeez_zz×eee_yyy v_p/o

v_p/o

v_p/o=vbeeexxx−ω^Leeex_xx

where ω is the angular velocity of the connecting rod (which may be positive or negative: the direction of rotation may not always be obvious). Once again, from geometry,

eee_xxx=sin(θc)eeexxx+cos(θc)eeeyyy

eee_yyy=−cos(θc)eeexxx+sin(θc)eeeyyy

e_z ee_zz=eeezzz

Substituting,

vvv_p/op/op/o=vbeeexxx−ω^Leeex_xx

=vbeeexxx−ω^Lsin(θc)eeexxx+cos(θc)eeeyyy

vvv_p/op/op/o=

vb−ω^Lsin(θc)

eeexxx−ω^Lcos(θc)eeeyyy

However, we already have an expression for vvv_p/op/op/o from our analysis of the crank- wheel:

v_p/o

vv_p/op/o=θ˙^Rcos(θ)eeexxx−θ˙^Rsin(θ)eeeyyy

Because both of these expressions give the velocity of the pointprelative to a fixed observer in terms of unit vectors which cannot change, we can equate them together, so that

θ˙^Rcos(θ)eeexxx−θ˙^Rsin(θ)eeeyyy=

vb−ω^Lsin(θc) ex

ex

ex−ω^Lcos(θc)eeeyyy

Note that the moving unit vectors were not the same during the analysis of the crank- wheel and the connecting rod, so the ‘prime’ coordinates from these two analyses could not be equated together. We may now use this expression to solve for the unknown ˙θ. Since we knowvb, let us first look at the coefficients of eeexxx.

θ˙Rcos(θ) =vb−ω^Lsin(θc) θ˙ = vb

Rcos(θ)−ω^L R

sin(θc) cos(θ)

Is is important to remember here that ˙θ is the angular velocity of the crank-wheel, whereasω is the angular velocity of the connecting rod. Next, we can equate the coefficients of eeeyyy, substitute the above result for ˙θand solve forω^.

θ˙^Rsin(θ) =ω^Lcos(θc) (2.20)

vb

Rcos(θ)−ω^L R

sin(θc) cos(θ)

Rsin(θ) =ω^Lcos(θc) vbsin(θ)

cos(θ)−ω^Lsin(θc)sin(θ)

cos(θ) =ω^Lcos(θc)

Noting that we have constrained ourselves here to examining cases for which cos(θ)=0; otherwise, the division would result in a singularity. Isolatingω^,

ω=v_b L

1

cos(θc)^cos(θ)_sin(θ)+sin(θc) (2.21) However, we still have the unknownθ^c in this expression. Thankfully, this can be obtained from the geometry.

Fig. 2.31 Crank-slider mechanism- determiningθc

Consider the triangle formed between the centre of the crank-wheel, the pointp and the pointq. The vertical line segment dividing the triangle under the pointphas a lengthRcos(θ). However, this length must also beLsin(θc). Consequently,

Rcos(θ) =Lsin(θc) sin(θc) = R

Lcos(θ) (2.22)

However, from the trigonometric identity,

sin²(θc) +cos²(θc) =1

cos(θc) = 1−sin²(θc)

where we are restricting ourselves to 0≤θc≤π/2. Substituting in the previous result,

cos(θc) =

1−R²

L²cos²(θ)

These results may now be substituted into Equation (2.21) to eliminateθ^{c, as}

ω =vb

L

1

cos(θc)^cos_sin(θ)^(θ)+sin(θc)

=vb

L

1 1−^R_L²2cos²(θ)

cos(θ) sin(θ)+

R Lcos(θ)

ω =vb

L

tan(θ)

1−^R_L²2cos²(θ) +^R_Lsin(θ)

where, for simplicity, we have multiplied through by tan(θ). Though cumbersome, this expression gives the value ofωas a function ofv_b, as required. Asωis positive for positivev_b, the connecting rod is rotating anti-clockwise when in the position illustrated.

It may also be interesting to notice here that, given the expressions which we have obtained above, it would equally be possible to solve for the velocity of the slider as a function of the angular velocity of the crank. From Equation (2.20),

θ˙^Rsin(θ) =ω^Lcos(θc) ω =θ˙^R

L sinθ cosθc

Substituting this into Equation (2.21),

ω =vb

L

1

cos(θc)^cos(θ)_sin(θ)+sin(θc) θ˙^R

L sinθ cosθc =vb

L

1

cos(θc)^cos(θ)_sin(θ)+sin(θc) vb=θ˙^Rsin(θ)

cos(θ)

sin(θ)+sin(θc) cos(θc)

Finally, from Equation (2.22),

sin(θc) = R Lcos(θ) cos(θc) =

1−R²

L²cos²(θ)

where we have again applied the trigonometric identity. Substituting, then,

v_b=θ˙^Rsin(θ)

cos(θ)

sin(θ)+sin(θc) cos(θc)

vb=θ˙^Rsin(θ)

cos(θ)

sin(θ)+ ^RLcos(θ) 1−^R_L²2cos²(θ)

The relationship between the angular velocity of the crank and the linear velocity of the slider is not straightforward; a plot of this function is included below. How- ever, Two interesting observations may be made of this result. First, the velocity vbbecomes nonreal (and therefore nonphysical) within the domain 0≤θ≤2π^for the case whenR/L>1. Furthermore, vb→∞within the domain whenR/L→1.

Examining the illustration of the mechanism in Figure 2.28, the reason for this sin- gularity becomes clear. One end of the connecting rod is constrained to remain on the horizontal centreline, and the other is constrained to follow the circumference of the crank. If the connecting rodCis shorter than the radius of the crankB(so thatR/L>1), the connecting rod won’t be able to ‘reach’ far enough to follow the circumference of the crank, and the mechanism will simply jam.

Fig. 2.32 Crank-slider mechanism- variation ofvbwithθ

It is also interesting to consider the case whenR/Lis very small. In this case, then,

vb≈θ˙Rsin(θ)

cos(θ)

sin(θ)+ (0)cos(θ) 1−^R_L2²cos²(θ)

=θ˙^Rsin(θ)

cos(θ) sin(θ)

vb≈θ˙Rcos(θ)

ForR/L→0, then, the motion becomes purely sinusoidal.

Continuing with our example, we next consider the acceleration of the connecting rod, repeating exactly the same procedure as for the velocity. We again begin with the relative acceleration equation,

aaappp///ooo=aaaooo///ooo+aaappp///ooo+2ωωωaxaxax×vvvppp///ooo+αααaxaxax×rrrppp///ooo+ωωωaxaxax×ωωωaxaxax×rrrppp///ooo

Because both the moving coordinate systemoand the pointpare stuck to the same bar, aaa_p/op/op/o =vvv_p/op/op/o =0. Again, sinceo andq are coincident, aaa_ooo/o/o/o=aaa_q/oq/oq/o, and we already know that

a_q/o a_q/o a_q/o=a_beeexxx

Substituting these values into the relative acceleration equation,

a_p/o a_p/o

a_p/o=abeeexxx+0+2ωωωaxaxax×0+α^eeez_zz×Leee_yyy+ω^eeez_zz×ω^eeez_zz×Leee_yyy

=abeeexxx+α^Leeez_zz×eeey_yy

+ω^eeez_zz×ω^Leeez_zz×eeey_yy

=abeeexxx−α^Leeex_xx+ω^eeez_zz× −ω^Leeex_xx

=a_beeexxx−α^Leeex_xx−ω^2Leeez_zz×eee_xxx a_p/o

a_p/o

a_p/o=abeeexxx−α^Leeex_xx−ω^2Leeey_yy

whereα is the magnitude of the angular acceleration of the connecting rod (which, likeω, may be negative). Once again, we express the moving unit vectors in terms of the fixed unit vectors, as

a_p/o

aa_p/op/o=abeeexxx−α^Lsin(θc)eeexxx+cos(θc)eeeyyy

−ω^2L−cos(θc)eeexxx+sin(θc)eeeyyy

=abeeexxx−α^Lsin(θc)eeexxx−α^Lcos(θc)eeeyyy+ω^2Lcos(θc)eeexxx−ω^2Lsin(θc)eeeyyy

a_p/o aa_p/op/o=

ab−α^Lsin(θc) +ω^2Lcos(θc) ex

ex

ex+

−α^Lcos(θc)−ω^2Lsin(θc) eeeyyy

This expression yielded the velocity of point p with respect to a fixed observer.

However, we have already obtained an expression for aaa_p/op/op/ofrom our analysis of the crank-wheel:

aaa_p/op/op/o=θ¨Rcos(θ)−θ˙^2Rsin(θ) ex

ex

ex+

−θ¨^Rsin(θ)−θ˙^2Rcos(θ) ey

eeyy

Since these are two independent expressions for the same quantity using the same coordinate system, we can equate them together, so that

θ¨^Rcos(θ)−θ˙^2Rsin(θ) ex

eexx+

−θ¨^Rsin(θ)−θ˙^2Rcos(θ) eeeyyy

=

a_b−α^Lsin(θc) +ω^2Lcos(θc) ex

ex

ex+

−α^Lcos(θc)−ω^2Lsin(θc) eeeyyy

(2.23) Once again, we can equate together the coefficients of eeexxxto solve for the unknown θ¨^{, as}

θ¨^Rcos(θ)−θ˙^2Rsin(θ) =a_b−α^Lsin(θc) +ω^2Lcos(θc) θ¨ = ab

Rcos(θ)−α^L R

sin(θc) cos(θ)+ω^2L

R cos(θc)

cos(θ) +θ˙^2sin(θ) cos(θ) Next, we shall equate the coefficients of eeeyyy.

−θ¨^Rsin(θ)−θ˙^2Rcos(θ) =−α^Lcos(θc)−ω^2Lsin(θc) We may now substitute in the previous result for ¨θ^{, as}

−

a_b

Rcos(θ)−α^L R

sin(θ^c) cos(θ)+ω^2L

R cos(θ^c)

cos(θ)+θ˙^2sin(θ) cos(θ)

Rsin(θ)

−θ˙^2Rcos(θ)

=−α^Lcos(θc)−ω^2Lsin(θc)

Distributing the multiplication,

−absin(θ)

cos(θ)+α^Lsin(θc)sin(θ)

cos(θ)−ω^2Lcos(θc)sin(θ) cos(θ)

−θ˙^2Rsin2(θ)

cos(θ) −θ˙^2Rcos(θ)

=−α^Lcos(θc)−ω^2Lsin(θc)

Simplifying and isolatingα^,

−absin(θ) cos(θ)+ω^2L

sin(θc)−cos(θc)sin(θ) cos(θ)

−θ˙^2Rsin2(θ) +cos²(θ) cos(θ)

=α^L

−cos(θc)−sin(θc)sin(θ) cos(θ)

Simplifying,

−ab

L sin(θ) cos(θ)+ω²

sin(θc)−cos(θc)sin(θ) cos(θ)

−θ˙^2R L

1 cos(θ)

=α

−cos(θc)−sin(θc)sin(θ) cos(θ)

Next, we can substitute the expressions for sin(θc)and cos(θc)we obtained earlier, so that

−a_b L

sin(θ) cos(θ)+ω²

R

Lcos(θ)−

1−R²

L²cos²(θ)sin(θ) cos(θ)

−θ˙^{2 R} Lcos(θ)

=α

−

1−R²

L²cos²(θ)−R

Lcos(θ)sin(θ) cos(θ)

Finally, we may expressα^as

α=

a_b

L tan(θ)−ω²

R

Lcos(θ)− 1−^R_L²2cos²(θ)tan(θ)

+θ˙²_Lcos^R_(θ) 1−^R_L²2cos²(θ) +^R_Lsin(θ)

where, as we had shown earlier, ω=v_b

L

tan(θ)

1−^R_L2²cos²(θ) +^R_Lsin(θ) and

θ˙ = vb

Rcos(θ)−ω^L R

sin(θc) cos(θ)

= vb

Rcos(θ)−ω^L R

R Lcos(θ)

cos(θ)

= vb

Rcos(θ)−ω θ˙ = v_b

Rcos(θ)−v_b L

tan(θ)

1−^R_L2²cos²(θ) +^R_Lsin(θ)

Though incredibly cumbersome, we have derived algebraic expressions forω ^and α (the angular velocity and angular acceleration of the connecting rod) which are functions ofθ^,R,L,vbandabonly, which are all known quantities.

As before, it would also be useful to determine the acceleration of the sliderab

as a function of the angular velocity of the crank. Let us consider the special case

when ¨θ=0, so that the crank is rotating with constant angular velocity ˙θ^{. Then,} recalling the combined relative acceleration equation (Equation 2.23),

θ¨^Rcos(θ)−θ˙²Rsin(θ) eeexxx+

−θ¨Rsin(θ)−θ˙^2Rcos(θ) eeeyyy

=

ab−α^Lsin(θc) +ω^2Lcos(θc) ex

ex

ex+

−α^Lcos(θc)−ω^2Lsin(θc) eeeyyy

and equating the scalar coefficients of eeeyyy,

−θ¨^Rsin(θ)−θ˙^2Rcos(θ) =−α^Lcos(θc)−ω^2Lsin(θc)

−(0)Rsin(θ)−θ˙^2Rcos(θ) =−α^Lcos(θc)−ω^2Lsin(θc) α =θ˙^2R

L cos(θ)

cos(θc)−ω^2sin(θc) cos(θc)

Next, we can equate the scalar coefficient of eeexxx(again, considering only the special case of ¨θ=0) to solve forab, as

θ¨^Rcos(θ)−θ˙^2Rsin(θ) =a_b−α^Lsin(θc) +ω^2Lcos(θc) (0)Rcos(θ)−θ˙^2Rsin(θ) =ab−α^Lsin(θc) +ω^2Lcos(θc)

ab=−θ˙²Rsin(θ) +α^Lsin(θc)−ω^2Lcos(θc)

Substituting in our earlier result forα^,

ab=−θ˙^2Rsin(θ) +

θ˙^2R L

cos(θ)

cos(θc)−ω^2sin(θc) cos(θc)

Lsin(θc)−ω^2Lcos(θc)

=−θ˙^2Rsin(θ) +θ˙^2Rcos(θ)sin(θc)

cos(θc)−ω^2Lsin2(θc)

cos(θc) −ω^2Lcos(θc)

=−θ˙^2Rsin(θ) +θ˙^2Rcos(θ)sin(θc)

cos(θc)− ω^2L cos(θc)

sin²(θc) +cos²(θc)

ab=−θ˙^2Rsin(θ) +θ˙^2Rcos(θ)sin(θc)

cos(θc)− ω^2L cos(θc) Next, we may use Equation (2.20) to eliminateω^{, as}

θ˙^Rsin(θ) =ω^Lcos(θc)

ω =θ˙^R L

sinθ cos(θc) Substituting this into the previous result forab,

a_b=−θ˙^2Rsin(θ) +θ˙^2Rcos(θ)sin(θc)

cos(θc)− ω^2L cos(θc)

=−θ˙^2Rsin(θ) +θ˙^2Rcos(θ)sin(θc) cos(θc)− L

cos(θc)

θ˙^2R2 L²

sin²θ cos²(θc)

ab=θ˙^2Rsin(θ)

−1+cos(θ) sin(θ)

sin(θc) cos(θc)−R

L sinθ cos³(θc)

From Equation (2.22), we already have shown

sin(θc) = R Lcos(θ) cos(θc) =

1−R²

L²cos²(θ) Substituting this into our equation fora_b, then,

a_b=θ˙^2Rsin(θ)

−1+cos(θ) sin(θ)

sin(θc) cos(θc)−R

L sinθ cos³(θc)

=θ˙^2Rsin(θ)

−1+cos(θ) sin(θ)

R Lcos(θ)

1−^R_L2²cos²(θ)− ^RLsinθ 1−^R_L²2cos²(θ)_3/2

ab=θ˙^2Rsin(θ)

−1+

R

Lcos²(θ)

sin(θ) 1−^R_L²2cos²(θ)−

R Lsinθ 1−^R_L²2cos²(θ)_3/2

This result has yielded the acceleration of the slider in terms of the crank angleθ for the special case of ¨θ=0. As with the slider velocity, the relationship between the slider acceleration andθis fairly complicated; a plot of this function is included below.

It is again interesting to note here that for the case ofR/L→1, the acceleration of the slider becomes infinite. This is to be expected, as we have already seen that the mechanism will no longer function ifL<R. Also, ifR/Lis very small, then

ab≈θ˙²Rsin(θ)

−1+0−0

Fig. 2.33 Crank-slider mechanism- variation ofabwithθ

ab≈ −θ˙^2Rsin(θ)

This is expected, as we saw that for smallR/L, v_b≈θ˙^Rcos(θ) and, therefore,

ab= d dtvb

≈ d

dtθ˙Rcos(θ)

ab≈θ¨^Rcos(θ) +θ˙^R˙cos(θ)−θ˙^2Rsin(θ)

where we have applied the product rule. Since the radius of the crank is constant, R˙=0; also, since we have only considered above the case where ¨θ=0,

ab≈(0)Rcos(θ) +θ˙(0)cos(θ)−θ˙^2Rsin(θ)

a_b≈ −θ˙^2Rsin(θ)

as expected.

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