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Question

A triangular block is moving in its own plane such that at any instant speeds of corner A and B are v each, as shown in figure. Then

A
velocity of end C is zero
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B
velocity of end C is v upwards
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C
velocity of mid-point of side BC is zero
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D
speed of mid point of side AC is v/2
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Solution

The correct options are
B velocity of end C is v upwards
C velocity of mid-point of side BC is zero
D speed of mid point of side AC is v/2


When we draw perpendiculars to the velocity vectors at A and B, they meet at D. Hence D is the Instantaneous Centre of rotation.
Velocity of D is zero.

From above figure ABD is equilateral triangle.
So, BD=a,D is the mid point of BC.

Hence, the frame will rotate about D clockwise direction.
Velocity of C will be perpendicular to CD, upwards.
Let angular velocity of rotation be ω
then VB=ω×DB=ωa=v
Now VC=ω×CD=ωaVC=v


From the triangle CED, d=a2
Hence VE=ω×d=ωa2
VE=(va)(a2)
VE=v2

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