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CONSTRAINT MOTIONLecture -1

When two or more bodies are connected & their motion are related to maintain connection. e.g. if wehave a block kept on incline plane and we want the block to maintain contact with it. The block can nothave 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 samevelocity and acceleration to maintain the contact

m2m1

21 vv

1v 2v

1a = 2a

1a 2a

If we keep a block on wedge which can move then again constraint is defined in refrence frame attachedto wedge. The block can not have any acceleration y direction in refrance frame attaced to wedge.

[ in ground frame] [in frame attached to wedge]

initial

later

y

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 onthe initial diagram.

x

y

Here there are three constraints involved (a) The wedge can move only horizontally (b) the rod canmove only vertically (c) the rod and wedge to have remain in contact thus their motion to be related usinggeometry.We can see that when wedge moves x along horizontal direction rod rises by y.

tan = xy

y = x tan

hence vR = vw tan differentiating and aR = aw tan

TEACHING NOTES (XIII)

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

A B

A B

x xA B

Although if we push B towards left there is no constraint relation as string will slack. If A & B areconnected by rod then A will have to move as B is moving in both the above cases.Along the string VA = VB.

Asking questionIf velocity of A is 2 ms1 downwards what is the velocity of B.

(a) A

B

(b) A

B (c) A B

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

r2

y

bx

r1

B

A

Clearly, the motion of B is the same as that of the centre of its pulley, so we establish position coordinatesx and y measured from a convenient fixed reference. The total length of the cable is

L = x + 2r2 + 2y + r1 + b

With L, r2, r1, and b all constant, the first and second time derivatives of the equation give0 = x + y2 or 0 = A + B

2

0 = x + y2 or 0 = aA + 2aBThe velocity and acceleration constraint equations indicate that, for the coordinates selected, the velocityof A must have a sign which is opposite to that of the velocity of B, and similarly for the accelerations.

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Ex. MA

x

y

xL

y

vR = vel. of ringvM = vel. of block

Method 1 : Total length = 22 yL + x

= 22 yL

y

dtady

+ dtdx

= cvR + vM

Rvdtdy

vM = + vR cMethod 2 : Wrong method

This point A has vel. equal to vM along string ring has vel. component along y axisvR = vM c

correct methodpoint A doesnt 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

m1V2 60

M2

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

m1

V2

60

V2

In refrence frame of attached to wedge wall will move horizontally towards right with speed V2 asshown.Thus it can be easily deduced that m1 will move with velocity v2 upwards. But this is velocity of m1 onframe attached to m2 using relation ship of net motion

1v = 12v

+ 2v

we can get 1v as shown

60V2

V12

Solving we get| 1v | = v2

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

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

m1

m2 m3

P

Sol. V2 VP = VP V3V3 = 0V1 = VP = 1 m/s

Discuss : Q.No. 13 Obj.-I of HCVQ.43 to 55 Obj. Q.Bank Single correctSheet Ex.-1 : Q.No. 19, 20