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
• 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 $v/2$

Answer 1

Answer: (B), (C), (D)

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.
$\Rightarrow$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 ${}^{\prime }{C}^{\prime }$ will be perpendicular to $CD$, upwards.
Let angular velocity of rotation be $\omega$
then $V_B=\omega \times DB=\omega a=v$
Now $V_C=\omega \times CD=\omega a\Rightarrow V_C=v$


From the triangle $CED,$ $d=\frac{a}{2}$
Hence $V_E=\omega \times d=\frac{\omega a}{2}$
$V_E=\left(\frac{v}{a}\right)\left(\frac{a}{2}\right)$
$V_E=\frac{v}{2}$