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

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