Constraint XIII

TEACHING NOTES (XIII) CONSTRAINT MOTION Lecture -1 When two or more bodies are connected & their motion are related to maintain connection. e.g. if we have a block kept on incline plane and we want the block to maintain contact with it. The block can not have velocity and acceleration in direction perpendicular to the incline.

If we have two block kept touching each other on horizontal surface as shown then they must have same velocity and acceleration to maintain the contact m1 m2

    v1  v 2 v1  v 2     a1 = a 2 a1  a 2 If we keep a block on wedge which can move then again constraint is defined in refrence frame attached to wedge. The block can not have any acceleration ‘ y’ direction in refrance frame attaced to wedge. [ in ground frame]

[in frame attached to wedge] y

initial

later

Ex.

Find relation between velocity and accln. of rod and wedge.

or say in normal direction VA = VB 

Lets imagine what happens when wedge is pushed towards left. We make a suporimposing diagram on the initial diagram. y  x

Here there are three constraints involved (a) The wedge can move only horizontally (b) the rod can move only vertically (c) the rod and wedge to have remain in contact thus their motion to be related using geometry. We can see that when wedge moves x along horizontal direction rod rises by y. y tan =  y = x tan  x hence vR = vw tan differentiating and aR = aw tan  Page # 1

Mathematical analysis (Constraint equation) When two bodies are connected by inextensible rope then their motion is interdependent if we want rope to remain taut. If we connect two block as shown in diagram and pull block B towards right the block A must cover same distance as B to keep string tight. A

B A

B

xA

xB

Although if we push B towards left there is no constraint relation as string will slack. If A & B are connected by rod then A will have to move as B is moving in both the above cases. Along the string VA= VB. Asking question If velocity of A is 2 ms–1 downwards what is the velocity of B. B

(a)

(b) A

B

(c)

A

A

B

Consider first the very simple system of two interconnected particles A and B shown in fig. Although it can be shown by inspection that the horizontal motion of A is twice the vertical motion of B, we will use this example to illustrate the method of analysis which will use for more complex situations where the results cannot be easily reached by inspection. x A

r2

b

r1

y

B

Clearly, the motion of B is the same as that of the centre of its pulley, so we establish position coordinates x and y measured from a convenient fixed reference. The total length of the cable is r L = x + 2 + 2y + r1 + b 2 With L, r2, r1, and b all constant, the first and second time derivatives of the equation give • • 0 = x + 2 y• or 0 = A + 2  B ••

••

0 = x + 2 y or 0 = aA + 2aB The velocity and acceleration constraint equations indicate that, for the coordinates selected, the velocity of A must have a sign which is opposite to that of the velocity of B, and similarly for the accelerations.

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x x

Ex.

L A



M

y y

vR = vel. of ring vM = vel. of block Method 1 : Total length =

L2  y 2 + x y

=

2

L y

dy dx + dta dt

2

 dy  = – cvR + vM    v R   dt  vM = + vR c Method 2 : Wrong method This point A has vel. equal to vM along string ring has vel. component along y axis vR = vM c correct method point A doesn’t has along string  vA = vR & its along string string is vR c = vM

Ex.

Find velocity vector of m1 if m2 is pulled with constant velocity v2 = 2 m/s

V2

Sol.

m1

M2 60°

This problem involves two constraints. One involves the rope and other is involves m1 and m2 remaining in constact with each other. These two constraints can be understood easily if we shift our refrence frame to the wedge. By doing this we will be able to simplify motion of block m1 and thus solve constraint of rope easily. V2

V2

m1 60º

In refrence frame of attached to wedge wall will move horizontally towards right with speed V2 as shown. Thus it can be easily deduced that m1 will move with velocity v2 upwards. But this is velocity of m1 on frame attached to m2 using relation ship of net motion    v1 = v12 + v2  we can get v1 as shown V12 60º V2

Solving we get  | v1 | = v2 Page # 3

We can check that the this velocity vector satisfies condition of m1 having no component of velocity perpendicular to incline with respect to the wedge. Eg.

If V2 = 2m/s upwards ; VP = 1 m/s upwards Find the velocity of block 1 and block 3 ?

m1

Sol.

V2 – VP = VP – V3 V3 = 0 V1 = – VP = 1 m/s 

Discuss :

P m2 m3

Q.No. 13 Obj.-I of HCV Q.43 to 55 Obj. Q.Bank Single correct Sheet Ex.-1 : Q.No. 19, 20

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